Convolution solutions (Sect. 6.6).

I Convolution of two functions.

I Properties of convolutions.

I Laplace Transform of a convolution.

I Impulse response solution.

I Solution decomposition theorem.

Convolution solutions (Sect. 6.6).

I Convolution of two functions.

I Properties of convolutions.

I Laplace Transform of a convolution.

I Impulse response solution.

I Solution decomposition theorem.

Convolution of two functions.

DefinitionThe convolution of piecewise continuous functions f , g : R → R isthe function f ∗ g : R → R given by

(f ∗ g)(t) =

∫ t

0

f (τ)g(t − τ) dτ.

Remarks:

I f ∗ g is also called the generalized product of f and g .

I The definition of convolution of two functions also holds inthe case that one of the functions is a generalized function,like Dirac’s delta.

Convolution of two functions.

DefinitionThe convolution of piecewise continuous functions f , g : R → R isthe function f ∗ g : R → R given by

(f ∗ g)(t) =

∫ t

0

f (τ)g(t − τ) dτ.

Remarks:

I f ∗ g is also called the generalized product of f and g .

I The definition of convolution of two functions also holds inthe case that one of the functions is a generalized function,like Dirac’s delta.

Convolution of two functions.

DefinitionThe convolution of piecewise continuous functions f , g : R → R isthe function f ∗ g : R → R given by

(f ∗ g)(t) =

∫ t

0

f (τ)g(t − τ) dτ.

Remarks:

I f ∗ g is also called the generalized product of f and g .

I The definition of convolution of two functions also holds inthe case that one of the functions is a generalized function,like Dirac’s delta.

Convolution of two functions.

Example

Find the convolution of f (t) = e−t and g(t) = sin(t).

Solution: By definition: (f ∗ g)(t) =

∫ t

0

e−τ sin(t − τ) dτ .

Integrate by parts twice:

∫ t

0

e−τ sin(t − τ) dτ =[e−τ cos(t − τ)

]∣∣∣t0

−[e−τ sin(t − τ)

]∣∣∣t0

−∫ t

0

e−τ sin(t − τ) dτ,

2

∫ t

0

e−τ sin(t − τ) dτ =[e−τ cos(t − τ)

]∣∣∣t0

−[e−τ sin(t − τ)

]∣∣∣t0

,

2(f ∗ g)(t) = e−t − cos(t)− 0 + sin(t).

We conclude: (f ∗ g)(t) =1

2

[e−t + sin(t)− cos(t)

]. C

Convolution of two functions.

Example

Find the convolution of f (t) = e−t and g(t) = sin(t).

Solution: By definition: (f ∗ g)(t) =

∫ t

0

e−τ sin(t − τ) dτ .

Integrate by parts twice:

∫ t

0

e−τ sin(t − τ) dτ =[e−τ cos(t − τ)

]∣∣∣t0

−[e−τ sin(t − τ)

]∣∣∣t0

−∫ t

0

e−τ sin(t − τ) dτ,

2

∫ t

0

e−τ sin(t − τ) dτ =[e−τ cos(t − τ)

]∣∣∣t0

−[e−τ sin(t − τ)

]∣∣∣t0

,

2(f ∗ g)(t) = e−t − cos(t)− 0 + sin(t).

We conclude: (f ∗ g)(t) =1

2

[e−t + sin(t)− cos(t)

]. C

Convolution of two functions.

Example

Find the convolution of f (t) = e−t and g(t) = sin(t).

Solution: By definition: (f ∗ g)(t) =

∫ t

0

e−τ sin(t − τ) dτ .

Integrate by parts twice:

∫ t

0

e−τ sin(t − τ) dτ =[e−τ cos(t − τ)

]∣∣∣t0

−[e−τ sin(t − τ)

]∣∣∣t0

−∫ t

0

e−τ sin(t − τ) dτ,

2

∫ t

0

e−τ sin(t − τ) dτ =[e−τ cos(t − τ)

]∣∣∣t0

−[e−τ sin(t − τ)

]∣∣∣t0

,

2(f ∗ g)(t) = e−t − cos(t)− 0 + sin(t).

We conclude: (f ∗ g)(t) =1

2

[e−t + sin(t)− cos(t)

]. C

Convolution of two functions.

Example

Find the convolution of f (t) = e−t and g(t) = sin(t).

Solution: By definition: (f ∗ g)(t) =

∫ t

0

e−τ sin(t − τ) dτ .

Integrate by parts twice:

∫ t

0

e−τ sin(t − τ) dτ =[e−τ cos(t − τ)

]∣∣∣t0

−[e−τ sin(t − τ)

]∣∣∣t0

−∫ t

0

e−τ sin(t − τ) dτ,

2

∫ t

0

e−τ sin(t − τ) dτ =[e−τ cos(t − τ)

]∣∣∣t0

−[e−τ sin(t − τ)

]∣∣∣t0

,

2(f ∗ g)(t) = e−t − cos(t)− 0 + sin(t).

We conclude: (f ∗ g)(t) =1

2

[e−t + sin(t)− cos(t)

]. C

Convolution of two functions.

Example

Find the convolution of f (t) = e−t and g(t) = sin(t).

Solution: By definition: (f ∗ g)(t) =

∫ t

0

e−τ sin(t − τ) dτ .

Integrate by parts twice:

∫ t

0

e−τ sin(t − τ) dτ =[e−τ cos(t − τ)

]∣∣∣t0

−[e−τ sin(t − τ)

]∣∣∣t0

−∫ t

0

e−τ sin(t − τ) dτ,

2

∫ t

0

e−τ sin(t − τ) dτ =[e−τ cos(t − τ)

]∣∣∣t0

−[e−τ sin(t − τ)

]∣∣∣t0

,

2(f ∗ g)(t) = e−t − cos(t)− 0 + sin(t).

We conclude: (f ∗ g)(t) =1

2

[e−t + sin(t)− cos(t)

]. C

Convolution of two functions.

Example

Find the convolution of f (t) = e−t and g(t) = sin(t).

Solution: By definition: (f ∗ g)(t) =

∫ t

0

e−τ sin(t − τ) dτ .

Integrate by parts twice:

∫ t

0

e−τ sin(t − τ) dτ =[e−τ cos(t − τ)

]∣∣∣t0

−[e−τ sin(t − τ)

]∣∣∣t0

−∫ t

0

e−τ sin(t − τ) dτ,

2

∫ t

0

e−τ sin(t − τ) dτ =[e−τ cos(t − τ)

]∣∣∣t0

−[e−τ sin(t − τ)

]∣∣∣t0

,

2(f ∗ g)(t) = e−t − cos(t)− 0 + sin(t).

We conclude: (f ∗ g)(t) =1

2

[e−t + sin(t)− cos(t)

]. C

Convolution of two functions.

Example

Find the convolution of f (t) = e−t and g(t) = sin(t).

Solution: By definition: (f ∗ g)(t) =

∫ t

0

e−τ sin(t − τ) dτ .

Integrate by parts twice:

∫ t

0

e−τ sin(t − τ) dτ =[e−τ cos(t − τ)

]∣∣∣t0

−[e−τ sin(t − τ)

]∣∣∣t0

−∫ t

0

e−τ sin(t − τ) dτ,

2

∫ t

0

e−τ sin(t − τ) dτ =[e−τ cos(t − τ)

]∣∣∣t0

−[e−τ sin(t − τ)

]∣∣∣t0

,

2(f ∗ g)(t) = e−t − cos(t)− 0 + sin(t).

We conclude: (f ∗ g)(t) =1

2

[e−t + sin(t)− cos(t)

]. C

Convolution solutions (Sect. 6.6).

I Convolution of two functions.

I Properties of convolutions.

I Laplace Transform of a convolution.

I Impulse response solution.

I Solution decomposition theorem.

Properties of convolutions.

Theorem (Properties)

For every piecewise continuous functions f , g , and h, hold:

(i) Commutativity: f ∗ g = g ∗ f ;

(ii) Associativity: f ∗ (g ∗ h) = (f ∗ g) ∗ h;

(iii) Distributivity: f ∗ (g + h) = f ∗ g + f ∗ h;

(iv) Neutral element: f ∗ 0 = 0;

(v) Identity element: f ∗ δ = f .

Proof:

(v): (f ∗ δ)(t) =

∫ t

0

f (τ) δ(t − τ) dτ = f (t).

Properties of convolutions.

Theorem (Properties)

For every piecewise continuous functions f , g , and h, hold:

(i) Commutativity: f ∗ g = g ∗ f ;

(ii) Associativity: f ∗ (g ∗ h) = (f ∗ g) ∗ h;

(iii) Distributivity: f ∗ (g + h) = f ∗ g + f ∗ h;

(iv) Neutral element: f ∗ 0 = 0;

(v) Identity element: f ∗ δ = f .

Proof:

(v): (f ∗ δ)(t) =

∫ t

0

f (τ) δ(t − τ) dτ

= f (t).

Properties of convolutions.

Theorem (Properties)

For every piecewise continuous functions f , g , and h, hold:

(i) Commutativity: f ∗ g = g ∗ f ;

(ii) Associativity: f ∗ (g ∗ h) = (f ∗ g) ∗ h;

(iii) Distributivity: f ∗ (g + h) = f ∗ g + f ∗ h;

(iv) Neutral element: f ∗ 0 = 0;

(v) Identity element: f ∗ δ = f .

Proof:

(v): (f ∗ δ)(t) =

∫ t

0

f (τ) δ(t − τ) dτ = f (t).

Properties of convolutions.

Proof:(1): Commutativity: f ∗ g = g ∗ f .

The definition of convolution is,

(f ∗ g)(t) =

∫ t

0

f (τ) g(t − τ) dτ.

Change the integration variable: τ̂ = t − τ , hence d τ̂ = −dτ ,

(f ∗ g)(t) =

∫ 0

tf (t − τ̂) g(τ̂)(−1) d τ̂

(f ∗ g)(t) =

∫ t

0

g(τ̂) f (t − τ̂) d τ̂

We conclude: (f ∗ g)(t) = (g ∗ f )(t).

Properties of convolutions.

Proof:(1): Commutativity: f ∗ g = g ∗ f .

The definition of convolution is,

(f ∗ g)(t) =

∫ t

0

f (τ) g(t − τ) dτ.

Change the integration variable: τ̂ = t − τ , hence d τ̂ = −dτ ,

(f ∗ g)(t) =

∫ 0

tf (t − τ̂) g(τ̂)(−1) d τ̂

(f ∗ g)(t) =

∫ t

0

g(τ̂) f (t − τ̂) d τ̂

We conclude: (f ∗ g)(t) = (g ∗ f )(t).

Properties of convolutions.

Proof:(1): Commutativity: f ∗ g = g ∗ f .

The definition of convolution is,

(f ∗ g)(t) =

∫ t

0

f (τ) g(t − τ) dτ.

Change the integration variable: τ̂ = t − τ ,

hence d τ̂ = −dτ ,

(f ∗ g)(t) =

∫ 0

tf (t − τ̂) g(τ̂)(−1) d τ̂

(f ∗ g)(t) =

∫ t

0

g(τ̂) f (t − τ̂) d τ̂

We conclude: (f ∗ g)(t) = (g ∗ f )(t).

Properties of convolutions.

Proof:(1): Commutativity: f ∗ g = g ∗ f .

The definition of convolution is,

(f ∗ g)(t) =

∫ t

0

f (τ) g(t − τ) dτ.

Change the integration variable: τ̂ = t − τ , hence d τ̂ = −dτ ,

(f ∗ g)(t) =

∫ 0

tf (t − τ̂) g(τ̂)(−1) d τ̂

(f ∗ g)(t) =

∫ t

0

g(τ̂) f (t − τ̂) d τ̂

We conclude: (f ∗ g)(t) = (g ∗ f )(t).

Properties of convolutions.

Proof:(1): Commutativity: f ∗ g = g ∗ f .

The definition of convolution is,

(f ∗ g)(t) =

∫ t

0

f (τ) g(t − τ) dτ.

Change the integration variable: τ̂ = t − τ , hence d τ̂ = −dτ ,

(f ∗ g)(t) =

∫ 0

tf (t − τ̂) g(τ̂)(−1) d τ̂

(f ∗ g)(t) =

∫ t

0

g(τ̂) f (t − τ̂) d τ̂

We conclude: (f ∗ g)(t) = (g ∗ f )(t).

Properties of convolutions.

Proof:(1): Commutativity: f ∗ g = g ∗ f .

The definition of convolution is,

(f ∗ g)(t) =

∫ t

0

f (τ) g(t − τ) dτ.

Change the integration variable: τ̂ = t − τ , hence d τ̂ = −dτ ,

(f ∗ g)(t) =

∫ 0

tf (t − τ̂) g(τ̂)(−1) d τ̂

(f ∗ g)(t) =

∫ t

0

g(τ̂) f (t − τ̂) d τ̂

We conclude: (f ∗ g)(t) = (g ∗ f )(t).

Properties of convolutions.

Proof:(1): Commutativity: f ∗ g = g ∗ f .

The definition of convolution is,

(f ∗ g)(t) =

∫ t

0

f (τ) g(t − τ) dτ.

Change the integration variable: τ̂ = t − τ , hence d τ̂ = −dτ ,

(f ∗ g)(t) =

∫ 0

tf (t − τ̂) g(τ̂)(−1) d τ̂

(f ∗ g)(t) =

∫ t

0

g(τ̂) f (t − τ̂) d τ̂

We conclude: (f ∗ g)(t) = (g ∗ f )(t).

Convolution solutions (Sect. 6.6).

I Convolution of two functions.

I Properties of convolutions.

I Laplace Transform of a convolution.

I Impulse response solution.

I Solution decomposition theorem.

Laplace Transform of a convolution.

Theorem (Laplace Transform)

If f , g have well-defined Laplace Transforms L[f ], L[g ], then

L[f ∗ g ] = L[f ]L[g ].

Proof: The key step is to interchange two integrals. We start wethe product of the Laplace transforms,

L[f ]L[g ] =[∫ ∞

0

e−st f (t) dt] [∫ ∞

0

e−st̃g(t̃) dt̃],

L[f ]L[g ] =

∫ ∞

0

e−st̃g(t̃)(∫ ∞

0

e−st f (t) dt)

dt̃,

L[f ]L[g ] =

∫ ∞

0

g(t̃)(∫ ∞

0

e−s(t+t̃)f (t) dt)

dt̃.

Laplace Transform of a convolution.

Theorem (Laplace Transform)

If f , g have well-defined Laplace Transforms L[f ], L[g ], then

L[f ∗ g ] = L[f ]L[g ].

Proof: The key step is to interchange two integrals.

We start wethe product of the Laplace transforms,

L[f ]L[g ] =[∫ ∞

0

e−st f (t) dt] [∫ ∞

0

e−st̃g(t̃) dt̃],

L[f ]L[g ] =

∫ ∞

0

e−st̃g(t̃)(∫ ∞

0

e−st f (t) dt)

dt̃,

L[f ]L[g ] =

∫ ∞

0

g(t̃)(∫ ∞

0

e−s(t+t̃)f (t) dt)

dt̃.

Laplace Transform of a convolution.

Theorem (Laplace Transform)

If f , g have well-defined Laplace Transforms L[f ], L[g ], then

L[f ∗ g ] = L[f ]L[g ].

Proof: The key step is to interchange two integrals. We start wethe product of the Laplace transforms,

L[f ]L[g ] =[∫ ∞

0

e−st f (t) dt] [∫ ∞

0

e−st̃g(t̃) dt̃],

L[f ]L[g ] =

∫ ∞

0

e−st̃g(t̃)(∫ ∞

0

e−st f (t) dt)

dt̃,

L[f ]L[g ] =

∫ ∞

0

g(t̃)(∫ ∞

0

e−s(t+t̃)f (t) dt)

dt̃.

Laplace Transform of a convolution.

Theorem (Laplace Transform)

If f , g have well-defined Laplace Transforms L[f ], L[g ], then

L[f ∗ g ] = L[f ]L[g ].

Proof: The key step is to interchange two integrals. We start wethe product of the Laplace transforms,

L[f ]L[g ] =[∫ ∞

0

e−st f (t) dt] [∫ ∞

0

e−st̃g(t̃) dt̃],

L[f ]L[g ] =

∫ ∞

0

e−st̃g(t̃)(∫ ∞

0

e−st f (t) dt)

dt̃,

L[f ]L[g ] =

∫ ∞

0

g(t̃)(∫ ∞

0

e−s(t+t̃)f (t) dt)

dt̃.

Laplace Transform of a convolution.

Theorem (Laplace Transform)

If f , g have well-defined Laplace Transforms L[f ], L[g ], then

L[f ∗ g ] = L[f ]L[g ].

Proof: The key step is to interchange two integrals. We start wethe product of the Laplace transforms,

L[f ]L[g ] =[∫ ∞

0

e−st f (t) dt] [∫ ∞

0

e−st̃g(t̃) dt̃],

L[f ]L[g ] =

∫ ∞

0

e−st̃g(t̃)(∫ ∞

0

e−st f (t) dt)

dt̃,

L[f ]L[g ] =

∫ ∞

0

g(t̃)(∫ ∞

0

e−s(t+t̃)f (t) dt)

dt̃.

Laplace Transform of a convolution.

Proof: Recall: L[f ]L[g ] =

∫ ∞

0

g(t̃)(∫ ∞

0

e−s(t+t̃)f (t) dt)

dt̃.

Change variables: τ = t + t̃, hence dτ = dt;

L[f ]L[g ] =

∫ ∞

0

g(t̃)(∫ ∞

t̃e−sτ f (τ − t̃) dτ

)dt̃.

L[f ]L[g ] =

∫ ∞

0

∫ ∞

t̃e−sτ g(t̃) f (τ − t̃) dτ dt̃.

The key step: Switch the order of integration.

t = tau

tau

t

0

L[f ]L[g ] =

∫ ∞

0

∫ τ

0e−sτ g(t̃) f (τ − t̃) dt̃ dτ.

Laplace Transform of a convolution.

Proof: Recall: L[f ]L[g ] =

∫ ∞

0

g(t̃)(∫ ∞

0

e−s(t+t̃)f (t) dt)

dt̃.

Change variables: τ = t + t̃,

hence dτ = dt;

L[f ]L[g ] =

∫ ∞

0

g(t̃)(∫ ∞

t̃e−sτ f (τ − t̃) dτ

)dt̃.

L[f ]L[g ] =

∫ ∞

0

∫ ∞

t̃e−sτ g(t̃) f (τ − t̃) dτ dt̃.

The key step: Switch the order of integration.

t = tau

tau

t

0

L[f ]L[g ] =

∫ ∞

0

∫ τ

0e−sτ g(t̃) f (τ − t̃) dt̃ dτ.

Laplace Transform of a convolution.

Proof: Recall: L[f ]L[g ] =

∫ ∞

0

g(t̃)(∫ ∞

0

e−s(t+t̃)f (t) dt)

dt̃.

Change variables: τ = t + t̃, hence dτ = dt;

L[f ]L[g ] =

∫ ∞

0

g(t̃)(∫ ∞

t̃e−sτ f (τ − t̃) dτ

)dt̃.

L[f ]L[g ] =

∫ ∞

0

∫ ∞

t̃e−sτ g(t̃) f (τ − t̃) dτ dt̃.

The key step: Switch the order of integration.

t = tau

tau

t

0

L[f ]L[g ] =

∫ ∞

0

∫ τ

0e−sτ g(t̃) f (τ − t̃) dt̃ dτ.

Laplace Transform of a convolution.

Proof: Recall: L[f ]L[g ] =

∫ ∞

0

g(t̃)(∫ ∞

0

e−s(t+t̃)f (t) dt)

dt̃.

Change variables: τ = t + t̃, hence dτ = dt;

L[f ]L[g ] =

∫ ∞

0

g(t̃)(∫ ∞

t̃e−sτ f (τ − t̃) dτ

)dt̃.

L[f ]L[g ] =

∫ ∞

0

∫ ∞

t̃e−sτ g(t̃) f (τ − t̃) dτ dt̃.

The key step: Switch the order of integration.

t = tau

tau

t

0

L[f ]L[g ] =

∫ ∞

0

∫ τ

0e−sτ g(t̃) f (τ − t̃) dt̃ dτ.

Laplace Transform of a convolution.

Proof: Recall: L[f ]L[g ] =

∫ ∞

0

g(t̃)(∫ ∞

0

e−s(t+t̃)f (t) dt)

dt̃.

Change variables: τ = t + t̃, hence dτ = dt;

L[f ]L[g ] =

∫ ∞

0

g(t̃)(∫ ∞

t̃e−sτ f (τ − t̃) dτ

)dt̃.

L[f ]L[g ] =

∫ ∞

0

∫ ∞

t̃e−sτ g(t̃) f (τ − t̃) dτ dt̃.

The key step: Switch the order of integration.

t = tau

tau

t

0

L[f ]L[g ] =

∫ ∞

0

∫ τ

0e−sτ g(t̃) f (τ − t̃) dt̃ dτ.

Laplace Transform of a convolution.

Proof: Recall: L[f ]L[g ] =

∫ ∞

0

g(t̃)(∫ ∞

0

e−s(t+t̃)f (t) dt)

dt̃.

Change variables: τ = t + t̃, hence dτ = dt;

L[f ]L[g ] =

∫ ∞

0

g(t̃)(∫ ∞

t̃e−sτ f (τ − t̃) dτ

)dt̃.

L[f ]L[g ] =

∫ ∞

0

∫ ∞

t̃e−sτ g(t̃) f (τ − t̃) dτ dt̃.

The key step: Switch the order of integration.

t = tau

tau

t

0

L[f ]L[g ] =

∫ ∞

0

∫ τ

0e−sτ g(t̃) f (τ − t̃) dt̃ dτ.

Laplace Transform of a convolution.

Proof: Recall: L[f ]L[g ] =

∫ ∞

0

g(t̃)(∫ ∞

0

e−s(t+t̃)f (t) dt)

dt̃.

Change variables: τ = t + t̃, hence dτ = dt;

L[f ]L[g ] =

∫ ∞

0

g(t̃)(∫ ∞

t̃e−sτ f (τ − t̃) dτ

)dt̃.

L[f ]L[g ] =

∫ ∞

0

∫ ∞

t̃e−sτ g(t̃) f (τ − t̃) dτ dt̃.

The key step: Switch the order of integration.

t = tau

tau

t

0

L[f ]L[g ] =

∫ ∞

0

∫ τ

0e−sτ g(t̃) f (τ − t̃) dt̃ dτ.

Laplace Transform of a convolution.

Proof: Recall: L[f ]L[g ] =

∫ ∞

0

g(t̃)(∫ ∞

0

e−s(t+t̃)f (t) dt)

dt̃.

Change variables: τ = t + t̃, hence dτ = dt;

L[f ]L[g ] =

∫ ∞

0

g(t̃)(∫ ∞

t̃e−sτ f (τ − t̃) dτ

)dt̃.

L[f ]L[g ] =

∫ ∞

0

∫ ∞

t̃e−sτ g(t̃) f (τ − t̃) dτ dt̃.

The key step: Switch the order of integration.

t = tau

tau

t

0

L[f ]L[g ] =

∫ ∞

0

∫ τ

0e−sτ g(t̃) f (τ − t̃) dt̃ dτ.

Laplace Transform of a convolution.

Proof: Recall: L[f ]L[g ] =

∫ ∞

0

∫ τ

0e−sτ g(t̃) f (τ − t̃) dt̃ dτ .

Then, is straightforward to check that

L[f ]L[g ] =

∫ ∞

0

e−sτ(∫ τ

0g(t̃) f (τ − t̃) dt̃

)dτ,

L[f ]L[g ] =

∫ ∞

0

e−sτ (g ∗ f )(τ) dτ

L[f ]L[g ] = L[g ∗ f ]

We conclude: L[f ∗ g ] = L[f ]L[g ].

Laplace Transform of a convolution.

Proof: Recall: L[f ]L[g ] =

∫ ∞

0

∫ τ

0e−sτ g(t̃) f (τ − t̃) dt̃ dτ .

Then, is straightforward to check that

L[f ]L[g ] =

∫ ∞

0

e−sτ(∫ τ

0g(t̃) f (τ − t̃) dt̃

)dτ,

L[f ]L[g ] =

∫ ∞

0

e−sτ (g ∗ f )(τ) dτ

L[f ]L[g ] = L[g ∗ f ]

We conclude: L[f ∗ g ] = L[f ]L[g ].

Laplace Transform of a convolution.

Proof: Recall: L[f ]L[g ] =

∫ ∞

0

∫ τ

0e−sτ g(t̃) f (τ − t̃) dt̃ dτ .

Then, is straightforward to check that

L[f ]L[g ] =

∫ ∞

0

e−sτ(∫ τ

0g(t̃) f (τ − t̃) dt̃

)dτ,

L[f ]L[g ] =

∫ ∞

0

e−sτ (g ∗ f )(τ) dτ

L[f ]L[g ] = L[g ∗ f ]

We conclude: L[f ∗ g ] = L[f ]L[g ].

Laplace Transform of a convolution.

Proof: Recall: L[f ]L[g ] =

∫ ∞

0

∫ τ

0e−sτ g(t̃) f (τ − t̃) dt̃ dτ .

Then, is straightforward to check that

L[f ]L[g ] =

∫ ∞

0

e−sτ(∫ τ

0g(t̃) f (τ − t̃) dt̃

)dτ,

L[f ]L[g ] =

∫ ∞

0

e−sτ (g ∗ f )(τ) dτ

L[f ]L[g ] = L[g ∗ f ]

We conclude: L[f ∗ g ] = L[f ]L[g ].

Laplace Transform of a convolution.

Proof: Recall: L[f ]L[g ] =

∫ ∞

0

∫ τ

0e−sτ g(t̃) f (τ − t̃) dt̃ dτ .

Then, is straightforward to check that

L[f ]L[g ] =

∫ ∞

0

e−sτ(∫ τ

0g(t̃) f (τ − t̃) dt̃

)dτ,

L[f ]L[g ] =

∫ ∞

0

e−sτ (g ∗ f )(τ) dτ

L[f ]L[g ] = L[g ∗ f ]

We conclude: L[f ∗ g ] = L[f ]L[g ].

Convolution solutions (Sect. 6.6).

I Convolution of two functions.

I Properties of convolutions.

I Laplace Transform of a convolution.

I Impulse response solution.

I Solution decomposition theorem.

Impulse response solution.

DefinitionThe impulse response solution is the function yδ solution of the IVP

y ′′δ + a1 y ′δ + a0 yδ = δ(t − c), yδ(0) = 0, y ′δ(0) = 0, c ∈ R.

Example

Find the impulse response solution of the IVP

y ′′δ + 2 y ′δ + 2 yδ = δ(t − c), yδ(0) = 0, y ′δ(0) = 0.

Solution: L[y ′′δ ] + 2L[y ′δ] + 2L[yδ] = L[δ(t − c)].

(s2 + 2s + 2)L[yδ] = e−cs ⇒ L[yδ] =e−cs

(s2 + 2s + 2).

Impulse response solution.

DefinitionThe impulse response solution is the function yδ solution of the IVP

y ′′δ + a1 y ′δ + a0 yδ = δ(t − c), yδ(0) = 0, y ′δ(0) = 0, c ∈ R.

Example

Find the impulse response solution of the IVP

y ′′δ + 2 y ′δ + 2 yδ = δ(t − c), yδ(0) = 0, y ′δ(0) = 0.

Solution: L[y ′′δ ] + 2L[y ′δ] + 2L[yδ] = L[δ(t − c)].

(s2 + 2s + 2)L[yδ] = e−cs ⇒ L[yδ] =e−cs

(s2 + 2s + 2).

Impulse response solution.

DefinitionThe impulse response solution is the function yδ solution of the IVP

y ′′δ + a1 y ′δ + a0 yδ = δ(t − c), yδ(0) = 0, y ′δ(0) = 0, c ∈ R.

Example

Find the impulse response solution of the IVP

y ′′δ + 2 y ′δ + 2 yδ = δ(t − c), yδ(0) = 0, y ′δ(0) = 0.

Solution: L[y ′′δ ] + 2L[y ′δ] + 2L[yδ] = L[δ(t − c)].

(s2 + 2s + 2)L[yδ] = e−cs ⇒ L[yδ] =e−cs

(s2 + 2s + 2).

Impulse response solution.

DefinitionThe impulse response solution is the function yδ solution of the IVP

y ′′δ + a1 y ′δ + a0 yδ = δ(t − c), yδ(0) = 0, y ′δ(0) = 0, c ∈ R.

Example

Find the impulse response solution of the IVP

y ′′δ + 2 y ′δ + 2 yδ = δ(t − c), yδ(0) = 0, y ′δ(0) = 0.

Solution: L[y ′′δ ] + 2L[y ′δ] + 2L[yδ] = L[δ(t − c)].

(s2 + 2s + 2)L[yδ] = e−cs

⇒ L[yδ] =e−cs

(s2 + 2s + 2).

Impulse response solution.

DefinitionThe impulse response solution is the function yδ solution of the IVP

y ′′δ + a1 y ′δ + a0 yδ = δ(t − c), yδ(0) = 0, y ′δ(0) = 0, c ∈ R.

Example

Find the impulse response solution of the IVP

y ′′δ + 2 y ′δ + 2 yδ = δ(t − c), yδ(0) = 0, y ′δ(0) = 0.

Solution: L[y ′′δ ] + 2L[y ′δ] + 2L[yδ] = L[δ(t − c)].

(s2 + 2s + 2)L[yδ] = e−cs ⇒ L[yδ] =e−cs

(s2 + 2s + 2).

Impulse response solution.

Example

Find the impulse response solution of the IVP

y ′′δ + 2 y ′δ + 2 yδ = δ(t − c), yδ(0) = 0, y ′δ(0) = 0, .

Solution: Recall: L[yδ] =e−cs

(s2 + 2s + 2).

Find the roots of the denominator,

s2 + 2s + 2 = 0 ⇒ s± =1

2

[−2±

√4− 8

]Complex roots. We complete the square:

s2 + 2s + 2 =[s2 + 2

(2

2

)s + 1

]− 1 + 2 = (s + 1)2 + 1.

Therefore, L[yδ] =e−cs

(s + 1)2 + 1.

Impulse response solution.

Example

Find the impulse response solution of the IVP

y ′′δ + 2 y ′δ + 2 yδ = δ(t − c), yδ(0) = 0, y ′δ(0) = 0, .

Solution: Recall: L[yδ] =e−cs

(s2 + 2s + 2).

Find the roots of the denominator,

s2 + 2s + 2 = 0

⇒ s± =1

2

[−2±

√4− 8

]Complex roots. We complete the square:

s2 + 2s + 2 =[s2 + 2

(2

2

)s + 1

]− 1 + 2 = (s + 1)2 + 1.

Therefore, L[yδ] =e−cs

(s + 1)2 + 1.

Impulse response solution.

Example

Find the impulse response solution of the IVP

y ′′δ + 2 y ′δ + 2 yδ = δ(t − c), yδ(0) = 0, y ′δ(0) = 0, .

Solution: Recall: L[yδ] =e−cs

(s2 + 2s + 2).

Find the roots of the denominator,

s2 + 2s + 2 = 0 ⇒ s± =1

2

[−2±

√4− 8

]

Complex roots. We complete the square:

s2 + 2s + 2 =[s2 + 2

(2

2

)s + 1

]− 1 + 2 = (s + 1)2 + 1.

Therefore, L[yδ] =e−cs

(s + 1)2 + 1.

Impulse response solution.

Example

Find the impulse response solution of the IVP

y ′′δ + 2 y ′δ + 2 yδ = δ(t − c), yδ(0) = 0, y ′δ(0) = 0, .

Solution: Recall: L[yδ] =e−cs

(s2 + 2s + 2).

Find the roots of the denominator,

s2 + 2s + 2 = 0 ⇒ s± =1

2

[−2±

√4− 8

]Complex roots.

We complete the square:

s2 + 2s + 2 =[s2 + 2

(2

2

)s + 1

]− 1 + 2 = (s + 1)2 + 1.

Therefore, L[yδ] =e−cs

(s + 1)2 + 1.

Impulse response solution.

Example

Find the impulse response solution of the IVP

y ′′δ + 2 y ′δ + 2 yδ = δ(t − c), yδ(0) = 0, y ′δ(0) = 0, .

Solution: Recall: L[yδ] =e−cs

(s2 + 2s + 2).

Find the roots of the denominator,

s2 + 2s + 2 = 0 ⇒ s± =1

2

[−2±

√4− 8

]Complex roots. We complete the square:

s2 + 2s + 2 =[s2 + 2

(2

2

)s + 1

]− 1 + 2 = (s + 1)2 + 1.

Therefore, L[yδ] =e−cs

(s + 1)2 + 1.

Impulse response solution.

Example

Find the impulse response solution of the IVP

y ′′δ + 2 y ′δ + 2 yδ = δ(t − c), yδ(0) = 0, y ′δ(0) = 0, .

Solution: Recall: L[yδ] =e−cs

(s2 + 2s + 2).

Find the roots of the denominator,

s2 + 2s + 2 = 0 ⇒ s± =1

2

[−2±

√4− 8

]Complex roots. We complete the square:

s2 + 2s + 2 =[s2 + 2

(2

2

)s + 1

]− 1 + 2

= (s + 1)2 + 1.

Therefore, L[yδ] =e−cs

(s + 1)2 + 1.

Impulse response solution.

Example

Find the impulse response solution of the IVP

y ′′δ + 2 y ′δ + 2 yδ = δ(t − c), yδ(0) = 0, y ′δ(0) = 0, .

Solution: Recall: L[yδ] =e−cs

(s2 + 2s + 2).

Find the roots of the denominator,

s2 + 2s + 2 = 0 ⇒ s± =1

2

[−2±

√4− 8

]Complex roots. We complete the square:

s2 + 2s + 2 =[s2 + 2

(2

2

)s + 1

]− 1 + 2 = (s + 1)2 + 1.

Therefore, L[yδ] =e−cs

(s + 1)2 + 1.

Impulse response solution.

Example

Find the impulse response solution of the IVP

y ′′δ + 2 y ′δ + 2 yδ = δ(t − c), yδ(0) = 0, y ′δ(0) = 0, .

Solution: Recall: L[yδ] =e−cs

(s2 + 2s + 2).

Find the roots of the denominator,

s2 + 2s + 2 = 0 ⇒ s± =1

2

[−2±

√4− 8

]Complex roots. We complete the square:

s2 + 2s + 2 =[s2 + 2

(2

2

)s + 1

]− 1 + 2 = (s + 1)2 + 1.

Therefore, L[yδ] =e−cs

(s + 1)2 + 1.

Impulse response solution.

Example

Find the impulse response solution of the IVP

y ′′δ + 2 y ′δ + 2 yδ = δ(t − c), yδ(0) = 0, y ′δ(0) = 0, .

Solution: Recall: L[yδ] =e−cs

(s + 1)2 + 1.

Recall: L[sin(t)] =1

s2 + 1, and L[f ](s − c) = L[ect f (t)].

1

(s + 1)2 + 1= L[e−t sin(t)] ⇒ L[yδ] = e−cs L[e−t sin(t)].

Since e−cs L[f ](s) = L[u(t − c) f (t − c)],

we conclude yδ(t) = u(t − c) e−(t−c) sin(t − c). C

Impulse response solution.

Example

Find the impulse response solution of the IVP

y ′′δ + 2 y ′δ + 2 yδ = δ(t − c), yδ(0) = 0, y ′δ(0) = 0, .

Solution: Recall: L[yδ] =e−cs

(s + 1)2 + 1.

Recall: L[sin(t)] =1

s2 + 1,

and L[f ](s − c) = L[ect f (t)].

1

(s + 1)2 + 1= L[e−t sin(t)] ⇒ L[yδ] = e−cs L[e−t sin(t)].

Since e−cs L[f ](s) = L[u(t − c) f (t − c)],

we conclude yδ(t) = u(t − c) e−(t−c) sin(t − c). C

Impulse response solution.

Example

Find the impulse response solution of the IVP

y ′′δ + 2 y ′δ + 2 yδ = δ(t − c), yδ(0) = 0, y ′δ(0) = 0, .

Solution: Recall: L[yδ] =e−cs

(s + 1)2 + 1.

Recall: L[sin(t)] =1

s2 + 1, and L[f ](s − c) = L[ect f (t)].

1

(s + 1)2 + 1= L[e−t sin(t)] ⇒ L[yδ] = e−cs L[e−t sin(t)].

Since e−cs L[f ](s) = L[u(t − c) f (t − c)],

we conclude yδ(t) = u(t − c) e−(t−c) sin(t − c). C

Impulse response solution.

Example

Find the impulse response solution of the IVP

y ′′δ + 2 y ′δ + 2 yδ = δ(t − c), yδ(0) = 0, y ′δ(0) = 0, .

Solution: Recall: L[yδ] =e−cs

(s + 1)2 + 1.

Recall: L[sin(t)] =1

s2 + 1, and L[f ](s − c) = L[ect f (t)].

1

(s + 1)2 + 1= L[e−t sin(t)]

⇒ L[yδ] = e−cs L[e−t sin(t)].

Since e−cs L[f ](s) = L[u(t − c) f (t − c)],

we conclude yδ(t) = u(t − c) e−(t−c) sin(t − c). C

Impulse response solution.

Example

Find the impulse response solution of the IVP

y ′′δ + 2 y ′δ + 2 yδ = δ(t − c), yδ(0) = 0, y ′δ(0) = 0, .

Solution: Recall: L[yδ] =e−cs

(s + 1)2 + 1.

Recall: L[sin(t)] =1

s2 + 1, and L[f ](s − c) = L[ect f (t)].

1

(s + 1)2 + 1= L[e−t sin(t)] ⇒ L[yδ] = e−cs L[e−t sin(t)].

Since e−cs L[f ](s) = L[u(t − c) f (t − c)],

we conclude yδ(t) = u(t − c) e−(t−c) sin(t − c). C

Impulse response solution.

Example

Find the impulse response solution of the IVP

y ′′δ + 2 y ′δ + 2 yδ = δ(t − c), yδ(0) = 0, y ′δ(0) = 0, .

Solution: Recall: L[yδ] =e−cs

(s + 1)2 + 1.

Recall: L[sin(t)] =1

s2 + 1, and L[f ](s − c) = L[ect f (t)].

1

(s + 1)2 + 1= L[e−t sin(t)] ⇒ L[yδ] = e−cs L[e−t sin(t)].

Since e−cs L[f ](s) = L[u(t − c) f (t − c)],

we conclude yδ(t) = u(t − c) e−(t−c) sin(t − c). C

Impulse response solution.

Example

Find the impulse response solution of the IVP

y ′′δ + 2 y ′δ + 2 yδ = δ(t − c), yδ(0) = 0, y ′δ(0) = 0, .

Solution: Recall: L[yδ] =e−cs

(s + 1)2 + 1.

Recall: L[sin(t)] =1

s2 + 1, and L[f ](s − c) = L[ect f (t)].

1

(s + 1)2 + 1= L[e−t sin(t)] ⇒ L[yδ] = e−cs L[e−t sin(t)].

Since e−cs L[f ](s) = L[u(t − c) f (t − c)],

we conclude yδ(t) = u(t − c) e−(t−c) sin(t − c). C

Convolution solutions (Sect. 6.6).

I Convolution of two functions.

I Properties of convolutions.

I Laplace Transform of a convolution.

I Impulse response solution.

I Solution decomposition theorem.

Solution decomposition theorem.

Theorem (Solution decomposition)

The solution y to the IVP

y ′′ + a1 y ′ + a0 y = g(t), y(0) = y0, y ′(0) = y1,

can be decomposed as

y(t) = yh(t) + (yδ ∗ g)(t),

where yh is the solution of the homogeneous IVP

y ′′h + a1 y ′h + a0 yh = 0, yh(0) = y0, y ′h(0) = y1,

and yδ is the impulse response solution, that is,

y ′′δ + a1 y ′δ + a0 yδ = δ(t), yδ(0) = 0, y ′δ(0) = 0.

Solution decomposition theorem.

Example

Use the Solution Decomposition Theorem to express the solution of

y ′′ + 2 y ′ + 2 y = sin(at), y(0) = 1, y ′(0) = −1.

Solution: L[y ′′] + 2L[y ′] + 2L[y ] = L[sin(at)], and recall,

L[y ′′] = s2 L[y ]− s (1)− (−1), L[y ′] = s L[y ]− 1.

(s2 + 2s + 2)L[y ]− s + 1− 2 = L[sin(at)].

L[y ] =(s + 1)

(s2 + 2s + 2)+

1

(s2 + 2s + 2)L[sin(at)].

Solution decomposition theorem.

Example

Use the Solution Decomposition Theorem to express the solution of

y ′′ + 2 y ′ + 2 y = sin(at), y(0) = 1, y ′(0) = −1.

Solution: L[y ′′] + 2L[y ′] + 2L[y ] = L[sin(at)],

and recall,

L[y ′′] = s2 L[y ]− s (1)− (−1), L[y ′] = s L[y ]− 1.

(s2 + 2s + 2)L[y ]− s + 1− 2 = L[sin(at)].

L[y ] =(s + 1)

(s2 + 2s + 2)+

1

(s2 + 2s + 2)L[sin(at)].

Solution decomposition theorem.

Example

Use the Solution Decomposition Theorem to express the solution of

y ′′ + 2 y ′ + 2 y = sin(at), y(0) = 1, y ′(0) = −1.

Solution: L[y ′′] + 2L[y ′] + 2L[y ] = L[sin(at)], and recall,

L[y ′′] = s2 L[y ]− s (1)− (−1),

L[y ′] = s L[y ]− 1.

(s2 + 2s + 2)L[y ]− s + 1− 2 = L[sin(at)].

L[y ] =(s + 1)

(s2 + 2s + 2)+

1

(s2 + 2s + 2)L[sin(at)].

Solution decomposition theorem.

Example

Use the Solution Decomposition Theorem to express the solution of

y ′′ + 2 y ′ + 2 y = sin(at), y(0) = 1, y ′(0) = −1.

Solution: L[y ′′] + 2L[y ′] + 2L[y ] = L[sin(at)], and recall,

L[y ′′] = s2 L[y ]− s (1)− (−1), L[y ′] = s L[y ]− 1.

(s2 + 2s + 2)L[y ]− s + 1− 2 = L[sin(at)].

L[y ] =(s + 1)

(s2 + 2s + 2)+

1

(s2 + 2s + 2)L[sin(at)].

Solution decomposition theorem.

Example

Use the Solution Decomposition Theorem to express the solution of

y ′′ + 2 y ′ + 2 y = sin(at), y(0) = 1, y ′(0) = −1.

Solution: L[y ′′] + 2L[y ′] + 2L[y ] = L[sin(at)], and recall,

L[y ′′] = s2 L[y ]− s (1)− (−1), L[y ′] = s L[y ]− 1.

(s2 + 2s + 2)L[y ]− s + 1− 2 = L[sin(at)].

L[y ] =(s + 1)

(s2 + 2s + 2)+

1

(s2 + 2s + 2)L[sin(at)].

Solution decomposition theorem.

Example

Use the Solution Decomposition Theorem to express the solution of

y ′′ + 2 y ′ + 2 y = sin(at), y(0) = 1, y ′(0) = −1.

Solution: L[y ′′] + 2L[y ′] + 2L[y ] = L[sin(at)], and recall,

L[y ′′] = s2 L[y ]− s (1)− (−1), L[y ′] = s L[y ]− 1.

(s2 + 2s + 2)L[y ]− s + 1− 2 = L[sin(at)].

L[y ] =(s + 1)

(s2 + 2s + 2)+

1

(s2 + 2s + 2)L[sin(at)].

Solution decomposition theorem.

Example

Use the Solution Decomposition Theorem to express the solution of

y ′′ + 2 y ′ + 2 y = sin(at), y(0) = 1, y ′(0) = −1.

Solution: Recall: L[y ] =(s + 1)

(s2 + 2s + 2)+

1

(s2 + 2s + 2)L[sin(at)].

But: L[yh] =(s + 1)

(s2 + 2s + 2)=

(s + 1)

(s + 1)2 + 1= L[e−t cos(t)],

and: L[yδ] =1

(s2 + 2s + 2)=

1

(s + 1)2 + 1= L[e−t sin(t)]. So,

L[y ] = L[yh] + L[yδ]L[g(t)] ⇒ y(t) = yh(t) + (yδ ∗ g)(t),

So: y(t) = e−t cos(t) +

∫ t

0e−τ sin(τ) sin[a(t − τ)] dτ . C

Solution decomposition theorem.

Example

Use the Solution Decomposition Theorem to express the solution of

y ′′ + 2 y ′ + 2 y = sin(at), y(0) = 1, y ′(0) = −1.

Solution: Recall: L[y ] =(s + 1)

(s2 + 2s + 2)+

1

(s2 + 2s + 2)L[sin(at)].

But: L[yh] =(s + 1)

(s2 + 2s + 2)

=(s + 1)

(s + 1)2 + 1= L[e−t cos(t)],

and: L[yδ] =1

(s2 + 2s + 2)=

1

(s + 1)2 + 1= L[e−t sin(t)]. So,

L[y ] = L[yh] + L[yδ]L[g(t)] ⇒ y(t) = yh(t) + (yδ ∗ g)(t),

So: y(t) = e−t cos(t) +

∫ t

0e−τ sin(τ) sin[a(t − τ)] dτ . C

Solution decomposition theorem.

Example

Use the Solution Decomposition Theorem to express the solution of

y ′′ + 2 y ′ + 2 y = sin(at), y(0) = 1, y ′(0) = −1.

Solution: Recall: L[y ] =(s + 1)

(s2 + 2s + 2)+

1

(s2 + 2s + 2)L[sin(at)].

But: L[yh] =(s + 1)

(s2 + 2s + 2)=

(s + 1)

(s + 1)2 + 1

= L[e−t cos(t)],

and: L[yδ] =1

(s2 + 2s + 2)=

1

(s + 1)2 + 1= L[e−t sin(t)]. So,

L[y ] = L[yh] + L[yδ]L[g(t)] ⇒ y(t) = yh(t) + (yδ ∗ g)(t),

So: y(t) = e−t cos(t) +

∫ t

0e−τ sin(τ) sin[a(t − τ)] dτ . C

Solution decomposition theorem.

Example

Use the Solution Decomposition Theorem to express the solution of

y ′′ + 2 y ′ + 2 y = sin(at), y(0) = 1, y ′(0) = −1.

Solution: Recall: L[y ] =(s + 1)

(s2 + 2s + 2)+

1

(s2 + 2s + 2)L[sin(at)].

But: L[yh] =(s + 1)

(s2 + 2s + 2)=

(s + 1)

(s + 1)2 + 1= L[e−t cos(t)],

and: L[yδ] =1

(s2 + 2s + 2)=

1

(s + 1)2 + 1= L[e−t sin(t)]. So,

L[y ] = L[yh] + L[yδ]L[g(t)] ⇒ y(t) = yh(t) + (yδ ∗ g)(t),

So: y(t) = e−t cos(t) +

∫ t

0e−τ sin(τ) sin[a(t − τ)] dτ . C

Solution decomposition theorem.

Example

Use the Solution Decomposition Theorem to express the solution of

y ′′ + 2 y ′ + 2 y = sin(at), y(0) = 1, y ′(0) = −1.

Solution: Recall: L[y ] =(s + 1)

(s2 + 2s + 2)+

1

(s2 + 2s + 2)L[sin(at)].

But: L[yh] =(s + 1)

(s2 + 2s + 2)=

(s + 1)

(s + 1)2 + 1= L[e−t cos(t)],

and: L[yδ] =1

(s2 + 2s + 2)

=1

(s + 1)2 + 1= L[e−t sin(t)]. So,

L[y ] = L[yh] + L[yδ]L[g(t)] ⇒ y(t) = yh(t) + (yδ ∗ g)(t),

So: y(t) = e−t cos(t) +

∫ t

0e−τ sin(τ) sin[a(t − τ)] dτ . C

Solution decomposition theorem.

Example

Use the Solution Decomposition Theorem to express the solution of

y ′′ + 2 y ′ + 2 y = sin(at), y(0) = 1, y ′(0) = −1.

Solution: Recall: L[y ] =(s + 1)

(s2 + 2s + 2)+

1

(s2 + 2s + 2)L[sin(at)].

But: L[yh] =(s + 1)

(s2 + 2s + 2)=

(s + 1)

(s + 1)2 + 1= L[e−t cos(t)],

and: L[yδ] =1

(s2 + 2s + 2)=

1

(s + 1)2 + 1

= L[e−t sin(t)]. So,

L[y ] = L[yh] + L[yδ]L[g(t)] ⇒ y(t) = yh(t) + (yδ ∗ g)(t),

So: y(t) = e−t cos(t) +

∫ t

0e−τ sin(τ) sin[a(t − τ)] dτ . C

Solution decomposition theorem.

Example

Use the Solution Decomposition Theorem to express the solution of

y ′′ + 2 y ′ + 2 y = sin(at), y(0) = 1, y ′(0) = −1.

Solution: Recall: L[y ] =(s + 1)

(s2 + 2s + 2)+

1

(s2 + 2s + 2)L[sin(at)].

But: L[yh] =(s + 1)

(s2 + 2s + 2)=

(s + 1)

(s + 1)2 + 1= L[e−t cos(t)],

and: L[yδ] =1

(s2 + 2s + 2)=

1

(s + 1)2 + 1= L[e−t sin(t)].

So,

L[y ] = L[yh] + L[yδ]L[g(t)] ⇒ y(t) = yh(t) + (yδ ∗ g)(t),

So: y(t) = e−t cos(t) +

∫ t

0e−τ sin(τ) sin[a(t − τ)] dτ . C

Solution decomposition theorem.

Example

Use the Solution Decomposition Theorem to express the solution of

y ′′ + 2 y ′ + 2 y = sin(at), y(0) = 1, y ′(0) = −1.

Solution: Recall: L[y ] =(s + 1)

(s2 + 2s + 2)+

1

(s2 + 2s + 2)L[sin(at)].

But: L[yh] =(s + 1)

(s2 + 2s + 2)=

(s + 1)

(s + 1)2 + 1= L[e−t cos(t)],

and: L[yδ] =1

(s2 + 2s + 2)=

1

(s + 1)2 + 1= L[e−t sin(t)]. So,

L[y ] = L[yh] + L[yδ]L[g(t)]

⇒ y(t) = yh(t) + (yδ ∗ g)(t),

So: y(t) = e−t cos(t) +

∫ t

0e−τ sin(τ) sin[a(t − τ)] dτ . C

Solution decomposition theorem.

Example

Use the Solution Decomposition Theorem to express the solution of

y ′′ + 2 y ′ + 2 y = sin(at), y(0) = 1, y ′(0) = −1.

Solution: Recall: L[y ] =(s + 1)

(s2 + 2s + 2)+

1

(s2 + 2s + 2)L[sin(at)].

But: L[yh] =(s + 1)

(s2 + 2s + 2)=

(s + 1)

(s + 1)2 + 1= L[e−t cos(t)],

and: L[yδ] =1

(s2 + 2s + 2)=

1

(s + 1)2 + 1= L[e−t sin(t)]. So,

L[y ] = L[yh] + L[yδ]L[g(t)] ⇒ y(t) = yh(t) + (yδ ∗ g)(t),

So: y(t) = e−t cos(t) +

∫ t

0e−τ sin(τ) sin[a(t − τ)] dτ . C

Solution decomposition theorem.

Example

Use the Solution Decomposition Theorem to express the solution of

y ′′ + 2 y ′ + 2 y = sin(at), y(0) = 1, y ′(0) = −1.

Solution: Recall: L[y ] =(s + 1)

(s2 + 2s + 2)+

1

(s2 + 2s + 2)L[sin(at)].

But: L[yh] =(s + 1)

(s2 + 2s + 2)=

(s + 1)

(s + 1)2 + 1= L[e−t cos(t)],

and: L[yδ] =1

(s2 + 2s + 2)=

1

(s + 1)2 + 1= L[e−t sin(t)]. So,

L[y ] = L[yh] + L[yδ]L[g(t)] ⇒ y(t) = yh(t) + (yδ ∗ g)(t),

So: y(t) = e−t cos(t) +

∫ t

0e−τ sin(τ) sin[a(t − τ)] dτ . C

Solution decomposition theorem.

Proof: Compute: L[y ′′] + a1 L[y ′] + a0 L[y ] = L[g(t)],

and recall,

L[y ′′] = s2 L[y ]− sy0 − y1, L[y ′] = s L[y ]− y0.

(s2 + a1s + a0)L[y ]− sy0 − y1 − a1y0 = L[g(t)].

L[y ] =(s + a1)y0 + y1

(s2 + a1s + a0)+

1

(s2 + a1s + a0)L[g(t)].

Recall: L[yh] =(s + a1)y0 + y1

(s2 + a1s + a0), and L[yδ] =

1

(s2 + a1s + a0).

Since, L[y ] = L[yh] + L[yδ]L[g(t)], so y(t) = yh(t) + (yδ ∗ g)(t).

Equivalently: y(t) = yh(t) +

∫ t

0yδ(τ)g(t − τ) dτ .

Solution decomposition theorem.

Proof: Compute: L[y ′′] + a1 L[y ′] + a0 L[y ] = L[g(t)], and recall,

L[y ′′] = s2 L[y ]− sy0 − y1,

L[y ′] = s L[y ]− y0.

(s2 + a1s + a0)L[y ]− sy0 − y1 − a1y0 = L[g(t)].

L[y ] =(s + a1)y0 + y1

(s2 + a1s + a0)+

1

(s2 + a1s + a0)L[g(t)].

Recall: L[yh] =(s + a1)y0 + y1

(s2 + a1s + a0), and L[yδ] =

1

(s2 + a1s + a0).

Since, L[y ] = L[yh] + L[yδ]L[g(t)], so y(t) = yh(t) + (yδ ∗ g)(t).

Equivalently: y(t) = yh(t) +

∫ t

0yδ(τ)g(t − τ) dτ .

Solution decomposition theorem.

Proof: Compute: L[y ′′] + a1 L[y ′] + a0 L[y ] = L[g(t)], and recall,

L[y ′′] = s2 L[y ]− sy0 − y1, L[y ′] = s L[y ]− y0.

(s2 + a1s + a0)L[y ]− sy0 − y1 − a1y0 = L[g(t)].

L[y ] =(s + a1)y0 + y1

(s2 + a1s + a0)+

1

(s2 + a1s + a0)L[g(t)].

Recall: L[yh] =(s + a1)y0 + y1

(s2 + a1s + a0), and L[yδ] =

1

(s2 + a1s + a0).

Since, L[y ] = L[yh] + L[yδ]L[g(t)], so y(t) = yh(t) + (yδ ∗ g)(t).

Equivalently: y(t) = yh(t) +

∫ t

0yδ(τ)g(t − τ) dτ .

Solution decomposition theorem.

Proof: Compute: L[y ′′] + a1 L[y ′] + a0 L[y ] = L[g(t)], and recall,

L[y ′′] = s2 L[y ]− sy0 − y1, L[y ′] = s L[y ]− y0.

(s2 + a1s + a0)L[y ]− sy0 − y1 − a1y0 = L[g(t)].

L[y ] =(s + a1)y0 + y1

(s2 + a1s + a0)+

1

(s2 + a1s + a0)L[g(t)].

Recall: L[yh] =(s + a1)y0 + y1

(s2 + a1s + a0), and L[yδ] =

1

(s2 + a1s + a0).

Since, L[y ] = L[yh] + L[yδ]L[g(t)], so y(t) = yh(t) + (yδ ∗ g)(t).

Equivalently: y(t) = yh(t) +

∫ t

0yδ(τ)g(t − τ) dτ .

Solution decomposition theorem.

Proof: Compute: L[y ′′] + a1 L[y ′] + a0 L[y ] = L[g(t)], and recall,

L[y ′′] = s2 L[y ]− sy0 − y1, L[y ′] = s L[y ]− y0.

(s2 + a1s + a0)L[y ]− sy0 − y1 − a1y0 = L[g(t)].

L[y ] =(s + a1)y0 + y1

(s2 + a1s + a0)+

1

(s2 + a1s + a0)L[g(t)].

Recall: L[yh] =(s + a1)y0 + y1

(s2 + a1s + a0), and L[yδ] =

1

(s2 + a1s + a0).

Since, L[y ] = L[yh] + L[yδ]L[g(t)], so y(t) = yh(t) + (yδ ∗ g)(t).

Equivalently: y(t) = yh(t) +

∫ t

0yδ(τ)g(t − τ) dτ .

Solution decomposition theorem.

Proof: Compute: L[y ′′] + a1 L[y ′] + a0 L[y ] = L[g(t)], and recall,

L[y ′′] = s2 L[y ]− sy0 − y1, L[y ′] = s L[y ]− y0.

(s2 + a1s + a0)L[y ]− sy0 − y1 − a1y0 = L[g(t)].

L[y ] =(s + a1)y0 + y1

(s2 + a1s + a0)+

1

(s2 + a1s + a0)L[g(t)].

Recall: L[yh] =(s + a1)y0 + y1

(s2 + a1s + a0),

and L[yδ] =1

(s2 + a1s + a0).

Since, L[y ] = L[yh] + L[yδ]L[g(t)], so y(t) = yh(t) + (yδ ∗ g)(t).

Equivalently: y(t) = yh(t) +

∫ t

0yδ(τ)g(t − τ) dτ .

Solution decomposition theorem.

Proof: Compute: L[y ′′] + a1 L[y ′] + a0 L[y ] = L[g(t)], and recall,

L[y ′′] = s2 L[y ]− sy0 − y1, L[y ′] = s L[y ]− y0.

(s2 + a1s + a0)L[y ]− sy0 − y1 − a1y0 = L[g(t)].

L[y ] =(s + a1)y0 + y1

(s2 + a1s + a0)+

1

(s2 + a1s + a0)L[g(t)].

Recall: L[yh] =(s + a1)y0 + y1

(s2 + a1s + a0), and L[yδ] =

1

(s2 + a1s + a0).

Since, L[y ] = L[yh] + L[yδ]L[g(t)], so y(t) = yh(t) + (yδ ∗ g)(t).

Equivalently: y(t) = yh(t) +

∫ t

0yδ(τ)g(t − τ) dτ .

Solution decomposition theorem.

Proof: Compute: L[y ′′] + a1 L[y ′] + a0 L[y ] = L[g(t)], and recall,

L[y ′′] = s2 L[y ]− sy0 − y1, L[y ′] = s L[y ]− y0.

(s2 + a1s + a0)L[y ]− sy0 − y1 − a1y0 = L[g(t)].

L[y ] =(s + a1)y0 + y1

(s2 + a1s + a0)+

1

(s2 + a1s + a0)L[g(t)].

Recall: L[yh] =(s + a1)y0 + y1

(s2 + a1s + a0), and L[yδ] =

1

(s2 + a1s + a0).

Since, L[y ] = L[yh] + L[yδ]L[g(t)],

so y(t) = yh(t) + (yδ ∗ g)(t).

Equivalently: y(t) = yh(t) +

∫ t

0yδ(τ)g(t − τ) dτ .

Solution decomposition theorem.

Proof: Compute: L[y ′′] + a1 L[y ′] + a0 L[y ] = L[g(t)], and recall,

L[y ′′] = s2 L[y ]− sy0 − y1, L[y ′] = s L[y ]− y0.

(s2 + a1s + a0)L[y ]− sy0 − y1 − a1y0 = L[g(t)].

L[y ] =(s + a1)y0 + y1

(s2 + a1s + a0)+

1

(s2 + a1s + a0)L[g(t)].

Recall: L[yh] =(s + a1)y0 + y1

(s2 + a1s + a0), and L[yδ] =

1

(s2 + a1s + a0).

Since, L[y ] = L[yh] + L[yδ]L[g(t)], so y(t) = yh(t) + (yδ ∗ g)(t).

Equivalently: y(t) = yh(t) +

∫ t

0yδ(τ)g(t − τ) dτ .

Solution decomposition theorem.

Proof: Compute: L[y ′′] + a1 L[y ′] + a0 L[y ] = L[g(t)], and recall,

L[y ′′] = s2 L[y ]− sy0 − y1, L[y ′] = s L[y ]− y0.

(s2 + a1s + a0)L[y ]− sy0 − y1 − a1y0 = L[g(t)].

L[y ] =(s + a1)y0 + y1

(s2 + a1s + a0)+

1

(s2 + a1s + a0)L[g(t)].

Recall: L[yh] =(s + a1)y0 + y1

(s2 + a1s + a0), and L[yδ] =

1

(s2 + a1s + a0).

Since, L[y ] = L[yh] + L[yδ]L[g(t)], so y(t) = yh(t) + (yδ ∗ g)(t).

Equivalently: y(t) = yh(t) +

∫ t

0yδ(τ)g(t − τ) dτ .

Systems of linear differential equations (Sect. 7.1).

I n × n systems of linear differential equations.

I Second order equations and first order systems.

I Main concepts from Linear Algebra.

n × n systems of linear differential equations.

Remark: Many physical systems must be described with morethan one differential equation.

Example

Newton’s law of motion for a particle of mass m moving in space.The unknown and the force are vector-valued functions,

x(t) =

x1(t)x2(t)x3(t)

, F(t) =

F1(t, x)F2(t, x)F3(t, x)

.

The equation of motion are: md2x

dt2= F

(t, x(t)

).

These are three differential equations,

md2x1

dt2= F1

(t, x(t)

), m

d2x2

dt2= F2

(t, x(t)

), m

d2x3

dt2= F3

(t, x(t)

).

C

n × n systems of linear differential equations.

Remark: Many physical systems must be described with morethan one differential equation.

Example

Newton’s law of motion for a particle of mass m moving in space.

The unknown and the force are vector-valued functions,

x(t) =

x1(t)x2(t)x3(t)

, F(t) =

F1(t, x)F2(t, x)F3(t, x)

.

The equation of motion are: md2x

dt2= F

(t, x(t)

).

These are three differential equations,

md2x1

dt2= F1

(t, x(t)

), m

d2x2

dt2= F2

(t, x(t)

), m

d2x3

dt2= F3

(t, x(t)

).

C

n × n systems of linear differential equations.

Remark: Many physical systems must be described with morethan one differential equation.

Example

Newton’s law of motion for a particle of mass m moving in space.The unknown and the force are vector-valued functions,

x(t) =

x1(t)x2(t)x3(t)

, F(t) =

F1(t, x)F2(t, x)F3(t, x)

.

The equation of motion are: md2x

dt2= F

(t, x(t)

).

These are three differential equations,

md2x1

dt2= F1

(t, x(t)

), m

d2x2

dt2= F2

(t, x(t)

), m

d2x3

dt2= F3

(t, x(t)

).

C

n × n systems of linear differential equations.

Remark: Many physical systems must be described with morethan one differential equation.

Example

Newton’s law of motion for a particle of mass m moving in space.The unknown and the force are vector-valued functions,

x(t) =

x1(t)x2(t)x3(t)

,

F(t) =

F1(t, x)F2(t, x)F3(t, x)

.

The equation of motion are: md2x

dt2= F

(t, x(t)

).

These are three differential equations,

md2x1

dt2= F1

(t, x(t)

), m

d2x2

dt2= F2

(t, x(t)

), m

d2x3

dt2= F3

(t, x(t)

).

C

n × n systems of linear differential equations.

Remark: Many physical systems must be described with morethan one differential equation.

Example

Newton’s law of motion for a particle of mass m moving in space.The unknown and the force are vector-valued functions,

x(t) =

x1(t)x2(t)x3(t)

, F(t) =

F1(t, x)F2(t, x)F3(t, x)

.

The equation of motion are: md2x

dt2= F

(t, x(t)

).

These are three differential equations,

md2x1

dt2= F1

(t, x(t)

), m

d2x2

dt2= F2

(t, x(t)

), m

d2x3

dt2= F3

(t, x(t)

).

C

n × n systems of linear differential equations.

Remark: Many physical systems must be described with morethan one differential equation.

Example

Newton’s law of motion for a particle of mass m moving in space.The unknown and the force are vector-valued functions,

x(t) =

x1(t)x2(t)x3(t)

, F(t) =

F1(t, x)F2(t, x)F3(t, x)

.

The equation of motion are: md2x

dt2= F

(t, x(t)

).

These are three differential equations,

md2x1

dt2= F1

(t, x(t)

), m

d2x2

dt2= F2

(t, x(t)

), m

d2x3

dt2= F3

(t, x(t)

).

C

n × n systems of linear differential equations.

Remark: Many physical systems must be described with morethan one differential equation.

Example

Newton’s law of motion for a particle of mass m moving in space.The unknown and the force are vector-valued functions,

x(t) =

x1(t)x2(t)x3(t)

, F(t) =

F1(t, x)F2(t, x)F3(t, x)

.

The equation of motion are: md2x

dt2= F

(t, x(t)

).

These are three differential equations,

md2x1

dt2= F1

(t, x(t)

), m

d2x2

dt2= F2

(t, x(t)

), m

d2x3

dt2= F3

(t, x(t)

).

C

n × n systems of linear differential equations.

DefinitionAn n × n system of linear first order differential equations is thefollowing: Given the functions aij , gi : [a, b] → R, wherei , j = 1, · · · , n, find n functions xj : [a, b] → R solutions of the nlinear differential equations

x ′1 = a11(t) x1 + · · ·+ a1n(t) xn + g1(t)

...

x ′n = an1(t) x1 + · · ·+ ann(t) xn + gn(t).

The system is called homogeneous iff the source functions satisfythat g1 = · · · = gn = 0.

n × n systems of linear differential equations.

Example

n = 1: Single differential equation: Find x1(t) solution of

x ′1 = a11(t) x1 + g1(t).

Example

n = 2: 2× 2 linear system: Find x1(t) and x2(t) solutions of

x ′1 = a11(t) x1 + a12(t) x2 + g1(t),

x ′2 = a21(t) x1 + a22(t) x2 + g2(t).

Example

n = 2: 2× 2 homogeneous linear system: Find x1(t) and x2(t),

x ′1 = a11(t) x1 + a12(t) x2

x ′2 = a21(t) x1 + a22(t) x2.

n × n systems of linear differential equations.

Example

n = 1: Single differential equation: Find x1(t) solution of

x ′1 = a11(t) x1 + g1(t).

Example

n = 2: 2× 2 linear system: Find x1(t) and x2(t) solutions of

x ′1 = a11(t) x1 + a12(t) x2 + g1(t),

x ′2 = a21(t) x1 + a22(t) x2 + g2(t).

Example

n = 2: 2× 2 homogeneous linear system: Find x1(t) and x2(t),

x ′1 = a11(t) x1 + a12(t) x2

x ′2 = a21(t) x1 + a22(t) x2.

n × n systems of linear differential equations.

Example

n = 1: Single differential equation: Find x1(t) solution of

x ′1 = a11(t) x1 + g1(t).

Example

n = 2: 2× 2 linear system: Find x1(t) and x2(t) solutions of

x ′1 = a11(t) x1 + a12(t) x2 + g1(t),

x ′2 = a21(t) x1 + a22(t) x2 + g2(t).

Example

n = 2: 2× 2 homogeneous linear system: Find x1(t) and x2(t),

x ′1 = a11(t) x1 + a12(t) x2

x ′2 = a21(t) x1 + a22(t) x2.

n × n systems of linear differential equations.

Example

Find x1(t), x2(t) solutions of the 2× 2,constant coefficients, homogeneous system

x ′1 = x1 − x2,

x ′2 = −x1 + x2.

Solution: Add up the equations, and subtract the equations,

(x1 + x2)′ = 0, (x1 − x2)

′ = 2(x1 − x2).

Introduce the unknowns v = x1 + x2, w = x1 − x2, then

v ′ = 0 ⇒ v = c1,

w ′ = 2w ⇒ w = c2e2t .

Back to x1 and x2: x1 =1

2(v + w), x2 =

1

2(v − w).

We conclude: x1(t) =1

2(c1 + c2e

2t), x2(t) =1

2(c1 − c2e

2t).C

n × n systems of linear differential equations.

Example

Find x1(t), x2(t) solutions of the 2× 2,constant coefficients, homogeneous system

x ′1 = x1 − x2,

x ′2 = −x1 + x2.

Solution: Add up the equations, and subtract the equations,

(x1 + x2)′ = 0, (x1 − x2)

′ = 2(x1 − x2).

Introduce the unknowns v = x1 + x2, w = x1 − x2, then

v ′ = 0 ⇒ v = c1,

w ′ = 2w ⇒ w = c2e2t .

Back to x1 and x2: x1 =1

2(v + w), x2 =

1

2(v − w).

We conclude: x1(t) =1

2(c1 + c2e

2t), x2(t) =1

2(c1 − c2e

2t).C

n × n systems of linear differential equations.

Example

Find x1(t), x2(t) solutions of the 2× 2,constant coefficients, homogeneous system

x ′1 = x1 − x2,

x ′2 = −x1 + x2.

Solution: Add up the equations, and subtract the equations,

(x1 + x2)′ = 0,

(x1 − x2)′ = 2(x1 − x2).

Introduce the unknowns v = x1 + x2, w = x1 − x2, then

v ′ = 0 ⇒ v = c1,

w ′ = 2w ⇒ w = c2e2t .

Back to x1 and x2: x1 =1

2(v + w), x2 =

1

2(v − w).

We conclude: x1(t) =1

2(c1 + c2e

2t), x2(t) =1

2(c1 − c2e

2t).C

n × n systems of linear differential equations.

Example

Find x1(t), x2(t) solutions of the 2× 2,constant coefficients, homogeneous system

x ′1 = x1 − x2,

x ′2 = −x1 + x2.

Solution: Add up the equations, and subtract the equations,

(x1 + x2)′ = 0, (x1 − x2)

′ = 2(x1 − x2).

Introduce the unknowns v = x1 + x2, w = x1 − x2, then

v ′ = 0 ⇒ v = c1,

w ′ = 2w ⇒ w = c2e2t .

Back to x1 and x2: x1 =1

2(v + w), x2 =

1

2(v − w).

We conclude: x1(t) =1

2(c1 + c2e

2t), x2(t) =1

2(c1 − c2e

2t).C

n × n systems of linear differential equations.

Example

Find x1(t), x2(t) solutions of the 2× 2,constant coefficients, homogeneous system

x ′1 = x1 − x2,

x ′2 = −x1 + x2.

Solution: Add up the equations, and subtract the equations,

(x1 + x2)′ = 0, (x1 − x2)

′ = 2(x1 − x2).

Introduce the unknowns v = x1 + x2,

w = x1 − x2, then

v ′ = 0 ⇒ v = c1,

w ′ = 2w ⇒ w = c2e2t .

Back to x1 and x2: x1 =1

2(v + w), x2 =

1

2(v − w).

We conclude: x1(t) =1

2(c1 + c2e

2t), x2(t) =1

2(c1 − c2e

2t).C

n × n systems of linear differential equations.

Example

Find x1(t), x2(t) solutions of the 2× 2,constant coefficients, homogeneous system

x ′1 = x1 − x2,

x ′2 = −x1 + x2.

Solution: Add up the equations, and subtract the equations,

(x1 + x2)′ = 0, (x1 − x2)

′ = 2(x1 − x2).

Introduce the unknowns v = x1 + x2, w = x1 − x2,

then

v ′ = 0 ⇒ v = c1,

w ′ = 2w ⇒ w = c2e2t .

Back to x1 and x2: x1 =1

2(v + w), x2 =

1

2(v − w).

We conclude: x1(t) =1

2(c1 + c2e

2t), x2(t) =1

2(c1 − c2e

2t).C

n × n systems of linear differential equations.

Example

Find x1(t), x2(t) solutions of the 2× 2,constant coefficients, homogeneous system

x ′1 = x1 − x2,

x ′2 = −x1 + x2.

Solution: Add up the equations, and subtract the equations,

(x1 + x2)′ = 0, (x1 − x2)

′ = 2(x1 − x2).

Introduce the unknowns v = x1 + x2, w = x1 − x2, then

v ′ = 0

⇒ v = c1,

w ′ = 2w ⇒ w = c2e2t .

Back to x1 and x2: x1 =1

2(v + w), x2 =

1

2(v − w).

We conclude: x1(t) =1

2(c1 + c2e

2t), x2(t) =1

2(c1 − c2e

2t).C

n × n systems of linear differential equations.

Example

Find x1(t), x2(t) solutions of the 2× 2,constant coefficients, homogeneous system

x ′1 = x1 − x2,

x ′2 = −x1 + x2.

Solution: Add up the equations, and subtract the equations,

(x1 + x2)′ = 0, (x1 − x2)

′ = 2(x1 − x2).

Introduce the unknowns v = x1 + x2, w = x1 − x2, then

v ′ = 0 ⇒ v = c1,

w ′ = 2w ⇒ w = c2e2t .

Back to x1 and x2: x1 =1

2(v + w), x2 =

1

2(v − w).

We conclude: x1(t) =1

2(c1 + c2e

2t), x2(t) =1

2(c1 − c2e

2t).C

n × n systems of linear differential equations.

Example

Find x1(t), x2(t) solutions of the 2× 2,constant coefficients, homogeneous system

x ′1 = x1 − x2,

x ′2 = −x1 + x2.

Solution: Add up the equations, and subtract the equations,

(x1 + x2)′ = 0, (x1 − x2)

′ = 2(x1 − x2).

Introduce the unknowns v = x1 + x2, w = x1 − x2, then

v ′ = 0 ⇒ v = c1,

w ′ = 2w

⇒ w = c2e2t .

Back to x1 and x2: x1 =1

2(v + w), x2 =

1

2(v − w).

We conclude: x1(t) =1

2(c1 + c2e

2t), x2(t) =1

2(c1 − c2e

2t).C

n × n systems of linear differential equations.

Example

Find x1(t), x2(t) solutions of the 2× 2,constant coefficients, homogeneous system

x ′1 = x1 − x2,

x ′2 = −x1 + x2.

Solution: Add up the equations, and subtract the equations,

(x1 + x2)′ = 0, (x1 − x2)

′ = 2(x1 − x2).

Introduce the unknowns v = x1 + x2, w = x1 − x2, then

v ′ = 0 ⇒ v = c1,

w ′ = 2w ⇒ w = c2e2t .

Back to x1 and x2: x1 =1

2(v + w), x2 =

1

2(v − w).

We conclude: x1(t) =1

2(c1 + c2e

2t), x2(t) =1

2(c1 − c2e

2t).C

n × n systems of linear differential equations.

Example

Find x1(t), x2(t) solutions of the 2× 2,constant coefficients, homogeneous system

x ′1 = x1 − x2,

x ′2 = −x1 + x2.

Solution: Add up the equations, and subtract the equations,

(x1 + x2)′ = 0, (x1 − x2)

′ = 2(x1 − x2).

Introduce the unknowns v = x1 + x2, w = x1 − x2, then

v ′ = 0 ⇒ v = c1,

w ′ = 2w ⇒ w = c2e2t .

Back to x1 and x2:

x1 =1

2(v + w), x2 =

1

2(v − w).

We conclude: x1(t) =1

2(c1 + c2e

2t), x2(t) =1

2(c1 − c2e

2t).C

n × n systems of linear differential equations.

Example

Find x1(t), x2(t) solutions of the 2× 2,constant coefficients, homogeneous system

x ′1 = x1 − x2,

x ′2 = −x1 + x2.

Solution: Add up the equations, and subtract the equations,

(x1 + x2)′ = 0, (x1 − x2)

′ = 2(x1 − x2).

Introduce the unknowns v = x1 + x2, w = x1 − x2, then

v ′ = 0 ⇒ v = c1,

w ′ = 2w ⇒ w = c2e2t .

Back to x1 and x2: x1 =1

2(v + w),

x2 =1

2(v − w).

We conclude: x1(t) =1

2(c1 + c2e

2t), x2(t) =1

2(c1 − c2e

2t).C

n × n systems of linear differential equations.

Example

Find x1(t), x2(t) solutions of the 2× 2,constant coefficients, homogeneous system

x ′1 = x1 − x2,

x ′2 = −x1 + x2.

Solution: Add up the equations, and subtract the equations,

(x1 + x2)′ = 0, (x1 − x2)

′ = 2(x1 − x2).

Introduce the unknowns v = x1 + x2, w = x1 − x2, then

v ′ = 0 ⇒ v = c1,

w ′ = 2w ⇒ w = c2e2t .

Back to x1 and x2: x1 =1

2(v + w), x2 =

1

2(v − w).

We conclude: x1(t) =1

2(c1 + c2e

2t), x2(t) =1

2(c1 − c2e

2t).C

n × n systems of linear differential equations.

Example

Find x1(t), x2(t) solutions of the 2× 2,constant coefficients, homogeneous system

x ′1 = x1 − x2,

x ′2 = −x1 + x2.

Solution: Add up the equations, and subtract the equations,

(x1 + x2)′ = 0, (x1 − x2)

′ = 2(x1 − x2).

Introduce the unknowns v = x1 + x2, w = x1 − x2, then

v ′ = 0 ⇒ v = c1,

w ′ = 2w ⇒ w = c2e2t .

Back to x1 and x2: x1 =1

2(v + w), x2 =

1

2(v − w).

We conclude: x1(t) =1

2(c1 + c2e

2t), x2(t) =1

2(c1 − c2e

2t).C

Systems of linear differential equations (Sect. 7.1).

I n × n systems of linear differential equations.

I Second order equations and first order systems.

I Main concepts from Linear Algebra.

Second order equations and first order systems.

Theorem (Reduction to first order)

Every solution y to the second order linear equation

y ′′ + p(t) y ′ + q(t) y = g(t), (1)

defines a solution x1 = y and x2 = y ′ of the 2× 2 first order lineardifferential system

x ′1 = x2, (2)

x ′2 = −q(t) x1 − p(t) x2 + g(t). (3)

Conversely, every solution x1, x2 of the 2× 2 first order linearsystem in Eqs. (2)-(3) defines a solution y = x1 of the second orderdifferential equation in (1).

Second order equations and first order systems.

Proof:(⇒) Given y solution of y ′′ + p(t) y ′ + q(t) y = g(t),

introduce x1 = y and x2 = y ′, hence x ′1 = y ′ = x2, that is,

x ′1 = x2.

Then, x ′2 = y ′′ = −q(t) y − p(t) y ′ + g(t). That is,

x ′2 = −q(t) x1 − p(t) x2 + g(t).

(⇐) Introduce x2 = x ′1 into x ′2 = −q(t) x1 − p(t) x2 + g(t).

x ′′1 = −q(t) x1 − p(t) x ′1 + g(t),

that isx ′′1 + p(t) x ′1 + q(t) x1 = g(t).

Second order equations and first order systems.

Proof:(⇒) Given y solution of y ′′ + p(t) y ′ + q(t) y = g(t),

introduce x1 = y and x2 = y ′,

hence x ′1 = y ′ = x2, that is,

x ′1 = x2.

Then, x ′2 = y ′′ = −q(t) y − p(t) y ′ + g(t). That is,

x ′2 = −q(t) x1 − p(t) x2 + g(t).

(⇐) Introduce x2 = x ′1 into x ′2 = −q(t) x1 − p(t) x2 + g(t).

x ′′1 = −q(t) x1 − p(t) x ′1 + g(t),

that isx ′′1 + p(t) x ′1 + q(t) x1 = g(t).

Second order equations and first order systems.

Proof:(⇒) Given y solution of y ′′ + p(t) y ′ + q(t) y = g(t),

introduce x1 = y and x2 = y ′, hence x ′1 = y ′ = x2,

that is,

x ′1 = x2.

Then, x ′2 = y ′′ = −q(t) y − p(t) y ′ + g(t). That is,

x ′2 = −q(t) x1 − p(t) x2 + g(t).

(⇐) Introduce x2 = x ′1 into x ′2 = −q(t) x1 − p(t) x2 + g(t).

x ′′1 = −q(t) x1 − p(t) x ′1 + g(t),

that isx ′′1 + p(t) x ′1 + q(t) x1 = g(t).

Second order equations and first order systems.

Proof:(⇒) Given y solution of y ′′ + p(t) y ′ + q(t) y = g(t),

introduce x1 = y and x2 = y ′, hence x ′1 = y ′ = x2, that is,

x ′1 = x2.

Then, x ′2 = y ′′ = −q(t) y − p(t) y ′ + g(t). That is,

x ′2 = −q(t) x1 − p(t) x2 + g(t).

(⇐) Introduce x2 = x ′1 into x ′2 = −q(t) x1 − p(t) x2 + g(t).

x ′′1 = −q(t) x1 − p(t) x ′1 + g(t),

that isx ′′1 + p(t) x ′1 + q(t) x1 = g(t).

Second order equations and first order systems.

Proof:(⇒) Given y solution of y ′′ + p(t) y ′ + q(t) y = g(t),

introduce x1 = y and x2 = y ′, hence x ′1 = y ′ = x2, that is,

x ′1 = x2.

Then, x ′2 = y ′′

= −q(t) y − p(t) y ′ + g(t). That is,

x ′2 = −q(t) x1 − p(t) x2 + g(t).

(⇐) Introduce x2 = x ′1 into x ′2 = −q(t) x1 − p(t) x2 + g(t).

x ′′1 = −q(t) x1 − p(t) x ′1 + g(t),

that isx ′′1 + p(t) x ′1 + q(t) x1 = g(t).

Second order equations and first order systems.

Proof:(⇒) Given y solution of y ′′ + p(t) y ′ + q(t) y = g(t),

introduce x1 = y and x2 = y ′, hence x ′1 = y ′ = x2, that is,

x ′1 = x2.

Then, x ′2 = y ′′ = −q(t) y − p(t) y ′ + g(t).

That is,

x ′2 = −q(t) x1 − p(t) x2 + g(t).

(⇐) Introduce x2 = x ′1 into x ′2 = −q(t) x1 − p(t) x2 + g(t).

x ′′1 = −q(t) x1 − p(t) x ′1 + g(t),

that isx ′′1 + p(t) x ′1 + q(t) x1 = g(t).

Second order equations and first order systems.

Proof:(⇒) Given y solution of y ′′ + p(t) y ′ + q(t) y = g(t),

introduce x1 = y and x2 = y ′, hence x ′1 = y ′ = x2, that is,

x ′1 = x2.

Then, x ′2 = y ′′ = −q(t) y − p(t) y ′ + g(t). That is,

x ′2 = −q(t) x1 − p(t) x2 + g(t).

(⇐) Introduce x2 = x ′1 into x ′2 = −q(t) x1 − p(t) x2 + g(t).

x ′′1 = −q(t) x1 − p(t) x ′1 + g(t),

that isx ′′1 + p(t) x ′1 + q(t) x1 = g(t).

Second order equations and first order systems.

Proof:(⇒) Given y solution of y ′′ + p(t) y ′ + q(t) y = g(t),

introduce x1 = y and x2 = y ′, hence x ′1 = y ′ = x2, that is,

x ′1 = x2.

Then, x ′2 = y ′′ = −q(t) y − p(t) y ′ + g(t). That is,

x ′2 = −q(t) x1 − p(t) x2 + g(t).

(⇐) Introduce x2 = x ′1 into x ′2 = −q(t) x1 − p(t) x2 + g(t).

x ′′1 = −q(t) x1 − p(t) x ′1 + g(t),

that isx ′′1 + p(t) x ′1 + q(t) x1 = g(t).

Second order equations and first order systems.

Proof:(⇒) Given y solution of y ′′ + p(t) y ′ + q(t) y = g(t),

introduce x1 = y and x2 = y ′, hence x ′1 = y ′ = x2, that is,

x ′1 = x2.

Then, x ′2 = y ′′ = −q(t) y − p(t) y ′ + g(t). That is,

x ′2 = −q(t) x1 − p(t) x2 + g(t).

(⇐) Introduce x2 = x ′1 into x ′2 = −q(t) x1 − p(t) x2 + g(t).

x ′′1 = −q(t) x1 − p(t) x ′1 + g(t),

that isx ′′1 + p(t) x ′1 + q(t) x1 = g(t).

Second order equations and first order systems.

Proof:(⇒) Given y solution of y ′′ + p(t) y ′ + q(t) y = g(t),

introduce x1 = y and x2 = y ′, hence x ′1 = y ′ = x2, that is,

x ′1 = x2.

Then, x ′2 = y ′′ = −q(t) y − p(t) y ′ + g(t). That is,

x ′2 = −q(t) x1 − p(t) x2 + g(t).

(⇐) Introduce x2 = x ′1 into x ′2 = −q(t) x1 − p(t) x2 + g(t).

x ′′1 = −q(t) x1 − p(t) x ′1 + g(t),

that isx ′′1 + p(t) x ′1 + q(t) x1 = g(t).

Second order equations and first order systems.

Example

Express as a first order system the equation

y ′′ + 2y ′ + 2y = sin(at).

Solution: Introduce the new unknowns

x1 = y , x2 = y ′ ⇒ x ′1 = x2.

Then, the differential equation can be written as

x ′2 + 2x2 + 2x1 = sin(at).

We conclude thatx ′1 = x2.

x ′2 = −2x1 − 2x2 + sin(at). C

Second order equations and first order systems.

Example

Express as a first order system the equation

y ′′ + 2y ′ + 2y = sin(at).

Solution: Introduce the new unknowns

x1 = y , x2 = y ′

⇒ x ′1 = x2.

Then, the differential equation can be written as

x ′2 + 2x2 + 2x1 = sin(at).

We conclude thatx ′1 = x2.

x ′2 = −2x1 − 2x2 + sin(at). C

Second order equations and first order systems.

Example

Express as a first order system the equation

y ′′ + 2y ′ + 2y = sin(at).

Solution: Introduce the new unknowns

x1 = y , x2 = y ′ ⇒ x ′1 = x2.

Then, the differential equation can be written as

x ′2 + 2x2 + 2x1 = sin(at).

We conclude thatx ′1 = x2.

x ′2 = −2x1 − 2x2 + sin(at). C

Second order equations and first order systems.

Example

Express as a first order system the equation

y ′′ + 2y ′ + 2y = sin(at).

Solution: Introduce the new unknowns

x1 = y , x2 = y ′ ⇒ x ′1 = x2.

Then, the differential equation can be written as

x ′2 + 2x2 + 2x1 = sin(at).

We conclude thatx ′1 = x2.

x ′2 = −2x1 − 2x2 + sin(at). C

Second order equations and first order systems.

Example

Express as a first order system the equation

y ′′ + 2y ′ + 2y = sin(at).

Solution: Introduce the new unknowns

x1 = y , x2 = y ′ ⇒ x ′1 = x2.

Then, the differential equation can be written as

x ′2 + 2x2 + 2x1 = sin(at).

We conclude thatx ′1 = x2.

x ′2 = −2x1 − 2x2 + sin(at). C

Second order equations and first order systems.

Remark: Systems of first order equations can, sometimes, betransformed into a second order single equation.

Example

Express as a single second order equationthe 2× 2 system and solve it,

x ′1 = −x1 + 3x2,

x ′2 = x1 − x2.

Solution: Compute x1 from the second equation: x1 = x ′2 + x2.Introduce this expression into the first equation,

(x ′2 + x2)′ = −(x ′2 + x2) + 3x2,

x ′′2 + x ′2 = −x ′2 − x2 + 3x2,

x ′′2 + 2x ′2 − 2x2 = 0.

Second order equations and first order systems.

Remark: Systems of first order equations can, sometimes, betransformed into a second order single equation.

Example

Express as a single second order equationthe 2× 2 system and solve it,

x ′1 = −x1 + 3x2,

x ′2 = x1 − x2.

Solution: Compute x1 from the second equation: x1 = x ′2 + x2.Introduce this expression into the first equation,

(x ′2 + x2)′ = −(x ′2 + x2) + 3x2,

x ′′2 + x ′2 = −x ′2 − x2 + 3x2,

x ′′2 + 2x ′2 − 2x2 = 0.

Second order equations and first order systems.

Remark: Systems of first order equations can, sometimes, betransformed into a second order single equation.

Example

Express as a single second order equationthe 2× 2 system and solve it,

x ′1 = −x1 + 3x2,

x ′2 = x1 − x2.

Solution: Compute x1 from the second equation:

x1 = x ′2 + x2.Introduce this expression into the first equation,

(x ′2 + x2)′ = −(x ′2 + x2) + 3x2,

x ′′2 + x ′2 = −x ′2 − x2 + 3x2,

x ′′2 + 2x ′2 − 2x2 = 0.

Second order equations and first order systems.

Remark: Systems of first order equations can, sometimes, betransformed into a second order single equation.

Example

Express as a single second order equationthe 2× 2 system and solve it,

x ′1 = −x1 + 3x2,

x ′2 = x1 − x2.

Solution: Compute x1 from the second equation: x1 = x ′2 + x2.

Introduce this expression into the first equation,

(x ′2 + x2)′ = −(x ′2 + x2) + 3x2,

x ′′2 + x ′2 = −x ′2 − x2 + 3x2,

x ′′2 + 2x ′2 − 2x2 = 0.

Second order equations and first order systems.

Remark: Systems of first order equations can, sometimes, betransformed into a second order single equation.

Example

Express as a single second order equationthe 2× 2 system and solve it,

x ′1 = −x1 + 3x2,

x ′2 = x1 − x2.

Solution: Compute x1 from the second equation: x1 = x ′2 + x2.Introduce this expression into the first equation,

(x ′2 + x2)′ = −(x ′2 + x2) + 3x2,

x ′′2 + x ′2 = −x ′2 − x2 + 3x2,

x ′′2 + 2x ′2 − 2x2 = 0.

Second order equations and first order systems.

Remark: Systems of first order equations can, sometimes, betransformed into a second order single equation.

Example

Express as a single second order equationthe 2× 2 system and solve it,

x ′1 = −x1 + 3x2,

x ′2 = x1 − x2.

Solution: Compute x1 from the second equation: x1 = x ′2 + x2.Introduce this expression into the first equation,

(x ′2 + x2)′ = −(x ′2 + x2) + 3x2,

x ′′2 + x ′2 = −x ′2 − x2 + 3x2,

x ′′2 + 2x ′2 − 2x2 = 0.

Second order equations and first order systems.

Remark: Systems of first order equations can, sometimes, betransformed into a second order single equation.

Example

Express as a single second order equationthe 2× 2 system and solve it,

x ′1 = −x1 + 3x2,

x ′2 = x1 − x2.

Solution: Compute x1 from the second equation: x1 = x ′2 + x2.Introduce this expression into the first equation,

(x ′2 + x2)′ = −(x ′2 + x2) + 3x2,

x ′′2 + x ′2 = −x ′2 − x2 + 3x2,

x ′′2 + 2x ′2 − 2x2 = 0.

Second order equations and first order systems.

Remark: Systems of first order equations can, sometimes, betransformed into a second order single equation.

Example

Express as a single second order equationthe 2× 2 system and solve it,

x ′1 = −x1 + 3x2,

x ′2 = x1 − x2.

Solution: Compute x1 from the second equation: x1 = x ′2 + x2.Introduce this expression into the first equation,

(x ′2 + x2)′ = −(x ′2 + x2) + 3x2,

x ′′2 + x ′2 = −x ′2 − x2 + 3x2,

x ′′2 + 2x ′2 − 2x2 = 0.

Second order equations and first order systems.

Example

Express as a single second order equationthe 2× 2 system and solve it,

x ′1 = −x1 + 3x2,

x ′2 = x1 − x2.

Solution: Recall: x ′′2 + 2x ′2 − 2x2 = 0.

r2+2r−2 = 0 ⇒ r± =1

2

[−2±

√4 + 8

]⇒ r± = −1±

√3.

Therefore, x2 = c1 er+ t + c2 er− t . Since x1 = x ′2 + x2,

x1 =(c1r+ er+ t + c2r− er− t

)+

(c1 er+ t + c2 er− t

),

We conclude: x1 = c1(1 + r+) er+ t + c2(1 + r−) er− t . C

Second order equations and first order systems.

Example

Express as a single second order equationthe 2× 2 system and solve it,

x ′1 = −x1 + 3x2,

x ′2 = x1 − x2.

Solution: Recall: x ′′2 + 2x ′2 − 2x2 = 0.

r2+2r−2 = 0

⇒ r± =1

2

[−2±

√4 + 8

]⇒ r± = −1±

√3.

Therefore, x2 = c1 er+ t + c2 er− t . Since x1 = x ′2 + x2,

x1 =(c1r+ er+ t + c2r− er− t

)+

(c1 er+ t + c2 er− t

),

We conclude: x1 = c1(1 + r+) er+ t + c2(1 + r−) er− t . C

Second order equations and first order systems.

Example

Express as a single second order equationthe 2× 2 system and solve it,

x ′1 = −x1 + 3x2,

x ′2 = x1 − x2.

Solution: Recall: x ′′2 + 2x ′2 − 2x2 = 0.

r2+2r−2 = 0 ⇒ r± =1

2

[−2±

√4 + 8

]

⇒ r± = −1±√

3.

Therefore, x2 = c1 er+ t + c2 er− t . Since x1 = x ′2 + x2,

x1 =(c1r+ er+ t + c2r− er− t

)+

(c1 er+ t + c2 er− t

),

We conclude: x1 = c1(1 + r+) er+ t + c2(1 + r−) er− t . C

Second order equations and first order systems.

Example

Express as a single second order equationthe 2× 2 system and solve it,

x ′1 = −x1 + 3x2,

x ′2 = x1 − x2.

Solution: Recall: x ′′2 + 2x ′2 − 2x2 = 0.

r2+2r−2 = 0 ⇒ r± =1

2

[−2±

√4 + 8

]⇒ r± = −1±

√3.

Therefore, x2 = c1 er+ t + c2 er− t . Since x1 = x ′2 + x2,

x1 =(c1r+ er+ t + c2r− er− t

)+

(c1 er+ t + c2 er− t

),

We conclude: x1 = c1(1 + r+) er+ t + c2(1 + r−) er− t . C

Second order equations and first order systems.

Example

Express as a single second order equationthe 2× 2 system and solve it,

x ′1 = −x1 + 3x2,

x ′2 = x1 − x2.

Solution: Recall: x ′′2 + 2x ′2 − 2x2 = 0.

r2+2r−2 = 0 ⇒ r± =1

2

[−2±

√4 + 8

]⇒ r± = −1±

√3.

Therefore, x2 = c1 er+ t + c2 er− t .

Since x1 = x ′2 + x2,

x1 =(c1r+ er+ t + c2r− er− t

)+

(c1 er+ t + c2 er− t

),

We conclude: x1 = c1(1 + r+) er+ t + c2(1 + r−) er− t . C

Second order equations and first order systems.

Example

Express as a single second order equationthe 2× 2 system and solve it,

x ′1 = −x1 + 3x2,

x ′2 = x1 − x2.

Solution: Recall: x ′′2 + 2x ′2 − 2x2 = 0.

r2+2r−2 = 0 ⇒ r± =1

2

[−2±

√4 + 8

]⇒ r± = −1±

√3.

Therefore, x2 = c1 er+ t + c2 er− t . Since x1 = x ′2 + x2,

x1 =(c1r+ er+ t + c2r− er− t

)+

(c1 er+ t + c2 er− t

),

We conclude: x1 = c1(1 + r+) er+ t + c2(1 + r−) er− t . C

Second order equations and first order systems.

Example

Express as a single second order equationthe 2× 2 system and solve it,

x ′1 = −x1 + 3x2,

x ′2 = x1 − x2.

Solution: Recall: x ′′2 + 2x ′2 − 2x2 = 0.

r2+2r−2 = 0 ⇒ r± =1

2

[−2±

√4 + 8

]⇒ r± = −1±

√3.

Therefore, x2 = c1 er+ t + c2 er− t . Since x1 = x ′2 + x2,

x1 =(c1r+ er+ t + c2r− er− t

)+

(c1 er+ t + c2 er− t

),

We conclude: x1 = c1(1 + r+) er+ t + c2(1 + r−) er− t . C

Second order equations and first order systems.

Example

Express as a single second order equationthe 2× 2 system and solve it,

x ′1 = −x1 + 3x2,

x ′2 = x1 − x2.

Solution: Recall: x ′′2 + 2x ′2 − 2x2 = 0.

r2+2r−2 = 0 ⇒ r± =1

2

[−2±

√4 + 8

]⇒ r± = −1±

√3.

Therefore, x2 = c1 er+ t + c2 er− t . Since x1 = x ′2 + x2,

x1 =(c1r+ er+ t + c2r− er− t

)+

(c1 er+ t + c2 er− t

),

We conclude: x1 = c1(1 + r+) er+ t + c2(1 + r−) er− t . C

Systems of linear differential equations (Sect. 7.1).

I n × n systems of linear differential equations.

I Second order equations and first order systems.

I Main concepts from Linear Algebra.

Main concepts from Linear Algebra.

Remark: Ideas from LinearAlgebra are useful to studysystems of linear differentialequations.

We review:

I Matrices m × n.

I Matrix operations.

I n-vectors, dot product.

I matrix-vector product.

DefinitionAn m × n matrix, A, is an array of numbers

A =

a11 · · · a1n...

...am1 · · · amn

,m rows,

n columns.

where aij ∈ C and i = 1, · · · ,m, and j = 1, · · · , n. An n × nmatrix is called a square matrix.

Main concepts from Linear Algebra.

Remark: Ideas from LinearAlgebra are useful to studysystems of linear differentialequations.

We review:

I Matrices m × n.

I Matrix operations.

I n-vectors, dot product.

I matrix-vector product.

DefinitionAn m × n matrix, A, is an array of numbers

A =

a11 · · · a1n...

...am1 · · · amn

,m rows,

n columns.

where aij ∈ C and i = 1, · · · ,m, and j = 1, · · · , n. An n × nmatrix is called a square matrix.

Main concepts from Linear Algebra.

Remark: Ideas from LinearAlgebra are useful to studysystems of linear differentialequations.

We review:

I Matrices m × n.

I Matrix operations.

I n-vectors, dot product.

I matrix-vector product.

DefinitionAn m × n matrix, A, is an array of numbers

A =

a11 · · · a1n...

...am1 · · · amn

,m rows,

n columns.

where aij ∈ C and i = 1, · · · ,m, and j = 1, · · · , n. An n × nmatrix is called a square matrix.

Main concepts from Linear Algebra.

Example

(a) 2× 2 matrix: A =

[1 23 4

].

(b) 2× 3 matrix: A =

[1 2 34 5 6

].

(c) 3× 2 matrix: A =

1 23 45 6

.

(d) 2× 2 complex-valued matrix: A =

[1 + i 2− i

3 4i

].

(e) The coefficients of a linear system can be grouped in a matrix,

x ′1 = −x1 + 3x2

x ′2 = x1 − x2

}⇒ A =

[−1 31 −1

].

Main concepts from Linear Algebra.

Example

(a) 2× 2 matrix: A =

[1 23 4

].

(b) 2× 3 matrix: A =

[1 2 34 5 6

].

(c) 3× 2 matrix: A =

1 23 45 6

.

(d) 2× 2 complex-valued matrix: A =

[1 + i 2− i

3 4i

].

(e) The coefficients of a linear system can be grouped in a matrix,

x ′1 = −x1 + 3x2

x ′2 = x1 − x2

}⇒ A =

[−1 31 −1

].

Main concepts from Linear Algebra.

Example

(a) 2× 2 matrix: A =

[1 23 4

].

(b) 2× 3 matrix: A =

[1 2 34 5 6

].

(c) 3× 2 matrix: A =

1 23 45 6

.

(d) 2× 2 complex-valued matrix: A =

[1 + i 2− i

3 4i

].

(e) The coefficients of a linear system can be grouped in a matrix,

x ′1 = −x1 + 3x2

x ′2 = x1 − x2

}⇒ A =

[−1 31 −1

].

Main concepts from Linear Algebra.

Example

(a) 2× 2 matrix: A =

[1 23 4

].

(b) 2× 3 matrix: A =

[1 2 34 5 6

].

(c) 3× 2 matrix: A =

1 23 45 6

.

(d) 2× 2 complex-valued matrix: A =

[1 + i 2− i

3 4i

].

(e) The coefficients of a linear system can be grouped in a matrix,

x ′1 = −x1 + 3x2

x ′2 = x1 − x2

}⇒ A =

[−1 31 −1

].

Main concepts from Linear Algebra.

Example

(a) 2× 2 matrix: A =

[1 23 4

].

(b) 2× 3 matrix: A =

[1 2 34 5 6

].

(c) 3× 2 matrix: A =

1 23 45 6

.

(d) 2× 2 complex-valued matrix: A =

[1 + i 2− i

3 4i

].

(e) The coefficients of a linear system can be grouped in a matrix,

x ′1 = −x1 + 3x2

x ′2 = x1 − x2

}⇒ A =

[−1 31 −1

].

Main concepts from Linear Algebra.

Example

(a) 2× 2 matrix: A =

[1 23 4

].

(b) 2× 3 matrix: A =

[1 2 34 5 6

].

(c) 3× 2 matrix: A =

1 23 45 6

.

(d) 2× 2 complex-valued matrix: A =

[1 + i 2− i

3 4i

].

(e) The coefficients of a linear system can be grouped in a matrix,

x ′1 = −x1 + 3x2

x ′2 = x1 − x2

}⇒ A =

[−1 31 −1

].

Main concepts from Linear Algebra.

Remark: An m × 1 matrix is called an m-vector.

Definition

An m-vector, v, is the array of numbers v =

v1...

vm

, where the

vector components vi ∈ C, with i = 1, · · · ,m.

Example

The unknowns of a 2× 2 linear system can be grouped in a2-vector, for example,

x ′1 = −x1 + 3x2

x ′2 = x1 − x2

}⇒ x =

[x1

x2

].

Main concepts from Linear Algebra.

Remark: An m × 1 matrix is called an m-vector.

Definition

An m-vector, v, is the array of numbers v =

v1...

vm

, where the

vector components vi ∈ C, with i = 1, · · · ,m.

Example

The unknowns of a 2× 2 linear system can be grouped in a2-vector, for example,

x ′1 = −x1 + 3x2

x ′2 = x1 − x2

}⇒ x =

[x1

x2

].

Main concepts from Linear Algebra.

Remark: An m × 1 matrix is called an m-vector.

Definition

An m-vector, v, is the array of numbers v =

v1...

vm

, where the

vector components vi ∈ C, with i = 1, · · · ,m.

Example

The unknowns of a 2× 2 linear system can be grouped in a2-vector,

for example,

x ′1 = −x1 + 3x2

x ′2 = x1 − x2

}⇒ x =

[x1

x2

].

Main concepts from Linear Algebra.

Remark: An m × 1 matrix is called an m-vector.

Definition

An m-vector, v, is the array of numbers v =

v1...

vm

, where the

vector components vi ∈ C, with i = 1, · · · ,m.

Example

The unknowns of a 2× 2 linear system can be grouped in a2-vector, for example,

x ′1 = −x1 + 3x2

x ′2 = x1 − x2

}⇒ x =

[x1

x2

].

Main concepts from Linear Algebra.

Remark: An m × 1 matrix is called an m-vector.

Definition

An m-vector, v, is the array of numbers v =

v1...

vm

, where the

vector components vi ∈ C, with i = 1, · · · ,m.

Example

The unknowns of a 2× 2 linear system can be grouped in a2-vector, for example,

x ′1 = −x1 + 3x2

x ′2 = x1 − x2

}⇒ x =

[x1

x2

].

Main concepts from Linear Algebra.

Remark: We present only examples of matrix operations.

Example

Consider a 2× 3 matrix A =

[1 2 + i −1 + 2i3i 2 1

].

(a) A-transpose: Interchange rows with columns:

AT =

1 3i2 + i 2−1 + 2i 1

. Notice that:(AT

)T= A.

(b) A-conjugate: Conjugate every matrix coefficient:

A =

[1 2− i −1− 2i−3i 2 1

]. Notice that:

(A

)= A.

Matrix A is real iff A = A. Matrix A is imaginary iff A = −A.

Main concepts from Linear Algebra.

Remark: We present only examples of matrix operations.

Example

Consider a 2× 3 matrix A =

[1 2 + i −1 + 2i3i 2 1

].

(a) A-transpose: Interchange rows with columns:

AT =

1 3i2 + i 2−1 + 2i 1

. Notice that:(AT

)T= A.

(b) A-conjugate: Conjugate every matrix coefficient:

A =

[1 2− i −1− 2i−3i 2 1

]. Notice that:

(A

)= A.

Matrix A is real iff A = A. Matrix A is imaginary iff A = −A.

Main concepts from Linear Algebra.

Remark: We present only examples of matrix operations.

Example

Consider a 2× 3 matrix A =

[1 2 + i −1 + 2i3i 2 1

].

(a) A-transpose: Interchange rows with columns:

AT =

1 3i2 + i 2−1 + 2i 1

. Notice that:(AT

)T= A.

(b) A-conjugate: Conjugate every matrix coefficient:

A =

[1 2− i −1− 2i−3i 2 1

]. Notice that:

(A

)= A.

Matrix A is real iff A = A. Matrix A is imaginary iff A = −A.

Main concepts from Linear Algebra.

Remark: We present only examples of matrix operations.

Example

Consider a 2× 3 matrix A =

[1 2 + i −1 + 2i3i 2 1

].

(a) A-transpose: Interchange rows with columns:

AT =

1 3i2 + i 2−1 + 2i 1

.

Notice that:(AT

)T= A.

(b) A-conjugate: Conjugate every matrix coefficient:

A =

[1 2− i −1− 2i−3i 2 1

]. Notice that:

(A

)= A.

Matrix A is real iff A = A. Matrix A is imaginary iff A = −A.

Main concepts from Linear Algebra.

Remark: We present only examples of matrix operations.

Example

Consider a 2× 3 matrix A =

[1 2 + i −1 + 2i3i 2 1

].

(a) A-transpose: Interchange rows with columns:

AT =

1 3i2 + i 2−1 + 2i 1

. Notice that:(AT

)T= A.

(b) A-conjugate: Conjugate every matrix coefficient:

A =

[1 2− i −1− 2i−3i 2 1

]. Notice that:

(A

)= A.

Matrix A is real iff A = A. Matrix A is imaginary iff A = −A.

Main concepts from Linear Algebra.

Remark: We present only examples of matrix operations.

Example

Consider a 2× 3 matrix A =

[1 2 + i −1 + 2i3i 2 1

].

(a) A-transpose: Interchange rows with columns:

AT =

1 3i2 + i 2−1 + 2i 1

. Notice that:(AT

)T= A.

(b) A-conjugate: Conjugate every matrix coefficient:

A =

[1 2− i −1− 2i−3i 2 1

]. Notice that:

(A

)= A.

Matrix A is real iff A = A. Matrix A is imaginary iff A = −A.

Main concepts from Linear Algebra.

Remark: We present only examples of matrix operations.

Example

Consider a 2× 3 matrix A =

[1 2 + i −1 + 2i3i 2 1

].

(a) A-transpose: Interchange rows with columns:

AT =

1 3i2 + i 2−1 + 2i 1

. Notice that:(AT

)T= A.

(b) A-conjugate: Conjugate every matrix coefficient:

A =

[1 2− i −1− 2i−3i 2 1

].

Notice that:(A

)= A.

Matrix A is real iff A = A. Matrix A is imaginary iff A = −A.

Main concepts from Linear Algebra.

Remark: We present only examples of matrix operations.

Example

Consider a 2× 3 matrix A =

[1 2 + i −1 + 2i3i 2 1

].

(a) A-transpose: Interchange rows with columns:

AT =

1 3i2 + i 2−1 + 2i 1

. Notice that:(AT

)T= A.

(b) A-conjugate: Conjugate every matrix coefficient:

A =

[1 2− i −1− 2i−3i 2 1

]. Notice that:

(A

)= A.

Matrix A is real iff A = A. Matrix A is imaginary iff A = −A.

Main concepts from Linear Algebra.

Remark: We present only examples of matrix operations.

Example

Consider a 2× 3 matrix A =

[1 2 + i −1 + 2i3i 2 1

].

(a) A-transpose: Interchange rows with columns:

AT =

1 3i2 + i 2−1 + 2i 1

. Notice that:(AT

)T= A.

(b) A-conjugate: Conjugate every matrix coefficient:

A =

[1 2− i −1− 2i−3i 2 1

]. Notice that:

(A

)= A.

Matrix A is real iff A = A.

Matrix A is imaginary iff A = −A.

Main concepts from Linear Algebra.

Remark: We present only examples of matrix operations.

Example

Consider a 2× 3 matrix A =

[1 2 + i −1 + 2i3i 2 1

].

(a) A-transpose: Interchange rows with columns:

AT =

1 3i2 + i 2−1 + 2i 1

. Notice that:(AT

)T= A.

(b) A-conjugate: Conjugate every matrix coefficient:

A =

[1 2− i −1− 2i−3i 2 1

]. Notice that:

(A

)= A.

Matrix A is real iff A = A. Matrix A is imaginary iff A = −A.

Main concepts from Linear Algebra.

Example

Consider a 2× 3 matrix A =

[1 2 + i −1 + 2i3i 2 1

].

(a) A-adjoint: Conjugate and transpose:

A∗ =

1 −3i2− i 2−1− 2i 1

. Notice that:(A∗

)∗= A.

(b) Addition of two m × n matrices is performed component-wise:[1 23 4

]+

[2 35 1

]=

[(1 + 2) (2 + 3)(3 + 5) (4 + 1)

]=

[3 58 5

].

The addition

[1 23 4

]+

[1 2 34 5 6

]is not defined.

Main concepts from Linear Algebra.

Example

Consider a 2× 3 matrix A =

[1 2 + i −1 + 2i3i 2 1

].

(a) A-adjoint: Conjugate and transpose:

A∗ =

1 −3i2− i 2−1− 2i 1

. Notice that:(A∗

)∗= A.

(b) Addition of two m × n matrices is performed component-wise:[1 23 4

]+

[2 35 1

]=

[(1 + 2) (2 + 3)(3 + 5) (4 + 1)

]=

[3 58 5

].

The addition

[1 23 4

]+

[1 2 34 5 6

]is not defined.

Main concepts from Linear Algebra.

Example

Consider a 2× 3 matrix A =

[1 2 + i −1 + 2i3i 2 1

].

(a) A-adjoint: Conjugate and transpose:

A∗ =

1 −3i2− i 2−1− 2i 1

.

Notice that:(A∗

)∗= A.

(b) Addition of two m × n matrices is performed component-wise:[1 23 4

]+

[2 35 1

]=

[(1 + 2) (2 + 3)(3 + 5) (4 + 1)

]=

[3 58 5

].

The addition

[1 23 4

]+

[1 2 34 5 6

]is not defined.

Main concepts from Linear Algebra.

Example

Consider a 2× 3 matrix A =

[1 2 + i −1 + 2i3i 2 1

].

(a) A-adjoint: Conjugate and transpose:

A∗ =

1 −3i2− i 2−1− 2i 1

. Notice that:(A∗

)∗= A.

(b) Addition of two m × n matrices is performed component-wise:[1 23 4

]+

[2 35 1

]=

[(1 + 2) (2 + 3)(3 + 5) (4 + 1)

]=

[3 58 5

].

The addition

[1 23 4

]+

[1 2 34 5 6

]is not defined.

Main concepts from Linear Algebra.

Example

Consider a 2× 3 matrix A =

[1 2 + i −1 + 2i3i 2 1

].

(a) A-adjoint: Conjugate and transpose:

A∗ =

1 −3i2− i 2−1− 2i 1

. Notice that:(A∗

)∗= A.

(b) Addition of two m × n matrices is performed component-wise:

[1 23 4

]+

[2 35 1

]=

[(1 + 2) (2 + 3)(3 + 5) (4 + 1)

]=

[3 58 5

].

The addition

[1 23 4

]+

[1 2 34 5 6

]is not defined.

Main concepts from Linear Algebra.

Example

Consider a 2× 3 matrix A =

[1 2 + i −1 + 2i3i 2 1

].

(a) A-adjoint: Conjugate and transpose:

A∗ =

1 −3i2− i 2−1− 2i 1

. Notice that:(A∗

)∗= A.

(b) Addition of two m × n matrices is performed component-wise:[1 23 4

]+

[2 35 1

]

=

[(1 + 2) (2 + 3)(3 + 5) (4 + 1)

]=

[3 58 5

].

The addition

[1 23 4

]+

[1 2 34 5 6

]is not defined.

Main concepts from Linear Algebra.

Example

Consider a 2× 3 matrix A =

[1 2 + i −1 + 2i3i 2 1

].

(a) A-adjoint: Conjugate and transpose:

A∗ =

1 −3i2− i 2−1− 2i 1

. Notice that:(A∗

)∗= A.

(b) Addition of two m × n matrices is performed component-wise:[1 23 4

]+

[2 35 1

]=

[(1 + 2) (2 + 3)(3 + 5) (4 + 1)

]

=

[3 58 5

].

The addition

[1 23 4

]+

[1 2 34 5 6

]is not defined.

Main concepts from Linear Algebra.

Example

Consider a 2× 3 matrix A =

[1 2 + i −1 + 2i3i 2 1

].

(a) A-adjoint: Conjugate and transpose:

A∗ =

1 −3i2− i 2−1− 2i 1

. Notice that:(A∗

)∗= A.

(b) Addition of two m × n matrices is performed component-wise:[1 23 4

]+

[2 35 1

]=

[(1 + 2) (2 + 3)(3 + 5) (4 + 1)

]=

[3 58 5

].

The addition

[1 23 4

]+

[1 2 34 5 6

]is not defined.

Main concepts from Linear Algebra.

Example

Consider a 2× 3 matrix A =

[1 2 + i −1 + 2i3i 2 1

].

(a) A-adjoint: Conjugate and transpose:

A∗ =

1 −3i2− i 2−1− 2i 1

. Notice that:(A∗

)∗= A.

(b) Addition of two m × n matrices is performed component-wise:[1 23 4

]+

[2 35 1

]=

[(1 + 2) (2 + 3)(3 + 5) (4 + 1)

]=

[3 58 5

].

The addition

[1 23 4

]+

[1 2 34 5 6

]is not defined.

Main concepts from Linear Algebra.

Example

Consider a 2× 3 matrix A =

[1 3 52 4 6

].

(a) Multiplication of a matrix by a number is performedcomponent-wise:

2A = 2

[1 3 52 4 6

]=

[2 6 104 8 12

],

[8 1216 20

]= 4

[2 34 5

].

Also:

A

3=

1

3

[1 3 52 4 6

]=

1

31

5

3

2

3

4

32

.

Main concepts from Linear Algebra.

Example

Consider a 2× 3 matrix A =

[1 3 52 4 6

].

(a) Multiplication of a matrix by a number is performedcomponent-wise:

2A = 2

[1 3 52 4 6

]=

[2 6 104 8 12

],

[8 1216 20

]= 4

[2 34 5

].

Also:

A

3=

1

3

[1 3 52 4 6

]=

1

31

5

3

2

3

4

32

.

Main concepts from Linear Algebra.

Example

Consider a 2× 3 matrix A =

[1 3 52 4 6

].

(a) Multiplication of a matrix by a number is performedcomponent-wise:

2A = 2

[1 3 52 4 6

]

=

[2 6 104 8 12

],

[8 1216 20

]= 4

[2 34 5

].

Also:

A

3=

1

3

[1 3 52 4 6

]=

1

31

5

3

2

3

4

32

.

Main concepts from Linear Algebra.

Example

Consider a 2× 3 matrix A =

[1 3 52 4 6

].

(a) Multiplication of a matrix by a number is performedcomponent-wise:

2A = 2

[1 3 52 4 6

]=

[2 6 104 8 12

],

[8 1216 20

]= 4

[2 34 5

].

Also:

A

3=

1

3

[1 3 52 4 6

]=

1

31

5

3

2

3

4

32

.

Main concepts from Linear Algebra.

Example

Consider a 2× 3 matrix A =

[1 3 52 4 6

].

(a) Multiplication of a matrix by a number is performedcomponent-wise:

2A = 2

[1 3 52 4 6

]=

[2 6 104 8 12

],

[8 1216 20

]

= 4

[2 34 5

].

Also:

A

3=

1

3

[1 3 52 4 6

]=

1

31

5

3

2

3

4

32

.

Main concepts from Linear Algebra.

Example

Consider a 2× 3 matrix A =

[1 3 52 4 6

].

(a) Multiplication of a matrix by a number is performedcomponent-wise:

2A = 2

[1 3 52 4 6

]=

[2 6 104 8 12

],

[8 1216 20

]= 4

[2 34 5

].

Also:

A

3=

1

3

[1 3 52 4 6

]=

1

31

5

3

2

3

4

32

.

Main concepts from Linear Algebra.

Example

Consider a 2× 3 matrix A =

[1 3 52 4 6

].

(a) Multiplication of a matrix by a number is performedcomponent-wise:

2A = 2

[1 3 52 4 6

]=

[2 6 104 8 12

],

[8 1216 20

]= 4

[2 34 5

].

Also:

A

3

=1

3

[1 3 52 4 6

]=

1

31

5

3

2

3

4

32

.

Main concepts from Linear Algebra.

Example

Consider a 2× 3 matrix A =

[1 3 52 4 6

].

(a) Multiplication of a matrix by a number is performedcomponent-wise:

2A = 2

[1 3 52 4 6

]=

[2 6 104 8 12

],

[8 1216 20

]= 4

[2 34 5

].

Also:

A

3=

1

3

[1 3 52 4 6

]

=

1

31

5

3

2

3

4

32

.

Main concepts from Linear Algebra.

Example

Consider a 2× 3 matrix A =

[1 3 52 4 6

].

(a) Multiplication of a matrix by a number is performedcomponent-wise:

2A = 2

[1 3 52 4 6

]=

[2 6 104 8 12

],

[8 1216 20

]= 4

[2 34 5

].

Also:

A

3=

1

3

[1 3 52 4 6

]=

1

31

5

3

2

3

4

32

.

Main concepts from Linear Algebra.

Example

(a) Matrix multiplication.

The matrix sizes is important:

Am × n

times Bn × `

defines ABm × `

Example: A is 2× 2, B is 2× 3, so AB is 2× 3:

AB =

[4 32 1

] [1 2 34 5 6

]=

[16 23 306 9 12

].

Notice B is 2× 3, A is 2× 2, so BA is not defined.

BA =

[1 2 34 5 6

] [4 32 1

]not defined.

Main concepts from Linear Algebra.

Example

(a) Matrix multiplication. The matrix sizes is important:

Am × n

times Bn × `

defines ABm × `

Example: A is 2× 2, B is 2× 3, so AB is 2× 3:

AB =

[4 32 1

] [1 2 34 5 6

]=

[16 23 306 9 12

].

Notice B is 2× 3, A is 2× 2, so BA is not defined.

BA =

[1 2 34 5 6

] [4 32 1

]not defined.

Main concepts from Linear Algebra.

Example

(a) Matrix multiplication. The matrix sizes is important:

Am × n

times Bn × `

defines ABm × `

Example: A is 2× 2, B is 2× 3, so AB is 2× 3:

AB =

[4 32 1

] [1 2 34 5 6

]=

[16 23 306 9 12

].

Notice B is 2× 3, A is 2× 2, so BA is not defined.

BA =

[1 2 34 5 6

] [4 32 1

]not defined.

Main concepts from Linear Algebra.

Example

(a) Matrix multiplication. The matrix sizes is important:

Am × n

times Bn × `

defines ABm × `

Example: A is 2× 2, B is 2× 3, so AB is 2× 3:

AB =

[4 32 1

] [1 2 34 5 6

]=

[16 23 306 9 12

].

Notice B is 2× 3, A is 2× 2, so BA is not defined.

BA =

[1 2 34 5 6

] [4 32 1

]not defined.

Main concepts from Linear Algebra.

Example

(a) Matrix multiplication. The matrix sizes is important:

Am × n

times Bn × `

defines ABm × `

Example: A is 2× 2, B is 2× 3, so AB is 2× 3:

AB =

[4 32 1

] [1 2 34 5 6

]=

[16 23 306 9 12

].

Notice B is 2× 3, A is 2× 2, so BA is not defined.

BA =

[1 2 34 5 6

] [4 32 1

]not defined.

Main concepts from Linear Algebra.

Example

(a) Matrix multiplication. The matrix sizes is important:

Am × n

times Bn × `

defines ABm × `

Example: A is 2× 2, B is 2× 3, so AB is 2× 3:

AB =

[4 32 1

] [1 2 34 5 6

]=

[16 23 306 9 12

].

Notice B is 2× 3, A is 2× 2, so BA is not defined.

BA =

[1 2 34 5 6

] [4 32 1

]not defined.

Main concepts from Linear Algebra.

Example

(a) Matrix multiplication. The matrix sizes is important:

Am × n

times Bn × `

defines ABm × `

Example: A is 2× 2, B is 2× 3, so AB is 2× 3:

AB =

[4 32 1

] [1 2 34 5 6

]=

[16 23 306 9 12

].

Notice B is 2× 3, A is 2× 2, so BA is not defined.

BA =

[1 2 34 5 6

] [4 32 1

]not defined.

Main concepts from Linear Algebra.

Remark: The matrix product is not commutative, that is, ingeneral holds AB 6= BA.

Example

Find AB and BA for A =

[2 −1−1 2

]and B =

[3 02 −1

].

Solution:

AB =

[2 −1−1 2

] [3 02 −1

]=

[(6− 2) (0 + 1)

(−3 + 4) (0− 2)

]=

[4 11 −2

].

BA =

[3 02 −1

] [2 −1−1 2

]=

[(6 + 0) (−3 + 0)(4 + 1) (−2− 2)

]=

[6 −35 −4

].

So AB 6= BA. C

Main concepts from Linear Algebra.

Remark: The matrix product is not commutative, that is, ingeneral holds AB 6= BA.

Example

Find AB and BA for A =

[2 −1−1 2

]and B =

[3 02 −1

].

Solution:

AB =

[2 −1−1 2

] [3 02 −1

]=

[(6− 2) (0 + 1)

(−3 + 4) (0− 2)

]=

[4 11 −2

].

BA =

[3 02 −1

] [2 −1−1 2

]=

[(6 + 0) (−3 + 0)(4 + 1) (−2− 2)

]=

[6 −35 −4

].

So AB 6= BA. C

Main concepts from Linear Algebra.

Remark: The matrix product is not commutative, that is, ingeneral holds AB 6= BA.

Example

Find AB and BA for A =

[2 −1−1 2

]and B =

[3 02 −1

].

Solution:

AB =

[2 −1−1 2

] [3 02 −1

]

=

[(6− 2) (0 + 1)

(−3 + 4) (0− 2)

]=

[4 11 −2

].

BA =

[3 02 −1

] [2 −1−1 2

]=

[(6 + 0) (−3 + 0)(4 + 1) (−2− 2)

]=

[6 −35 −4

].

So AB 6= BA. C

Main concepts from Linear Algebra.

Remark: The matrix product is not commutative, that is, ingeneral holds AB 6= BA.

Example

Find AB and BA for A =

[2 −1−1 2

]and B =

[3 02 −1

].

Solution:

AB =

[2 −1−1 2

] [3 02 −1

]=

[(6− 2) (0 + 1)

(−3 + 4) (0− 2)

]

=

[4 11 −2

].

BA =

[3 02 −1

] [2 −1−1 2

]=

[(6 + 0) (−3 + 0)(4 + 1) (−2− 2)

]=

[6 −35 −4

].

So AB 6= BA. C

Main concepts from Linear Algebra.

Remark: The matrix product is not commutative, that is, ingeneral holds AB 6= BA.

Example

Find AB and BA for A =

[2 −1−1 2

]and B =

[3 02 −1

].

Solution:

AB =

[2 −1−1 2

] [3 02 −1

]=

[(6− 2) (0 + 1)

(−3 + 4) (0− 2)

]=

[4 11 −2

].

BA =

[3 02 −1

] [2 −1−1 2

]=

[(6 + 0) (−3 + 0)(4 + 1) (−2− 2)

]=

[6 −35 −4

].

So AB 6= BA. C

Main concepts from Linear Algebra.

Remark: The matrix product is not commutative, that is, ingeneral holds AB 6= BA.

Example

Find AB and BA for A =

[2 −1−1 2

]and B =

[3 02 −1

].

Solution:

AB =

[2 −1−1 2

] [3 02 −1

]=

[(6− 2) (0 + 1)

(−3 + 4) (0− 2)

]=

[4 11 −2

].

BA =

[3 02 −1

] [2 −1−1 2

]

=

[(6 + 0) (−3 + 0)(4 + 1) (−2− 2)

]=

[6 −35 −4

].

So AB 6= BA. C

Main concepts from Linear Algebra.

Remark: The matrix product is not commutative, that is, ingeneral holds AB 6= BA.

Example

Find AB and BA for A =

[2 −1−1 2

]and B =

[3 02 −1

].

Solution:

AB =

[2 −1−1 2

] [3 02 −1

]=

[(6− 2) (0 + 1)

(−3 + 4) (0− 2)

]=

[4 11 −2

].

BA =

[3 02 −1

] [2 −1−1 2

]=

[(6 + 0) (−3 + 0)(4 + 1) (−2− 2)

]

=

[6 −35 −4

].

So AB 6= BA. C

Main concepts from Linear Algebra.

Remark: The matrix product is not commutative, that is, ingeneral holds AB 6= BA.

Example

Find AB and BA for A =

[2 −1−1 2

]and B =

[3 02 −1

].

Solution:

AB =

[2 −1−1 2

] [3 02 −1

]=

[(6− 2) (0 + 1)

(−3 + 4) (0− 2)

]=

[4 11 −2

].

BA =

[3 02 −1

] [2 −1−1 2

]=

[(6 + 0) (−3 + 0)(4 + 1) (−2− 2)

]=

[6 −35 −4

].

So AB 6= BA. C

Main concepts from Linear Algebra.

Remark: The matrix product is not commutative, that is, ingeneral holds AB 6= BA.

Example

Find AB and BA for A =

[2 −1−1 2

]and B =

[3 02 −1

].

Solution:

AB =

[2 −1−1 2

] [3 02 −1

]=

[(6− 2) (0 + 1)

(−3 + 4) (0− 2)

]=

[4 11 −2

].

BA =

[3 02 −1

] [2 −1−1 2

]=

[(6 + 0) (−3 + 0)(4 + 1) (−2− 2)

]=

[6 −35 −4

].

So AB 6= BA. C

Main concepts from Linear Algebra.

Remark: There exist matrices A 6= 0 and B 6= 0 with AB = 0.

Example

Find AB for A =

[1 −1−1 1

]and B =

[1 −11 −1

].

Solution:

AB =

[1 −1−1 1

] [1 −11 −1

]=

[(1− 1) (−1 + 1)

(−1 + 1) (1− 1)

]=

[0 00 0

].

C

Recall: If a, b ∈ R and ab = 0, then either a = 0 or b = 0.

We have just shown that this statement is not true for matrices.

Main concepts from Linear Algebra.

Remark: There exist matrices A 6= 0 and B 6= 0 with AB = 0.

Example

Find AB for A =

[1 −1−1 1

]and B =

[1 −11 −1

].

Solution:

AB =

[1 −1−1 1

] [1 −11 −1

]=

[(1− 1) (−1 + 1)

(−1 + 1) (1− 1)

]=

[0 00 0

].

C

Recall: If a, b ∈ R and ab = 0, then either a = 0 or b = 0.

We have just shown that this statement is not true for matrices.

Main concepts from Linear Algebra.

Remark: There exist matrices A 6= 0 and B 6= 0 with AB = 0.

Example

Find AB for A =

[1 −1−1 1

]and B =

[1 −11 −1

].

Solution:

AB =

[1 −1−1 1

] [1 −11 −1

]

=

[(1− 1) (−1 + 1)

(−1 + 1) (1− 1)

]=

[0 00 0

].

C

Recall: If a, b ∈ R and ab = 0, then either a = 0 or b = 0.

We have just shown that this statement is not true for matrices.

Main concepts from Linear Algebra.

Remark: There exist matrices A 6= 0 and B 6= 0 with AB = 0.

Example

Find AB for A =

[1 −1−1 1

]and B =

[1 −11 −1

].

Solution:

AB =

[1 −1−1 1

] [1 −11 −1

]=

[(1− 1) (−1 + 1)

(−1 + 1) (1− 1)

]

=

[0 00 0

].

C

Recall: If a, b ∈ R and ab = 0, then either a = 0 or b = 0.

We have just shown that this statement is not true for matrices.

Main concepts from Linear Algebra.

Remark: There exist matrices A 6= 0 and B 6= 0 with AB = 0.

Example

Find AB for A =

[1 −1−1 1

]and B =

[1 −11 −1

].

Solution:

AB =

[1 −1−1 1

] [1 −11 −1

]=

[(1− 1) (−1 + 1)

(−1 + 1) (1− 1)

]=

[0 00 0

].

C

Recall: If a, b ∈ R and ab = 0, then either a = 0 or b = 0.

We have just shown that this statement is not true for matrices.

Main concepts from Linear Algebra.

Remark: There exist matrices A 6= 0 and B 6= 0 with AB = 0.

Example

Find AB for A =

[1 −1−1 1

]and B =

[1 −11 −1

].

Solution:

AB =

[1 −1−1 1

] [1 −11 −1

]=

[(1− 1) (−1 + 1)

(−1 + 1) (1− 1)

]=

[0 00 0

].

C

Recall: If a, b ∈ R and ab = 0, then either a = 0 or b = 0.

We have just shown that this statement is not true for matrices.

Main concepts from Linear Algebra.

Remark: There exist matrices A 6= 0 and B 6= 0 with AB = 0.

Example

Find AB for A =

[1 −1−1 1

]and B =

[1 −11 −1

].

Solution:

AB =

[1 −1−1 1

] [1 −11 −1

]=

[(1− 1) (−1 + 1)

(−1 + 1) (1− 1)

]=

[0 00 0

].

C

Recall: If a, b ∈ R and ab = 0, then either a = 0 or b = 0.

We have just shown that this statement is not true for matrices.

Review Exam 3.

I Sections 6.1-6.6.

I 5 or 6 problems.

I 50 minutes.

I Laplace Transform table included.

Exam: November 11, 2008. Problem 2

Example

Use Laplace Transform to find y solution of

y ′′ − 2y ′ + 2y = δ(t − 2), y(0) = 1, y ′(0) = 3.

Solution: Compute the LT of the equation,

L[y ′′]− 2L[y ′] + 2L[y ] = L[δ(t − 2)] = e−2s

L[y ′′] = s2 L[y ]− s y(0)− y ′(0), L[y ′] = s L[y ]− y(0).

(s2 − 2s + 2)L[y ]− s y(0)− y ′(0) + 2 y(0) = e−2s

(s2 − 2s + 2)L[y ]− s − 1 = e−2s

L[y ] =(s + 1)

(s2 − 2s + 2)+

1

(s2 − 2s + 2)e−2s .

Exam: November 11, 2008. Problem 2

Example

Use Laplace Transform to find y solution of

y ′′ − 2y ′ + 2y = δ(t − 2), y(0) = 1, y ′(0) = 3.

Solution: Compute the LT of the equation,

L[y ′′]− 2L[y ′] + 2L[y ] = L[δ(t − 2)] = e−2s

L[y ′′] = s2 L[y ]− s y(0)− y ′(0), L[y ′] = s L[y ]− y(0).

(s2 − 2s + 2)L[y ]− s y(0)− y ′(0) + 2 y(0) = e−2s

(s2 − 2s + 2)L[y ]− s − 1 = e−2s

L[y ] =(s + 1)

(s2 − 2s + 2)+

1

(s2 − 2s + 2)e−2s .

Exam: November 11, 2008. Problem 2

Example

Use Laplace Transform to find y solution of

y ′′ − 2y ′ + 2y = δ(t − 2), y(0) = 1, y ′(0) = 3.

Solution: Compute the LT of the equation,

L[y ′′]− 2L[y ′] + 2L[y ] = L[δ(t − 2)]

= e−2s

L[y ′′] = s2 L[y ]− s y(0)− y ′(0), L[y ′] = s L[y ]− y(0).

(s2 − 2s + 2)L[y ]− s y(0)− y ′(0) + 2 y(0) = e−2s

(s2 − 2s + 2)L[y ]− s − 1 = e−2s

L[y ] =(s + 1)

(s2 − 2s + 2)+

1

(s2 − 2s + 2)e−2s .

Exam: November 11, 2008. Problem 2

Example

Use Laplace Transform to find y solution of

y ′′ − 2y ′ + 2y = δ(t − 2), y(0) = 1, y ′(0) = 3.

Solution: Compute the LT of the equation,

L[y ′′]− 2L[y ′] + 2L[y ] = L[δ(t − 2)] = e−2s

L[y ′′] = s2 L[y ]− s y(0)− y ′(0), L[y ′] = s L[y ]− y(0).

(s2 − 2s + 2)L[y ]− s y(0)− y ′(0) + 2 y(0) = e−2s

(s2 − 2s + 2)L[y ]− s − 1 = e−2s

L[y ] =(s + 1)

(s2 − 2s + 2)+

1

(s2 − 2s + 2)e−2s .

Exam: November 11, 2008. Problem 2

Example

Use Laplace Transform to find y solution of

y ′′ − 2y ′ + 2y = δ(t − 2), y(0) = 1, y ′(0) = 3.

Solution: Compute the LT of the equation,

L[y ′′]− 2L[y ′] + 2L[y ] = L[δ(t − 2)] = e−2s

L[y ′′] = s2 L[y ]− s y(0)− y ′(0),

L[y ′] = s L[y ]− y(0).

(s2 − 2s + 2)L[y ]− s y(0)− y ′(0) + 2 y(0) = e−2s

(s2 − 2s + 2)L[y ]− s − 1 = e−2s

L[y ] =(s + 1)

(s2 − 2s + 2)+

1

(s2 − 2s + 2)e−2s .

Exam: November 11, 2008. Problem 2

Example

Use Laplace Transform to find y solution of

y ′′ − 2y ′ + 2y = δ(t − 2), y(0) = 1, y ′(0) = 3.

Solution: Compute the LT of the equation,

L[y ′′]− 2L[y ′] + 2L[y ] = L[δ(t − 2)] = e−2s

L[y ′′] = s2 L[y ]− s y(0)− y ′(0), L[y ′] = s L[y ]− y(0).

(s2 − 2s + 2)L[y ]− s y(0)− y ′(0) + 2 y(0) = e−2s

(s2 − 2s + 2)L[y ]− s − 1 = e−2s

L[y ] =(s + 1)

(s2 − 2s + 2)+

1

(s2 − 2s + 2)e−2s .

Exam: November 11, 2008. Problem 2

Example

Use Laplace Transform to find y solution of

y ′′ − 2y ′ + 2y = δ(t − 2), y(0) = 1, y ′(0) = 3.

Solution: Compute the LT of the equation,

L[y ′′]− 2L[y ′] + 2L[y ] = L[δ(t − 2)] = e−2s

L[y ′′] = s2 L[y ]− s y(0)− y ′(0), L[y ′] = s L[y ]− y(0).

(s2 − 2s + 2)L[y ]− s y(0)− y ′(0) + 2 y(0) = e−2s

(s2 − 2s + 2)L[y ]− s − 1 = e−2s

L[y ] =(s + 1)

(s2 − 2s + 2)+

1

(s2 − 2s + 2)e−2s .

Exam: November 11, 2008. Problem 2

Example

Use Laplace Transform to find y solution of

y ′′ − 2y ′ + 2y = δ(t − 2), y(0) = 1, y ′(0) = 3.

Solution: Compute the LT of the equation,

L[y ′′]− 2L[y ′] + 2L[y ] = L[δ(t − 2)] = e−2s

L[y ′′] = s2 L[y ]− s y(0)− y ′(0), L[y ′] = s L[y ]− y(0).

(s2 − 2s + 2)L[y ]− s y(0)− y ′(0) + 2 y(0) = e−2s

(s2 − 2s + 2)L[y ]− s − 1 = e−2s

L[y ] =(s + 1)

(s2 − 2s + 2)+

1

(s2 − 2s + 2)e−2s .

Exam: November 11, 2008. Problem 2

Example

Use Laplace Transform to find y solution of

y ′′ − 2y ′ + 2y = δ(t − 2), y(0) = 1, y ′(0) = 3.

Solution: Compute the LT of the equation,

L[y ′′]− 2L[y ′] + 2L[y ] = L[δ(t − 2)] = e−2s

L[y ′′] = s2 L[y ]− s y(0)− y ′(0), L[y ′] = s L[y ]− y(0).

(s2 − 2s + 2)L[y ]− s y(0)− y ′(0) + 2 y(0) = e−2s

(s2 − 2s + 2)L[y ]− s − 1 = e−2s

L[y ] =(s + 1)

(s2 − 2s + 2)+

1

(s2 − 2s + 2)e−2s .

Exam: November 11, 2008. Problem 2

Example

Use Laplace Transform to find y solution of

y ′′ − 2y ′ + 2y = δ(t − 2), y(0) = 1, y ′(0) = 3.

Solution: Recall: L[y ] =(s + 1)

(s2 − 2s + 2)+

1

(s2 − 2s + 2)e−2s .

s2 − 2s + 2 = 0 ⇒ s± =1

2

[2±

√4− 8

], complex roots.

s2 − 2s + 2 = (s2 − 2s + 1)− 1 + 2 = (s − 1)2 + 1.

L[y ] =s + 1

(s − 1)2 + 1+

1

(s − 1)2 + 1e−2s

L[y ] =(s − 1 + 1) + 1

(s − 1)2 + 1+

1

(s − 1)2 + 1e−2s

Exam: November 11, 2008. Problem 2

Example

Use Laplace Transform to find y solution of

y ′′ − 2y ′ + 2y = δ(t − 2), y(0) = 1, y ′(0) = 3.

Solution: Recall: L[y ] =(s + 1)

(s2 − 2s + 2)+

1

(s2 − 2s + 2)e−2s .

s2 − 2s + 2 = 0

⇒ s± =1

2

[2±

√4− 8

], complex roots.

s2 − 2s + 2 = (s2 − 2s + 1)− 1 + 2 = (s − 1)2 + 1.

L[y ] =s + 1

(s − 1)2 + 1+

1

(s − 1)2 + 1e−2s

L[y ] =(s − 1 + 1) + 1

(s − 1)2 + 1+

1

(s − 1)2 + 1e−2s

Exam: November 11, 2008. Problem 2

Example

Use Laplace Transform to find y solution of

y ′′ − 2y ′ + 2y = δ(t − 2), y(0) = 1, y ′(0) = 3.

Solution: Recall: L[y ] =(s + 1)

(s2 − 2s + 2)+

1

(s2 − 2s + 2)e−2s .

s2 − 2s + 2 = 0 ⇒ s± =1

2

[2±

√4− 8

],

complex roots.

s2 − 2s + 2 = (s2 − 2s + 1)− 1 + 2 = (s − 1)2 + 1.

L[y ] =s + 1

(s − 1)2 + 1+

1

(s − 1)2 + 1e−2s

L[y ] =(s − 1 + 1) + 1

(s − 1)2 + 1+

1

(s − 1)2 + 1e−2s

Exam: November 11, 2008. Problem 2

Example

Use Laplace Transform to find y solution of

y ′′ − 2y ′ + 2y = δ(t − 2), y(0) = 1, y ′(0) = 3.

Solution: Recall: L[y ] =(s + 1)

(s2 − 2s + 2)+

1

(s2 − 2s + 2)e−2s .

s2 − 2s + 2 = 0 ⇒ s± =1

2

[2±

√4− 8

], complex roots.

s2 − 2s + 2 = (s2 − 2s + 1)− 1 + 2 = (s − 1)2 + 1.

L[y ] =s + 1

(s − 1)2 + 1+

1

(s − 1)2 + 1e−2s

L[y ] =(s − 1 + 1) + 1

(s − 1)2 + 1+

1

(s − 1)2 + 1e−2s

Exam: November 11, 2008. Problem 2

Example

Use Laplace Transform to find y solution of

y ′′ − 2y ′ + 2y = δ(t − 2), y(0) = 1, y ′(0) = 3.

Solution: Recall: L[y ] =(s + 1)

(s2 − 2s + 2)+

1

(s2 − 2s + 2)e−2s .

s2 − 2s + 2 = 0 ⇒ s± =1

2

[2±

√4− 8

], complex roots.

s2 − 2s + 2 = (s2 − 2s + 1)− 1 + 2

= (s − 1)2 + 1.

L[y ] =s + 1

(s − 1)2 + 1+

1

(s − 1)2 + 1e−2s

L[y ] =(s − 1 + 1) + 1

(s − 1)2 + 1+

1

(s − 1)2 + 1e−2s

Exam: November 11, 2008. Problem 2

Example

Use Laplace Transform to find y solution of

y ′′ − 2y ′ + 2y = δ(t − 2), y(0) = 1, y ′(0) = 3.

Solution: Recall: L[y ] =(s + 1)

(s2 − 2s + 2)+

1

(s2 − 2s + 2)e−2s .

s2 − 2s + 2 = 0 ⇒ s± =1

2

[2±

√4− 8

], complex roots.

s2 − 2s + 2 = (s2 − 2s + 1)− 1 + 2 = (s − 1)2 + 1.

L[y ] =s + 1

(s − 1)2 + 1+

1

(s − 1)2 + 1e−2s

L[y ] =(s − 1 + 1) + 1

(s − 1)2 + 1+

1

(s − 1)2 + 1e−2s

Exam: November 11, 2008. Problem 2

Example

Use Laplace Transform to find y solution of

y ′′ − 2y ′ + 2y = δ(t − 2), y(0) = 1, y ′(0) = 3.

Solution: Recall: L[y ] =(s + 1)

(s2 − 2s + 2)+

1

(s2 − 2s + 2)e−2s .

s2 − 2s + 2 = 0 ⇒ s± =1

2

[2±

√4− 8

], complex roots.

s2 − 2s + 2 = (s2 − 2s + 1)− 1 + 2 = (s − 1)2 + 1.

L[y ] =s + 1

(s − 1)2 + 1+

1

(s − 1)2 + 1e−2s

L[y ] =(s − 1 + 1) + 1

(s − 1)2 + 1+

1

(s − 1)2 + 1e−2s

Exam: November 11, 2008. Problem 2

Example

Use Laplace Transform to find y solution of

y ′′ − 2y ′ + 2y = δ(t − 2), y(0) = 1, y ′(0) = 3.

Solution: Recall: L[y ] =(s + 1)

(s2 − 2s + 2)+

1

(s2 − 2s + 2)e−2s .

s2 − 2s + 2 = 0 ⇒ s± =1

2

[2±

√4− 8

], complex roots.

s2 − 2s + 2 = (s2 − 2s + 1)− 1 + 2 = (s − 1)2 + 1.

L[y ] =s + 1

(s − 1)2 + 1+

1

(s − 1)2 + 1e−2s

L[y ] =(s − 1 + 1) + 1

(s − 1)2 + 1+

1

(s − 1)2 + 1e−2s

Exam: November 11, 2008. Problem 2

Example

Use Laplace Transform to find y solution of

y ′′ − 2y ′ + 2y = δ(t − 2), y(0) = 1, y ′(0) = 3.

Solution: Recall: L[y ] =(s − 1) + 2

(s − 1)2 + 1+

1

(s − 1)2 + 1e−2s ,

L[y ] =(s − 1)

(s − 1)2 + 1+ 2

1

(s − 1)2 + 1+ e−2s 1

(s − 1)2 + 1,

L[cos(at)] =s

s2 + a2, L[sin(at)] =

a

s2 + a2,

L[y ] = L[cos(t)]∣∣(s−1)

+ 2L[sin(t)]∣∣(s−1)

+ e−2s L[sin(t)]∣∣(s−1)

.

Exam: November 11, 2008. Problem 2

Example

Use Laplace Transform to find y solution of

y ′′ − 2y ′ + 2y = δ(t − 2), y(0) = 1, y ′(0) = 3.

Solution: Recall: L[y ] =(s − 1) + 2

(s − 1)2 + 1+

1

(s − 1)2 + 1e−2s ,

L[y ] =(s − 1)

(s − 1)2 + 1+ 2

1

(s − 1)2 + 1+ e−2s 1

(s − 1)2 + 1,

L[cos(at)] =s

s2 + a2, L[sin(at)] =

a

s2 + a2,

L[y ] = L[cos(t)]∣∣(s−1)

+ 2L[sin(t)]∣∣(s−1)

+ e−2s L[sin(t)]∣∣(s−1)

.

Exam: November 11, 2008. Problem 2

Example

Use Laplace Transform to find y solution of

y ′′ − 2y ′ + 2y = δ(t − 2), y(0) = 1, y ′(0) = 3.

Solution: Recall: L[y ] =(s − 1) + 2

(s − 1)2 + 1+

1

(s − 1)2 + 1e−2s ,

L[y ] =(s − 1)

(s − 1)2 + 1+ 2

1

(s − 1)2 + 1+ e−2s 1

(s − 1)2 + 1,

L[cos(at)] =s

s2 + a2,

L[sin(at)] =a

s2 + a2,

L[y ] = L[cos(t)]∣∣(s−1)

+ 2L[sin(t)]∣∣(s−1)

+ e−2s L[sin(t)]∣∣(s−1)

.

Exam: November 11, 2008. Problem 2

Example

Use Laplace Transform to find y solution of

y ′′ − 2y ′ + 2y = δ(t − 2), y(0) = 1, y ′(0) = 3.

Solution: Recall: L[y ] =(s − 1) + 2

(s − 1)2 + 1+

1

(s − 1)2 + 1e−2s ,

L[y ] =(s − 1)

(s − 1)2 + 1+ 2

1

(s − 1)2 + 1+ e−2s 1

(s − 1)2 + 1,

L[cos(at)] =s

s2 + a2, L[sin(at)] =

a

s2 + a2,

L[y ] = L[cos(t)]∣∣(s−1)

+ 2L[sin(t)]∣∣(s−1)

+ e−2s L[sin(t)]∣∣(s−1)

.

Exam: November 11, 2008. Problem 2

Example

Use Laplace Transform to find y solution of

y ′′ − 2y ′ + 2y = δ(t − 2), y(0) = 1, y ′(0) = 3.

Solution: Recall: L[y ] =(s − 1) + 2

(s − 1)2 + 1+

1

(s − 1)2 + 1e−2s ,

L[y ] =(s − 1)

(s − 1)2 + 1+ 2

1

(s − 1)2 + 1+ e−2s 1

(s − 1)2 + 1,

L[cos(at)] =s

s2 + a2, L[sin(at)] =

a

s2 + a2,

L[y ] = L[cos(t)]∣∣(s−1)

+ 2L[sin(t)]∣∣(s−1)

+ e−2s L[sin(t)]∣∣(s−1)

.

Exam: November 11, 2008. Problem 2

Example

Use Laplace Transform to find y solution of

y ′′ − 2y ′ + 2y = δ(t − 2), y(0) = 1, y ′(0) = 3.

Solution: Recall: L[y ] =(s − 1) + 2

(s − 1)2 + 1+

1

(s − 1)2 + 1e−2s ,

L[y ] =(s − 1)

(s − 1)2 + 1+ 2

1

(s − 1)2 + 1+ e−2s 1

(s − 1)2 + 1,

L[cos(at)] =s

s2 + a2, L[sin(at)] =

a

s2 + a2,

L[y ] = L[cos(t)]∣∣(s−1)

+ 2L[sin(t)]∣∣(s−1)

+ e−2s L[sin(t)]∣∣(s−1)

.

Exam: November 11, 2008. Problem 2

Example

Use Laplace Transform to find y solution of

y ′′ − 2y ′ + 2y = δ(t − 2), y(0) = 1, y ′(0) = 3.

Solution: Recall: L[y ] =(s − 1) + 2

(s − 1)2 + 1+

1

(s − 1)2 + 1e−2s ,

L[y ] =(s − 1)

(s − 1)2 + 1+ 2

1

(s − 1)2 + 1+ e−2s 1

(s − 1)2 + 1,

L[cos(at)] =s

s2 + a2, L[sin(at)] =

a

s2 + a2,

L[y ] = L[cos(t)]∣∣(s−1)

+ 2L[sin(t)]∣∣(s−1)

+ e−2s L[sin(t)]∣∣(s−1)

.

Exam: November 11, 2008. Problem 2

Example

Use Laplace Transform to find y solution of

y ′′ − 2y ′ + 2y = δ(t − 2), y(0) = 1, y ′(0) = 3.

Solution: Recall:

L[y ] = L[cos(t)]∣∣(s−1)

+ 2L[sin(t)]∣∣(s−1)

+ e−2s L[sin(t)]∣∣(s−1)

and L[f (t)]∣∣(s−c)

= L[ect f (t)]. Therefore,

L[y ] = L[et cos(t)] + 2L[et sin(t)] + e−2s L[et sin(t)].

Also recall: e−cs L[f (t)] = L[uc(t) f (t − c)]. Therefore,

L[y ] = L[et cos(t)] + 2L[et sin(t)] + L[u2(t) e(t−2) sin(t − 2)].

y(t) =[cos(t) + 2 sin(t)

]et + u2(t) sin(t − 2) e(t−2). C

Exam: November 11, 2008. Problem 2

Example

Use Laplace Transform to find y solution of

y ′′ − 2y ′ + 2y = δ(t − 2), y(0) = 1, y ′(0) = 3.

Solution: Recall:

L[y ] = L[cos(t)]∣∣(s−1)

+ 2L[sin(t)]∣∣(s−1)

+ e−2s L[sin(t)]∣∣(s−1)

and L[f (t)]∣∣(s−c)

= L[ect f (t)].

Therefore,

L[y ] = L[et cos(t)] + 2L[et sin(t)] + e−2s L[et sin(t)].

Also recall: e−cs L[f (t)] = L[uc(t) f (t − c)]. Therefore,

L[y ] = L[et cos(t)] + 2L[et sin(t)] + L[u2(t) e(t−2) sin(t − 2)].

y(t) =[cos(t) + 2 sin(t)

]et + u2(t) sin(t − 2) e(t−2). C

Exam: November 11, 2008. Problem 2

Example

Use Laplace Transform to find y solution of

y ′′ − 2y ′ + 2y = δ(t − 2), y(0) = 1, y ′(0) = 3.

Solution: Recall:

L[y ] = L[cos(t)]∣∣(s−1)

+ 2L[sin(t)]∣∣(s−1)

+ e−2s L[sin(t)]∣∣(s−1)

and L[f (t)]∣∣(s−c)

= L[ect f (t)]. Therefore,

L[y ] = L[et cos(t)]

+ 2L[et sin(t)] + e−2s L[et sin(t)].

Also recall: e−cs L[f (t)] = L[uc(t) f (t − c)]. Therefore,

L[y ] = L[et cos(t)] + 2L[et sin(t)] + L[u2(t) e(t−2) sin(t − 2)].

y(t) =[cos(t) + 2 sin(t)

]et + u2(t) sin(t − 2) e(t−2). C

Exam: November 11, 2008. Problem 2

Example

Use Laplace Transform to find y solution of

y ′′ − 2y ′ + 2y = δ(t − 2), y(0) = 1, y ′(0) = 3.

Solution: Recall:

L[y ] = L[cos(t)]∣∣(s−1)

+ 2L[sin(t)]∣∣(s−1)

+ e−2s L[sin(t)]∣∣(s−1)

and L[f (t)]∣∣(s−c)

= L[ect f (t)]. Therefore,

L[y ] = L[et cos(t)] + 2L[et sin(t)]

+ e−2s L[et sin(t)].

Also recall: e−cs L[f (t)] = L[uc(t) f (t − c)]. Therefore,

L[y ] = L[et cos(t)] + 2L[et sin(t)] + L[u2(t) e(t−2) sin(t − 2)].

y(t) =[cos(t) + 2 sin(t)

]et + u2(t) sin(t − 2) e(t−2). C

Exam: November 11, 2008. Problem 2

Example

Use Laplace Transform to find y solution of

y ′′ − 2y ′ + 2y = δ(t − 2), y(0) = 1, y ′(0) = 3.

Solution: Recall:

L[y ] = L[cos(t)]∣∣(s−1)

+ 2L[sin(t)]∣∣(s−1)

+ e−2s L[sin(t)]∣∣(s−1)

and L[f (t)]∣∣(s−c)

= L[ect f (t)]. Therefore,

L[y ] = L[et cos(t)] + 2L[et sin(t)] + e−2s L[et sin(t)].

Also recall: e−cs L[f (t)] = L[uc(t) f (t − c)]. Therefore,

L[y ] = L[et cos(t)] + 2L[et sin(t)] + L[u2(t) e(t−2) sin(t − 2)].

y(t) =[cos(t) + 2 sin(t)

]et + u2(t) sin(t − 2) e(t−2). C

Exam: November 11, 2008. Problem 2

Example

Use Laplace Transform to find y solution of

y ′′ − 2y ′ + 2y = δ(t − 2), y(0) = 1, y ′(0) = 3.

Solution: Recall:

L[y ] = L[cos(t)]∣∣(s−1)

+ 2L[sin(t)]∣∣(s−1)

+ e−2s L[sin(t)]∣∣(s−1)

and L[f (t)]∣∣(s−c)

= L[ect f (t)]. Therefore,

L[y ] = L[et cos(t)] + 2L[et sin(t)] + e−2s L[et sin(t)].

Also recall: e−cs L[f (t)] = L[uc(t) f (t − c)].

Therefore,

L[y ] = L[et cos(t)] + 2L[et sin(t)] + L[u2(t) e(t−2) sin(t − 2)].

y(t) =[cos(t) + 2 sin(t)

]et + u2(t) sin(t − 2) e(t−2). C

Exam: November 11, 2008. Problem 2

Example

Use Laplace Transform to find y solution of

y ′′ − 2y ′ + 2y = δ(t − 2), y(0) = 1, y ′(0) = 3.

Solution: Recall:

L[y ] = L[cos(t)]∣∣(s−1)

+ 2L[sin(t)]∣∣(s−1)

+ e−2s L[sin(t)]∣∣(s−1)

and L[f (t)]∣∣(s−c)

= L[ect f (t)]. Therefore,

L[y ] = L[et cos(t)] + 2L[et sin(t)] + e−2s L[et sin(t)].

Also recall: e−cs L[f (t)] = L[uc(t) f (t − c)]. Therefore,

L[y ] = L[et cos(t)] + 2L[et sin(t)] + L[u2(t) e(t−2) sin(t − 2)].

y(t) =[cos(t) + 2 sin(t)

]et + u2(t) sin(t − 2) e(t−2). C

Exam: November 11, 2008. Problem 2

Example

Use Laplace Transform to find y solution of

y ′′ − 2y ′ + 2y = δ(t − 2), y(0) = 1, y ′(0) = 3.

Solution: Recall:

L[y ] = L[cos(t)]∣∣(s−1)

+ 2L[sin(t)]∣∣(s−1)

+ e−2s L[sin(t)]∣∣(s−1)

and L[f (t)]∣∣(s−c)

= L[ect f (t)]. Therefore,

L[y ] = L[et cos(t)] + 2L[et sin(t)] + e−2s L[et sin(t)].

Also recall: e−cs L[f (t)] = L[uc(t) f (t − c)]. Therefore,

L[y ] = L[et cos(t)] + 2L[et sin(t)] + L[u2(t) e(t−2) sin(t − 2)].

y(t) =[cos(t) + 2 sin(t)

]et + u2(t) sin(t − 2) e(t−2). C

Exam: November 11, 2008. Problem 3

Example

Sketch the graph of g and use LT to find y solution of

y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =

{0, t < 2,

e(t−2), t > 2.

Solution:

t

u ( t − 2 )1

t2

g ( t )

e

Express g using step functions,

g(t) = u2(t) e(t−2).

L[uc(t) f (t − c)] = e−cs L[f (t)].

Therefore,

L[g(t)] = e−2sL[et ].

We obtain: L[g(t)] =e−2s

(s − 1).

Exam: November 11, 2008. Problem 3

Example

Sketch the graph of g and use LT to find y solution of

y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =

{0, t < 2,

e(t−2), t > 2.

Solution:

t

u ( t − 2 )1

t2

g ( t )

e

Express g using step functions,

g(t) = u2(t) e(t−2).

L[uc(t) f (t − c)] = e−cs L[f (t)].

Therefore,

L[g(t)] = e−2sL[et ].

We obtain: L[g(t)] =e−2s

(s − 1).

Exam: November 11, 2008. Problem 3

Example

Sketch the graph of g and use LT to find y solution of

y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =

{0, t < 2,

e(t−2), t > 2.

Solution:

t

u ( t − 2 )1

t2

g ( t )

e

Express g using step functions,

g(t) = u2(t) e(t−2).

L[uc(t) f (t − c)] = e−cs L[f (t)].

Therefore,

L[g(t)] = e−2sL[et ].

We obtain: L[g(t)] =e−2s

(s − 1).

Exam: November 11, 2008. Problem 3

Example

Sketch the graph of g and use LT to find y solution of

y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =

{0, t < 2,

e(t−2), t > 2.

Solution:

t

u ( t − 2 )1

t2

g ( t )

e

Express g using step functions,

g(t) = u2(t) e(t−2).

L[uc(t) f (t − c)] = e−cs L[f (t)].

Therefore,

L[g(t)] = e−2sL[et ].

We obtain: L[g(t)] =e−2s

(s − 1).

Exam: November 11, 2008. Problem 3

Example

Sketch the graph of g and use LT to find y solution of

y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =

{0, t < 2,

e(t−2), t > 2.

Solution:

t

u ( t − 2 )1

t2

g ( t )

e

Express g using step functions,

g(t) = u2(t) e(t−2).

L[uc(t) f (t − c)] = e−cs L[f (t)].

Therefore,

L[g(t)] = e−2sL[et ].

We obtain: L[g(t)] =e−2s

(s − 1).

Exam: November 11, 2008. Problem 3

Example

Sketch the graph of g and use LT to find y solution of

y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =

{0, t < 2,

e(t−2), t > 2.

Solution:

t

u ( t − 2 )1

t2

g ( t )

e

Express g using step functions,

g(t) = u2(t) e(t−2).

L[uc(t) f (t − c)] = e−cs L[f (t)].

Therefore,

L[g(t)] = e−2sL[et ].

We obtain: L[g(t)] =e−2s

(s − 1).

Exam: November 11, 2008. Problem 3

Example

Sketch the graph of g and use LT to find y solution of

y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =

{0, t < 2,

e(t−2), t > 2.

Solution:

t

u ( t − 2 )1

t2

g ( t )

e

Express g using step functions,

g(t) = u2(t) e(t−2).

L[uc(t) f (t − c)] = e−cs L[f (t)].

Therefore,

L[g(t)] = e−2sL[et ].

We obtain: L[g(t)] =e−2s

(s − 1).

Exam: November 11, 2008. Problem 3

Example

Sketch the graph of g and use LT to find y solution of

y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =

{0, t < 2,

e(t−2), t > 2.

Solution: Recall: L[g(t)] =e−2s

(s − 1).

L[y ′′] + 3L[y ] = L[g(t)] =e−2s

(s − 1).

(s2 + 3)L[y ] =e−2s

(s − 1)⇒ L[y ] = e−2s 1

(s − 1)(s2 + 3).

H(s) =1

(s − 1)(s2 + 3)=

a

(s − 1)+

(bs + c)

(s2 + 3)

1 = a(s2 + 3) + (bs + c)(s − 1)

Exam: November 11, 2008. Problem 3

Example

Sketch the graph of g and use LT to find y solution of

y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =

{0, t < 2,

e(t−2), t > 2.

Solution: Recall: L[g(t)] =e−2s

(s − 1).

L[y ′′] + 3L[y ] = L[g(t)]

=e−2s

(s − 1).

(s2 + 3)L[y ] =e−2s

(s − 1)⇒ L[y ] = e−2s 1

(s − 1)(s2 + 3).

H(s) =1

(s − 1)(s2 + 3)=

a

(s − 1)+

(bs + c)

(s2 + 3)

1 = a(s2 + 3) + (bs + c)(s − 1)

Exam: November 11, 2008. Problem 3

Example

Sketch the graph of g and use LT to find y solution of

y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =

{0, t < 2,

e(t−2), t > 2.

Solution: Recall: L[g(t)] =e−2s

(s − 1).

L[y ′′] + 3L[y ] = L[g(t)] =e−2s

(s − 1).

(s2 + 3)L[y ] =e−2s

(s − 1)⇒ L[y ] = e−2s 1

(s − 1)(s2 + 3).

H(s) =1

(s − 1)(s2 + 3)=

a

(s − 1)+

(bs + c)

(s2 + 3)

1 = a(s2 + 3) + (bs + c)(s − 1)

Exam: November 11, 2008. Problem 3

Example

Sketch the graph of g and use LT to find y solution of

y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =

{0, t < 2,

e(t−2), t > 2.

Solution: Recall: L[g(t)] =e−2s

(s − 1).

L[y ′′] + 3L[y ] = L[g(t)] =e−2s

(s − 1).

(s2 + 3)L[y ] =e−2s

(s − 1)

⇒ L[y ] = e−2s 1

(s − 1)(s2 + 3).

H(s) =1

(s − 1)(s2 + 3)=

a

(s − 1)+

(bs + c)

(s2 + 3)

1 = a(s2 + 3) + (bs + c)(s − 1)

Exam: November 11, 2008. Problem 3

Example

Sketch the graph of g and use LT to find y solution of

y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =

{0, t < 2,

e(t−2), t > 2.

Solution: Recall: L[g(t)] =e−2s

(s − 1).

L[y ′′] + 3L[y ] = L[g(t)] =e−2s

(s − 1).

(s2 + 3)L[y ] =e−2s

(s − 1)⇒ L[y ] = e−2s 1

(s − 1)(s2 + 3).

H(s) =1

(s − 1)(s2 + 3)=

a

(s − 1)+

(bs + c)

(s2 + 3)

1 = a(s2 + 3) + (bs + c)(s − 1)

Exam: November 11, 2008. Problem 3

Example

Sketch the graph of g and use LT to find y solution of

y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =

{0, t < 2,

e(t−2), t > 2.

Solution: Recall: L[g(t)] =e−2s

(s − 1).

L[y ′′] + 3L[y ] = L[g(t)] =e−2s

(s − 1).

(s2 + 3)L[y ] =e−2s

(s − 1)⇒ L[y ] = e−2s 1

(s − 1)(s2 + 3).

H(s) =1

(s − 1)(s2 + 3)

=a

(s − 1)+

(bs + c)

(s2 + 3)

1 = a(s2 + 3) + (bs + c)(s − 1)

Exam: November 11, 2008. Problem 3

Example

Sketch the graph of g and use LT to find y solution of

y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =

{0, t < 2,

e(t−2), t > 2.

Solution: Recall: L[g(t)] =e−2s

(s − 1).

L[y ′′] + 3L[y ] = L[g(t)] =e−2s

(s − 1).

(s2 + 3)L[y ] =e−2s

(s − 1)⇒ L[y ] = e−2s 1

(s − 1)(s2 + 3).

H(s) =1

(s − 1)(s2 + 3)=

a

(s − 1)+

(bs + c)

(s2 + 3)

1 = a(s2 + 3) + (bs + c)(s − 1)

Exam: November 11, 2008. Problem 3

Example

Sketch the graph of g and use LT to find y solution of

y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =

{0, t < 2,

e(t−2), t > 2.

Solution: Recall: L[g(t)] =e−2s

(s − 1).

L[y ′′] + 3L[y ] = L[g(t)] =e−2s

(s − 1).

(s2 + 3)L[y ] =e−2s

(s − 1)⇒ L[y ] = e−2s 1

(s − 1)(s2 + 3).

H(s) =1

(s − 1)(s2 + 3)=

a

(s − 1)+

(bs + c)

(s2 + 3)

1 = a(s2 + 3) + (bs + c)(s − 1)

Exam: November 11, 2008. Problem 3

Example

Sketch the graph of g and use LT to find y solution of

y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =

{0, t < 2,

e(t−2), t > 2.

Solution: Recall: 1 = a(s2 + 3) + (bs + c)(s − 1).

1 = as2 + 3a + bs2 + cs − bs − c

1 = (a + b) s2 + (c − b) s + (3a− c)

a + b = 0, c − b = 0, 3a− c = 1.

a = −b, c = b, −3b−b = 1 ⇒ b = −1

4, a =

1

4, c = −1

4.

H(s) =1

4

[ 1

s − 1− s + 1

s2 + 3

].

Exam: November 11, 2008. Problem 3

Example

Sketch the graph of g and use LT to find y solution of

y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =

{0, t < 2,

e(t−2), t > 2.

Solution: Recall: 1 = a(s2 + 3) + (bs + c)(s − 1).

1 = as2 + 3a + bs2 + cs − bs − c

1 = (a + b) s2 + (c − b) s + (3a− c)

a + b = 0, c − b = 0, 3a− c = 1.

a = −b, c = b, −3b−b = 1 ⇒ b = −1

4, a =

1

4, c = −1

4.

H(s) =1

4

[ 1

s − 1− s + 1

s2 + 3

].

Exam: November 11, 2008. Problem 3

Example

Sketch the graph of g and use LT to find y solution of

y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =

{0, t < 2,

e(t−2), t > 2.

Solution: Recall: 1 = a(s2 + 3) + (bs + c)(s − 1).

1 = as2 + 3a + bs2 + cs − bs − c

1 = (a + b) s2 + (c − b) s + (3a− c)

a + b = 0, c − b = 0, 3a− c = 1.

a = −b, c = b, −3b−b = 1 ⇒ b = −1

4, a =

1

4, c = −1

4.

H(s) =1

4

[ 1

s − 1− s + 1

s2 + 3

].

Exam: November 11, 2008. Problem 3

Example

Sketch the graph of g and use LT to find y solution of

y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =

{0, t < 2,

e(t−2), t > 2.

Solution: Recall: 1 = a(s2 + 3) + (bs + c)(s − 1).

1 = as2 + 3a + bs2 + cs − bs − c

1 = (a + b) s2 + (c − b) s + (3a− c)

a + b = 0, c − b = 0, 3a− c = 1.

a = −b, c = b, −3b−b = 1 ⇒ b = −1

4, a =

1

4, c = −1

4.

H(s) =1

4

[ 1

s − 1− s + 1

s2 + 3

].

Exam: November 11, 2008. Problem 3

Example

Sketch the graph of g and use LT to find y solution of

y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =

{0, t < 2,

e(t−2), t > 2.

Solution: Recall: 1 = a(s2 + 3) + (bs + c)(s − 1).

1 = as2 + 3a + bs2 + cs − bs − c

1 = (a + b) s2 + (c − b) s + (3a− c)

a + b = 0, c − b = 0, 3a− c = 1.

a = −b,

c = b, −3b−b = 1 ⇒ b = −1

4, a =

1

4, c = −1

4.

H(s) =1

4

[ 1

s − 1− s + 1

s2 + 3

].

Exam: November 11, 2008. Problem 3

Example

Sketch the graph of g and use LT to find y solution of

y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =

{0, t < 2,

e(t−2), t > 2.

Solution: Recall: 1 = a(s2 + 3) + (bs + c)(s − 1).

1 = as2 + 3a + bs2 + cs − bs − c

1 = (a + b) s2 + (c − b) s + (3a− c)

a + b = 0, c − b = 0, 3a− c = 1.

a = −b, c = b,

−3b−b = 1 ⇒ b = −1

4, a =

1

4, c = −1

4.

H(s) =1

4

[ 1

s − 1− s + 1

s2 + 3

].

Exam: November 11, 2008. Problem 3

Example

Sketch the graph of g and use LT to find y solution of

y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =

{0, t < 2,

e(t−2), t > 2.

Solution: Recall: 1 = a(s2 + 3) + (bs + c)(s − 1).

1 = as2 + 3a + bs2 + cs − bs − c

1 = (a + b) s2 + (c − b) s + (3a− c)

a + b = 0, c − b = 0, 3a− c = 1.

a = −b, c = b, −3b−b = 1

⇒ b = −1

4, a =

1

4, c = −1

4.

H(s) =1

4

[ 1

s − 1− s + 1

s2 + 3

].

Exam: November 11, 2008. Problem 3

Example

Sketch the graph of g and use LT to find y solution of

y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =

{0, t < 2,

e(t−2), t > 2.

Solution: Recall: 1 = a(s2 + 3) + (bs + c)(s − 1).

1 = as2 + 3a + bs2 + cs − bs − c

1 = (a + b) s2 + (c − b) s + (3a− c)

a + b = 0, c − b = 0, 3a− c = 1.

a = −b, c = b, −3b−b = 1 ⇒ b = −1

4,

a =1

4, c = −1

4.

H(s) =1

4

[ 1

s − 1− s + 1

s2 + 3

].

Exam: November 11, 2008. Problem 3

Example

Sketch the graph of g and use LT to find y solution of

y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =

{0, t < 2,

e(t−2), t > 2.

Solution: Recall: 1 = a(s2 + 3) + (bs + c)(s − 1).

1 = as2 + 3a + bs2 + cs − bs − c

1 = (a + b) s2 + (c − b) s + (3a− c)

a + b = 0, c − b = 0, 3a− c = 1.

a = −b, c = b, −3b−b = 1 ⇒ b = −1

4, a =

1

4,

c = −1

4.

H(s) =1

4

[ 1

s − 1− s + 1

s2 + 3

].

Exam: November 11, 2008. Problem 3

Example

Sketch the graph of g and use LT to find y solution of

y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =

{0, t < 2,

e(t−2), t > 2.

Solution: Recall: 1 = a(s2 + 3) + (bs + c)(s − 1).

1 = as2 + 3a + bs2 + cs − bs − c

1 = (a + b) s2 + (c − b) s + (3a− c)

a + b = 0, c − b = 0, 3a− c = 1.

a = −b, c = b, −3b−b = 1 ⇒ b = −1

4, a =

1

4, c = −1

4.

H(s) =1

4

[ 1

s − 1− s + 1

s2 + 3

].

Exam: November 11, 2008. Problem 3

Example

Sketch the graph of g and use LT to find y solution of

y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =

{0, t < 2,

e(t−2), t > 2.

Solution: Recall: 1 = a(s2 + 3) + (bs + c)(s − 1).

1 = as2 + 3a + bs2 + cs − bs − c

1 = (a + b) s2 + (c − b) s + (3a− c)

a + b = 0, c − b = 0, 3a− c = 1.

a = −b, c = b, −3b−b = 1 ⇒ b = −1

4, a =

1

4, c = −1

4.

H(s) =1

4

[ 1

s − 1− s + 1

s2 + 3

].

Exam: November 11, 2008. Problem 3

Example

Sketch the graph of g and use LT to find y solution of

y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =

{0, t < 2,

e(t−2), t > 2.

Solution: Recall: H(s) =1

4

[ 1

s − 1− s + 1

s2 + 3

], L[y ] = e−2s H(s).

H(s) =1

4

[ 1

s − 1− s

s2 + 3− 1√

3

√3

s2 + 3

],

H(s) =1

4

[L[et ]− L

[cos

(√3 t

)]− 1√

3L

[sin

(√3 t

)]].

H(s) = L[1

4

(et − cos

(√3 t

)− 1√

3sin

(√3 t

))].

Exam: November 11, 2008. Problem 3

Example

Sketch the graph of g and use LT to find y solution of

y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =

{0, t < 2,

e(t−2), t > 2.

Solution: Recall: H(s) =1

4

[ 1

s − 1− s + 1

s2 + 3

], L[y ] = e−2s H(s).

H(s) =1

4

[ 1

s − 1− s

s2 + 3− 1√

3

√3

s2 + 3

],

H(s) =1

4

[L[et ]− L

[cos

(√3 t

)]− 1√

3L

[sin

(√3 t

)]].

H(s) = L[1

4

(et − cos

(√3 t

)− 1√

3sin

(√3 t

))].

Exam: November 11, 2008. Problem 3

Example

Sketch the graph of g and use LT to find y solution of

y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =

{0, t < 2,

e(t−2), t > 2.

Solution: Recall: H(s) =1

4

[ 1

s − 1− s + 1

s2 + 3

], L[y ] = e−2s H(s).

H(s) =1

4

[ 1

s − 1− s

s2 + 3− 1√

3

√3

s2 + 3

],

H(s) =1

4

[L[et ]

− L[cos

(√3 t

)]− 1√

3L

[sin

(√3 t

)]].

H(s) = L[1

4

(et − cos

(√3 t

)− 1√

3sin

(√3 t

))].

Exam: November 11, 2008. Problem 3

Example

Sketch the graph of g and use LT to find y solution of

y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =

{0, t < 2,

e(t−2), t > 2.

Solution: Recall: H(s) =1

4

[ 1

s − 1− s + 1

s2 + 3

], L[y ] = e−2s H(s).

H(s) =1

4

[ 1

s − 1− s

s2 + 3− 1√

3

√3

s2 + 3

],

H(s) =1

4

[L[et ]− L

[cos

(√3 t

)]

− 1√3L

[sin

(√3 t

)]].

H(s) = L[1

4

(et − cos

(√3 t

)− 1√

3sin

(√3 t

))].

Exam: November 11, 2008. Problem 3

Example

Sketch the graph of g and use LT to find y solution of

y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =

{0, t < 2,

e(t−2), t > 2.

Solution: Recall: H(s) =1

4

[ 1

s − 1− s + 1

s2 + 3

], L[y ] = e−2s H(s).

H(s) =1

4

[ 1

s − 1− s

s2 + 3− 1√

3

√3

s2 + 3

],

H(s) =1

4

[L[et ]− L

[cos

(√3 t

)]− 1√

3L

[sin

(√3 t

)]].

H(s) = L[1

4

(et − cos

(√3 t

)− 1√

3sin

(√3 t

))].

Exam: November 11, 2008. Problem 3

Example

Sketch the graph of g and use LT to find y solution of

y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =

{0, t < 2,

e(t−2), t > 2.

Solution: Recall: H(s) =1

4

[ 1

s − 1− s + 1

s2 + 3

], L[y ] = e−2s H(s).

H(s) =1

4

[ 1

s − 1− s

s2 + 3− 1√

3

√3

s2 + 3

],

H(s) =1

4

[L[et ]− L

[cos

(√3 t

)]− 1√

3L

[sin

(√3 t

)]].

H(s) = L[1

4

(et − cos

(√3 t

)− 1√

3sin

(√3 t

))].

Exam: November 11, 2008. Problem 3

Example

Sketch the graph of g and use LT to find y solution of

y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =

{0, t < 2,

e(t−2), t > 2.

Solution: Recall: H(s) = L[1

4

(et − cos

(√3 t

)− 1√

3sin

(√3 t

))].

h(t) =1

4

(et − cos

(√3 t

)− 1√

3sin

(√3 t

)), H(s) = L[h(t)].

L[y ] = e−2s H(s) = e−2s L[h(t)] = L[u2(t) h(t − 2)].

We conclude: y(t) = u2(t) h(t − 2). Equivalently,

y(t) =u2(t)

4

[e(t−2) − cos

(√3 (t − 2)

)− 1√

3sin

(√3 (t − 2)

)].C

Exam: November 11, 2008. Problem 3

Example

Sketch the graph of g and use LT to find y solution of

y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =

{0, t < 2,

e(t−2), t > 2.

Solution: Recall: H(s) = L[1

4

(et − cos

(√3 t

)− 1√

3sin

(√3 t

))].

h(t) =1

4

(et − cos

(√3 t

)− 1√

3sin

(√3 t

)),

H(s) = L[h(t)].

L[y ] = e−2s H(s) = e−2s L[h(t)] = L[u2(t) h(t − 2)].

We conclude: y(t) = u2(t) h(t − 2). Equivalently,

y(t) =u2(t)

4

[e(t−2) − cos

(√3 (t − 2)

)− 1√

3sin

(√3 (t − 2)

)].C

Exam: November 11, 2008. Problem 3

Example

Sketch the graph of g and use LT to find y solution of

y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =

{0, t < 2,

e(t−2), t > 2.

Solution: Recall: H(s) = L[1

4

(et − cos

(√3 t

)− 1√

3sin

(√3 t

))].

h(t) =1

4

(et − cos

(√3 t

)− 1√

3sin

(√3 t

)), H(s) = L[h(t)].

L[y ] = e−2s H(s) = e−2s L[h(t)] = L[u2(t) h(t − 2)].

We conclude: y(t) = u2(t) h(t − 2). Equivalently,

y(t) =u2(t)

4

[e(t−2) − cos

(√3 (t − 2)

)− 1√

3sin

(√3 (t − 2)

)].C

Exam: November 11, 2008. Problem 3

Example

Sketch the graph of g and use LT to find y solution of

y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =

{0, t < 2,

e(t−2), t > 2.

Solution: Recall: H(s) = L[1

4

(et − cos

(√3 t

)− 1√

3sin

(√3 t

))].

h(t) =1

4

(et − cos

(√3 t

)− 1√

3sin

(√3 t

)), H(s) = L[h(t)].

L[y ] = e−2s H(s)

= e−2s L[h(t)] = L[u2(t) h(t − 2)].

We conclude: y(t) = u2(t) h(t − 2). Equivalently,

y(t) =u2(t)

4

[e(t−2) − cos

(√3 (t − 2)

)− 1√

3sin

(√3 (t − 2)

)].C

Exam: November 11, 2008. Problem 3

Example

Sketch the graph of g and use LT to find y solution of

y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =

{0, t < 2,

e(t−2), t > 2.

Solution: Recall: H(s) = L[1

4

(et − cos

(√3 t

)− 1√

3sin

(√3 t

))].

h(t) =1

4

(et − cos

(√3 t

)− 1√

3sin

(√3 t

)), H(s) = L[h(t)].

L[y ] = e−2s H(s) = e−2s L[h(t)]

= L[u2(t) h(t − 2)].

We conclude: y(t) = u2(t) h(t − 2). Equivalently,

y(t) =u2(t)

4

[e(t−2) − cos

(√3 (t − 2)

)− 1√

3sin

(√3 (t − 2)

)].C

Exam: November 11, 2008. Problem 3

Example

Sketch the graph of g and use LT to find y solution of

y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =

{0, t < 2,

e(t−2), t > 2.

Solution: Recall: H(s) = L[1

4

(et − cos

(√3 t

)− 1√

3sin

(√3 t

))].

h(t) =1

4

(et − cos

(√3 t

)− 1√

3sin

(√3 t

)), H(s) = L[h(t)].

L[y ] = e−2s H(s) = e−2s L[h(t)] = L[u2(t) h(t − 2)].

We conclude: y(t) = u2(t) h(t − 2). Equivalently,

y(t) =u2(t)

4

[e(t−2) − cos

(√3 (t − 2)

)− 1√

3sin

(√3 (t − 2)

)].C

Exam: November 11, 2008. Problem 3

Example

Sketch the graph of g and use LT to find y solution of

y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =

{0, t < 2,

e(t−2), t > 2.

Solution: Recall: H(s) = L[1

4

(et − cos

(√3 t

)− 1√

3sin

(√3 t

))].

h(t) =1

4

(et − cos

(√3 t

)− 1√

3sin

(√3 t

)), H(s) = L[h(t)].

L[y ] = e−2s H(s) = e−2s L[h(t)] = L[u2(t) h(t − 2)].

We conclude: y(t) = u2(t) h(t − 2).

Equivalently,

y(t) =u2(t)

4

[e(t−2) − cos

(√3 (t − 2)

)− 1√

3sin

(√3 (t − 2)

)].C

Exam: November 11, 2008. Problem 3

Example

Sketch the graph of g and use LT to find y solution of

y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =

{0, t < 2,

e(t−2), t > 2.

Solution: Recall: H(s) = L[1

4

(et − cos

(√3 t

)− 1√

3sin

(√3 t

))].

h(t) =1

4

(et − cos

(√3 t

)− 1√

3sin

(√3 t

)), H(s) = L[h(t)].

L[y ] = e−2s H(s) = e−2s L[h(t)] = L[u2(t) h(t − 2)].

We conclude: y(t) = u2(t) h(t − 2). Equivalently,

y(t) =u2(t)

4

[e(t−2) − cos

(√3 (t − 2)

)− 1√

3sin

(√3 (t − 2)

)].C

Extra problem

Example

Use convolutions to find f satisfying L[f (t)] =e−2s

(s − 1)(s2 + 3).

Solution: One way to solve this is with the splitting

L[f (t)] = e−2s 1

(s2 + 3)

1

(s − 1)= e−2s 1√

3

√3

(s2 + 3)

1

(s − 1),

L[f (t)] = e−2s 1√3L[sin

(√3 t

)]L[et ]

L[f (t)] =1√3L[u2(t) sin

(√3 (t − 2)

)]L[et ].

f (t) =1√3

∫ t

0u2(τ) sin

(√3 (τ − 2)

)e(t−τ) dτ. C

Extra problem

Example

Use convolutions to find f satisfying L[f (t)] =e−2s

(s − 1)(s2 + 3).

Solution: One way to solve this is with the splitting

L[f (t)] = e−2s 1

(s2 + 3)

1

(s − 1)

= e−2s 1√3

√3

(s2 + 3)

1

(s − 1),

L[f (t)] = e−2s 1√3L[sin

(√3 t

)]L[et ]

L[f (t)] =1√3L[u2(t) sin

(√3 (t − 2)

)]L[et ].

f (t) =1√3

∫ t

0u2(τ) sin

(√3 (τ − 2)

)e(t−τ) dτ. C

Extra problem

Example

Use convolutions to find f satisfying L[f (t)] =e−2s

(s − 1)(s2 + 3).

Solution: One way to solve this is with the splitting

L[f (t)] = e−2s 1

(s2 + 3)

1

(s − 1)= e−2s 1√

3

√3

(s2 + 3)

1

(s − 1),

L[f (t)] = e−2s 1√3L[sin

(√3 t

)]L[et ]

L[f (t)] =1√3L[u2(t) sin

(√3 (t − 2)

)]L[et ].

f (t) =1√3

∫ t

0u2(τ) sin

(√3 (τ − 2)

)e(t−τ) dτ. C

Extra problem

Example

Use convolutions to find f satisfying L[f (t)] =e−2s

(s − 1)(s2 + 3).

Solution: One way to solve this is with the splitting

L[f (t)] = e−2s 1

(s2 + 3)

1

(s − 1)= e−2s 1√

3

√3

(s2 + 3)

1

(s − 1),

L[f (t)] = e−2s 1√3L[sin

(√3 t

)]L[et ]

L[f (t)] =1√3L[u2(t) sin

(√3 (t − 2)

)]L[et ].

f (t) =1√3

∫ t

0u2(τ) sin

(√3 (τ − 2)

)e(t−τ) dτ. C

Extra problem

Example

Use convolutions to find f satisfying L[f (t)] =e−2s

(s − 1)(s2 + 3).

Solution: One way to solve this is with the splitting

L[f (t)] = e−2s 1

(s2 + 3)

1

(s − 1)= e−2s 1√

3

√3

(s2 + 3)

1

(s − 1),

L[f (t)] = e−2s 1√3L[sin

(√3 t

)]L[et ]

L[f (t)] =1√3L[u2(t) sin

(√3 (t − 2)

)]L[et ].

f (t) =1√3

∫ t

0u2(τ) sin

(√3 (τ − 2)

)e(t−τ) dτ. C

Extra problem

Example

Use convolutions to find f satisfying L[f (t)] =e−2s

(s − 1)(s2 + 3).

Solution: One way to solve this is with the splitting

L[f (t)] = e−2s 1

(s2 + 3)

1

(s − 1)= e−2s 1√

3

√3

(s2 + 3)

1

(s − 1),

L[f (t)] = e−2s 1√3L[sin

(√3 t

)]L[et ]

L[f (t)] =1√3L[u2(t) sin

(√3 (t − 2)

)]L[et ].

f (t) =1√3

∫ t

0u2(τ) sin

(√3 (τ − 2)

)e(t−τ) dτ. C

Exam: November 12, 2008. Problem 3

ExampleSketch the graph of g and use LT to find y solution of

y ′′ − 6y = g(t), y(0) = y ′(0) = 0, g(t) =

{0, t < π,

sin(t − π), t > π.

Solution:

1

g ( t )

u ( t − pi )sin ( t )

pit

Express g using step functions,

g(t) = uπ(t) sin(t − π).

L[uc(t) f (t − c)] = e−cs L[f (t)].

Therefore,

L[g(t)] = e−πsL[sin(t)].

We obtain: L[g(t)] =e−πs

s2 + 1.

Exam: November 12, 2008. Problem 3

ExampleSketch the graph of g and use LT to find y solution of

y ′′ − 6y = g(t), y(0) = y ′(0) = 0, g(t) =

{0, t < π,

sin(t − π), t > π.

Solution:

1

g ( t )

u ( t − pi )sin ( t )

pit

Express g using step functions,

g(t) = uπ(t) sin(t − π).

L[uc(t) f (t − c)] = e−cs L[f (t)].

Therefore,

L[g(t)] = e−πsL[sin(t)].

We obtain: L[g(t)] =e−πs

s2 + 1.

Exam: November 12, 2008. Problem 3

ExampleSketch the graph of g and use LT to find y solution of

y ′′ − 6y = g(t), y(0) = y ′(0) = 0, g(t) =

{0, t < π,

sin(t − π), t > π.

Solution:

1

g ( t )

u ( t − pi )sin ( t )

pit

Express g using step functions,

g(t) = uπ(t) sin(t − π).

L[uc(t) f (t − c)] = e−cs L[f (t)].

Therefore,

L[g(t)] = e−πsL[sin(t)].

We obtain: L[g(t)] =e−πs

s2 + 1.

Exam: November 12, 2008. Problem 3

ExampleSketch the graph of g and use LT to find y solution of

y ′′ − 6y = g(t), y(0) = y ′(0) = 0, g(t) =

{0, t < π,

sin(t − π), t > π.

Solution:

1

g ( t )

u ( t − pi )sin ( t )

pit

Express g using step functions,

g(t) = uπ(t) sin(t − π).

L[uc(t) f (t − c)] = e−cs L[f (t)].

Therefore,

L[g(t)] = e−πsL[sin(t)].

We obtain: L[g(t)] =e−πs

s2 + 1.

Exam: November 12, 2008. Problem 3

ExampleSketch the graph of g and use LT to find y solution of

y ′′ − 6y = g(t), y(0) = y ′(0) = 0, g(t) =

{0, t < π,

sin(t − π), t > π.

Solution:

1

g ( t )

u ( t − pi )sin ( t )

pit

Express g using step functions,

g(t) = uπ(t) sin(t − π).

L[uc(t) f (t − c)] = e−cs L[f (t)].

Therefore,

L[g(t)] = e−πsL[sin(t)].

We obtain: L[g(t)] =e−πs

s2 + 1.

Exam: November 12, 2008. Problem 3

ExampleSketch the graph of g and use LT to find y solution of

y ′′ − 6y = g(t), y(0) = y ′(0) = 0, g(t) =

{0, t < π,

sin(t − π), t > π.

Solution:

1

g ( t )

u ( t − pi )sin ( t )

pit

Express g using step functions,

g(t) = uπ(t) sin(t − π).

L[uc(t) f (t − c)] = e−cs L[f (t)].

Therefore,

L[g(t)] = e−πsL[sin(t)].

We obtain: L[g(t)] =e−πs

s2 + 1.

Exam: November 12, 2008. Problem 3

ExampleSketch the graph of g and use LT to find y solution of

y ′′ − 6y = g(t), y(0) = y ′(0) = 0, g(t) =

{0, t < π,

sin(t − π), t > π.

Solution:

1

g ( t )

u ( t − pi )sin ( t )

pit

Express g using step functions,

g(t) = uπ(t) sin(t − π).

L[uc(t) f (t − c)] = e−cs L[f (t)].

Therefore,

L[g(t)] = e−πsL[sin(t)].

We obtain: L[g(t)] =e−πs

s2 + 1.

Exam: November 12, 2008. Problem 3ExampleSketch the graph of g and use LT to find y solution of

y ′′ − 6y = g(t), y(0) = y ′(0) = 0, g(t) =

{0, t < π,

sin(t − π), t > π.

Solution: L[g(t)] =e−πs

s2 + 1.

L[y ′′]− 6L[y ] = L[g(t)] =e−πs

s2 + 1.

(s2 − 6)L[y ] =e−πs

s2 + 1⇒ L[y ] = e−πs 1

(s2 + 1)(s2 − 6).

H(s) =1

(s2 + 1)(s2 − 6)=

1

(s2 + 1)(s +√

6)(s −√

6)

H(s) =a

(s +√

6)+

b

(s −√

6)+

(cs + d)

(s2 + 1).

Exam: November 12, 2008. Problem 3ExampleSketch the graph of g and use LT to find y solution of

y ′′ − 6y = g(t), y(0) = y ′(0) = 0, g(t) =

{0, t < π,

sin(t − π), t > π.

Solution: L[g(t)] =e−πs

s2 + 1.

L[y ′′]− 6L[y ] = L[g(t)]

=e−πs

s2 + 1.

(s2 − 6)L[y ] =e−πs

s2 + 1⇒ L[y ] = e−πs 1

(s2 + 1)(s2 − 6).

H(s) =1

(s2 + 1)(s2 − 6)=

1

(s2 + 1)(s +√

6)(s −√

6)

H(s) =a

(s +√

6)+

b

(s −√

6)+

(cs + d)

(s2 + 1).

Exam: November 12, 2008. Problem 3ExampleSketch the graph of g and use LT to find y solution of

y ′′ − 6y = g(t), y(0) = y ′(0) = 0, g(t) =

{0, t < π,

sin(t − π), t > π.

Solution: L[g(t)] =e−πs

s2 + 1.

L[y ′′]− 6L[y ] = L[g(t)] =e−πs

s2 + 1.

(s2 − 6)L[y ] =e−πs

s2 + 1⇒ L[y ] = e−πs 1

(s2 + 1)(s2 − 6).

H(s) =1

(s2 + 1)(s2 − 6)=

1

(s2 + 1)(s +√

6)(s −√

6)

H(s) =a

(s +√

6)+

b

(s −√

6)+

(cs + d)

(s2 + 1).

Exam: November 12, 2008. Problem 3ExampleSketch the graph of g and use LT to find y solution of

y ′′ − 6y = g(t), y(0) = y ′(0) = 0, g(t) =

{0, t < π,

sin(t − π), t > π.

Solution: L[g(t)] =e−πs

s2 + 1.

L[y ′′]− 6L[y ] = L[g(t)] =e−πs

s2 + 1.

(s2 − 6)L[y ] =e−πs

s2 + 1

⇒ L[y ] = e−πs 1

(s2 + 1)(s2 − 6).

H(s) =1

(s2 + 1)(s2 − 6)=

1

(s2 + 1)(s +√

6)(s −√

6)

H(s) =a

(s +√

6)+

b

(s −√

6)+

(cs + d)

(s2 + 1).

Exam: November 12, 2008. Problem 3ExampleSketch the graph of g and use LT to find y solution of

y ′′ − 6y = g(t), y(0) = y ′(0) = 0, g(t) =

{0, t < π,

sin(t − π), t > π.

Solution: L[g(t)] =e−πs

s2 + 1.

L[y ′′]− 6L[y ] = L[g(t)] =e−πs

s2 + 1.

(s2 − 6)L[y ] =e−πs

s2 + 1⇒ L[y ] = e−πs 1

(s2 + 1)(s2 − 6).

H(s) =1

(s2 + 1)(s2 − 6)=

1

(s2 + 1)(s +√

6)(s −√

6)

H(s) =a

(s +√

6)+

b

(s −√

6)+

(cs + d)

(s2 + 1).

Exam: November 12, 2008. Problem 3ExampleSketch the graph of g and use LT to find y solution of

y ′′ − 6y = g(t), y(0) = y ′(0) = 0, g(t) =

{0, t < π,

sin(t − π), t > π.

Solution: L[g(t)] =e−πs

s2 + 1.

L[y ′′]− 6L[y ] = L[g(t)] =e−πs

s2 + 1.

(s2 − 6)L[y ] =e−πs

s2 + 1⇒ L[y ] = e−πs 1

(s2 + 1)(s2 − 6).

H(s) =1

(s2 + 1)(s2 − 6)

=1

(s2 + 1)(s +√

6)(s −√

6)

H(s) =a

(s +√

6)+

b

(s −√

6)+

(cs + d)

(s2 + 1).

Exam: November 12, 2008. Problem 3ExampleSketch the graph of g and use LT to find y solution of

y ′′ − 6y = g(t), y(0) = y ′(0) = 0, g(t) =

{0, t < π,

sin(t − π), t > π.

Solution: L[g(t)] =e−πs

s2 + 1.

L[y ′′]− 6L[y ] = L[g(t)] =e−πs

s2 + 1.

(s2 − 6)L[y ] =e−πs

s2 + 1⇒ L[y ] = e−πs 1

(s2 + 1)(s2 − 6).

H(s) =1

(s2 + 1)(s2 − 6)=

1

(s2 + 1)(s +√

6)(s −√

6)

H(s) =a

(s +√

6)+

b

(s −√

6)+

(cs + d)

(s2 + 1).

Exam: November 12, 2008. Problem 3ExampleSketch the graph of g and use LT to find y solution of

y ′′ − 6y = g(t), y(0) = y ′(0) = 0, g(t) =

{0, t < π,

sin(t − π), t > π.

Solution: L[g(t)] =e−πs

s2 + 1.

L[y ′′]− 6L[y ] = L[g(t)] =e−πs

s2 + 1.

(s2 − 6)L[y ] =e−πs

s2 + 1⇒ L[y ] = e−πs 1

(s2 + 1)(s2 − 6).

H(s) =1

(s2 + 1)(s2 − 6)=

1

(s2 + 1)(s +√

6)(s −√

6)

H(s) =a

(s +√

6)+

b

(s −√

6)+

(cs + d)

(s2 + 1).

Exam: November 12, 2008. Problem 3

ExampleSketch the graph of g and use LT to find y solution of

y ′′ − 6y = g(t), y(0) = y ′(0) = 0, g(t) =

{0, t < π,

sin(t − π), t > π.

Solution: H(s) =a

(s +√

6)+

b

(s −√

6)+

(cs + d)

(s2 + 1).

1

(s2 + 1)(s +√

6)(s −√

6)=

a

(s +√

6)+

b

(s −√

6)+

(cs + d)

(s2 + 1)

1 = a(s −√

6)(s2 + 1) + b(s +√

6)(s2 + 1) + (cs + d)(s2 − 6).

The solution is: a = − 1

14√

6, b =

1

14√

6, c = 0, d = −1

7.

Exam: November 12, 2008. Problem 3

ExampleSketch the graph of g and use LT to find y solution of

y ′′ − 6y = g(t), y(0) = y ′(0) = 0, g(t) =

{0, t < π,

sin(t − π), t > π.

Solution: H(s) =a

(s +√

6)+

b

(s −√

6)+

(cs + d)

(s2 + 1).

1

(s2 + 1)(s +√

6)(s −√

6)=

a

(s +√

6)+

b

(s −√

6)+

(cs + d)

(s2 + 1)

1 = a(s −√

6)(s2 + 1) + b(s +√

6)(s2 + 1) + (cs + d)(s2 − 6).

The solution is: a = − 1

14√

6, b =

1

14√

6, c = 0, d = −1

7.

Exam: November 12, 2008. Problem 3

ExampleSketch the graph of g and use LT to find y solution of

y ′′ − 6y = g(t), y(0) = y ′(0) = 0, g(t) =

{0, t < π,

sin(t − π), t > π.

Solution: H(s) =a

(s +√

6)+

b

(s −√

6)+

(cs + d)

(s2 + 1).

1

(s2 + 1)(s +√

6)(s −√

6)=

a

(s +√

6)+

b

(s −√

6)+

(cs + d)

(s2 + 1)

1 = a(s −√

6)(s2 + 1) + b(s +√

6)(s2 + 1) + (cs + d)(s2 − 6).

The solution is: a = − 1

14√

6, b =

1

14√

6, c = 0, d = −1

7.

Exam: November 12, 2008. Problem 3

ExampleSketch the graph of g and use LT to find y solution of

y ′′ − 6y = g(t), y(0) = y ′(0) = 0, g(t) =

{0, t < π,

sin(t − π), t > π.

Solution: H(s) =a

(s +√

6)+

b

(s −√

6)+

(cs + d)

(s2 + 1).

1

(s2 + 1)(s +√

6)(s −√

6)=

a

(s +√

6)+

b

(s −√

6)+

(cs + d)

(s2 + 1)

1 = a(s −√

6)(s2 + 1) + b(s +√

6)(s2 + 1) + (cs + d)(s2 − 6).

The solution is: a = − 1

14√

6, b =

1

14√

6, c = 0, d = −1

7.

Exam: November 12, 2008. Problem 3ExampleSketch the graph of g and use LT to find y solution of

y ′′ − 6y = g(t), y(0) = y ′(0) = 0, g(t) =

{0, t < π,

sin(t − π), t > π.

Solution: H(s) =1

14√

6

[− 1

(s +√

6)+

1

(s −√

6)− 2

√6

(s2 + 1)

].

H(s) =1

14√

6

[−L

[e−

√6 t

]+ L

[e√

6 t]− 2√

6L[sin(t)]]

H(s) = L[ 1

14√

6

(−e−

√6 t + e

√6 t − 2

√6 sin(t)

)].

h(t) =1

14√

6

[−e−

√6 t + e

√6 t − 2

√6 sin(t)

]⇒ H(s) = L[h(t)].

Exam: November 12, 2008. Problem 3ExampleSketch the graph of g and use LT to find y solution of

y ′′ − 6y = g(t), y(0) = y ′(0) = 0, g(t) =

{0, t < π,

sin(t − π), t > π.

Solution: H(s) =1

14√

6

[− 1

(s +√

6)+

1

(s −√

6)− 2

√6

(s2 + 1)

].

H(s) =1

14√

6

[−L

[e−

√6 t

]+ L

[e√

6 t]− 2√

6L[sin(t)]]

H(s) = L[ 1

14√

6

(−e−

√6 t + e

√6 t − 2

√6 sin(t)

)].

h(t) =1

14√

6

[−e−

√6 t + e

√6 t − 2

√6 sin(t)

]⇒ H(s) = L[h(t)].

Exam: November 12, 2008. Problem 3ExampleSketch the graph of g and use LT to find y solution of

y ′′ − 6y = g(t), y(0) = y ′(0) = 0, g(t) =

{0, t < π,

sin(t − π), t > π.

Solution: H(s) =1

14√

6

[− 1

(s +√

6)+

1

(s −√

6)− 2

√6

(s2 + 1)

].

H(s) =1

14√

6

[−L

[e−

√6 t

]+ L

[e√

6 t]− 2√

6L[sin(t)]]

H(s) = L[ 1

14√

6

(−e−

√6 t + e

√6 t − 2

√6 sin(t)

)].

h(t) =1

14√

6

[−e−

√6 t + e

√6 t − 2

√6 sin(t)

]⇒ H(s) = L[h(t)].

Exam: November 12, 2008. Problem 3ExampleSketch the graph of g and use LT to find y solution of

y ′′ − 6y = g(t), y(0) = y ′(0) = 0, g(t) =

{0, t < π,

sin(t − π), t > π.

Solution: H(s) =1

14√

6

[− 1

(s +√

6)+

1

(s −√

6)− 2

√6

(s2 + 1)

].

H(s) =1

14√

6

[−L

[e−

√6 t

]+ L

[e√

6 t]− 2√

6L[sin(t)]]

H(s) = L[ 1

14√

6

(−e−

√6 t + e

√6 t − 2

√6 sin(t)

)].

h(t) =1

14√

6

[−e−

√6 t + e

√6 t − 2

√6 sin(t)

]

⇒ H(s) = L[h(t)].

Exam: November 12, 2008. Problem 3ExampleSketch the graph of g and use LT to find y solution of

y ′′ − 6y = g(t), y(0) = y ′(0) = 0, g(t) =

{0, t < π,

sin(t − π), t > π.

Solution: H(s) =1

14√

6

[− 1

(s +√

6)+

1

(s −√

6)− 2

√6

(s2 + 1)

].

H(s) =1

14√

6

[−L

[e−

√6 t

]+ L

[e√

6 t]− 2√

6L[sin(t)]]

H(s) = L[ 1

14√

6

(−e−

√6 t + e

√6 t − 2

√6 sin(t)

)].

h(t) =1

14√

6

[−e−

√6 t + e

√6 t − 2

√6 sin(t)

]⇒ H(s) = L[h(t)].

Exam: November 12, 2008. Problem 3

ExampleSketch the graph of g and use LT to find y solution of

y ′′ − 6y = g(t), y(0) = y ′(0) = 0, g(t) =

{0, t < π,

sin(t − π), t > π.

Solution: Recall: L[y ] = e−πs H(s), where H(s) = L[h(t)], and

h(t) =1

14√

6

[−e−

√6 t + e

√6 t − 2

√6 sin(t)

].

L[y ] = e−πs L[h(t)] = L[uπ(t) h(t−π)] ⇒ y(t) = uπ(t) h(t−π).

Equivalently:

y(t) =uπ(t)

14√

6

[−e−

√6 (t−π) + e

√6 (t−π) − 2

√6 sin(t − π)

]. C

Exam: November 12, 2008. Problem 3

ExampleSketch the graph of g and use LT to find y solution of

y ′′ − 6y = g(t), y(0) = y ′(0) = 0, g(t) =

{0, t < π,

sin(t − π), t > π.

Solution: Recall: L[y ] = e−πs H(s), where H(s) = L[h(t)], and

h(t) =1

14√

6

[−e−

√6 t + e

√6 t − 2

√6 sin(t)

].

L[y ] = e−πs L[h(t)]

= L[uπ(t) h(t−π)] ⇒ y(t) = uπ(t) h(t−π).

Equivalently:

y(t) =uπ(t)

14√

6

[−e−

√6 (t−π) + e

√6 (t−π) − 2

√6 sin(t − π)

]. C

Exam: November 12, 2008. Problem 3

ExampleSketch the graph of g and use LT to find y solution of

y ′′ − 6y = g(t), y(0) = y ′(0) = 0, g(t) =

{0, t < π,

sin(t − π), t > π.

Solution: Recall: L[y ] = e−πs H(s), where H(s) = L[h(t)], and

h(t) =1

14√

6

[−e−

√6 t + e

√6 t − 2

√6 sin(t)

].

L[y ] = e−πs L[h(t)] = L[uπ(t) h(t−π)]

⇒ y(t) = uπ(t) h(t−π).

Equivalently:

y(t) =uπ(t)

14√

6

[−e−

√6 (t−π) + e

√6 (t−π) − 2

√6 sin(t − π)

]. C

Exam: November 12, 2008. Problem 3

ExampleSketch the graph of g and use LT to find y solution of

y ′′ − 6y = g(t), y(0) = y ′(0) = 0, g(t) =

{0, t < π,

sin(t − π), t > π.

Solution: Recall: L[y ] = e−πs H(s), where H(s) = L[h(t)], and

h(t) =1

14√

6

[−e−

√6 t + e

√6 t − 2

√6 sin(t)

].

L[y ] = e−πs L[h(t)] = L[uπ(t) h(t−π)] ⇒ y(t) = uπ(t) h(t−π).

Equivalently:

y(t) =uπ(t)

14√

6

[−e−

√6 (t−π) + e

√6 (t−π) − 2

√6 sin(t − π)

]. C

Exam: November 12, 2008. Problem 3

ExampleSketch the graph of g and use LT to find y solution of

y ′′ − 6y = g(t), y(0) = y ′(0) = 0, g(t) =

{0, t < π,

sin(t − π), t > π.

Solution: Recall: L[y ] = e−πs H(s), where H(s) = L[h(t)], and

h(t) =1

14√

6

[−e−

√6 t + e

√6 t − 2

√6 sin(t)

].

L[y ] = e−πs L[h(t)] = L[uπ(t) h(t−π)] ⇒ y(t) = uπ(t) h(t−π).

Equivalently:

y(t) =uπ(t)

14√

6

[−e−

√6 (t−π) + e

√6 (t−π) − 2

√6 sin(t − π)

]. C