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Second Order Nonhomogeneous Linear

Differential Equations with Constant

Coefficients:

the method of undetermined coefficients

Xu-Yan Chen

Second Order Nonhomogeneous Linear DifferentialEquations with Constant Coefficients:

a2y(t) + a1y

(t) + a0y(t) = f(t),

where a2 6= 0, a1, a0 are constants, and f(t) is a given function(called the nonhomogeneous term).

General solution structure:

y(t) = yp(t) + yc(t)

where yp(t) is a particular solution of the nonhomog equation,and yc(t) are solutions of the homogeneous equation:

a2y

c (t) + a1y

c(t) + a0yc(t) = 0.

The characteristic roots: a22 + a1 + a0 = 0

The complementary solutions yc(t).

Second Order Nonhomogeneous Linear DifferentialEquations with Constant Coefficients:

a2y(t) + a1y

(t) + a0y(t) = f(t),

where a2 6= 0, a1, a0 are constants, and f(t) is a given function(called the nonhomogeneous term).

General solution structure:

y(t) = yp(t) + yc(t)

where yp(t) is a particular solution of the nonhomog equation,and yc(t) are solutions of the homogeneous equation:

a2y

c (t) + a1y

c(t) + a0yc(t) = 0.

The characteristic roots: a22 + a1 + a0 = 0

The complementary solutions yc(t).

Second Order Nonhomogeneous Linear DifferentialEquations with Constant Coefficients:

a2y(t) + a1y

(t) + a0y(t) = f(t),

where a2 6= 0, a1, a0 are constants, and f(t) is a given function(called the nonhomogeneous term).

General solution structure:

y(t) = yp(t) + yc(t)

where yp(t) is a particular solution of the nonhomog equation,and yc(t) are solutions of the homogeneous equation:

a2y

c (t) + a1y

c(t) + a0yc(t) = 0.

The characteristic roots: a22 + a1 + a0 = 0

The complementary solutions yc(t).

What is this note about? The Method ofUndetermined Coefficients:

Second Order Nonhomogeneous Linear DifferentialEquations with Constant Coefficients:

a2y(t) + a1y

(t) + a0y(t) = f(t),

where a2 6= 0, a1, a0 are constants, and f(t) is a given function(called the nonhomogeneous term).

General solution structure:

y(t) = yp(t) + yc(t)

a2y

c (t) + a1y

c(t) + a0yc(t) = 0.

The characteristic roots: a22 + a1 + a0 = 0

The complementary solutions yc(t).

What is this note about? The Method ofUndetermined Coefficients: a method of finding yp(t), whenthe nonhomog term f(t) belongs a simple class.

Second Order Nonhomogeneous Linear DifferentialEquations with Constant Coefficients:

a2y(t) + a1y

(t) + a0y(t) = f(t),

where a2 6= 0, a1, a0 are constants, and f(t) is a given function(called the nonhomogeneous term).

General solution structure:

y(t) = yp(t) + yc(t)

a2y

c (t) + a1y

c(t) + a0yc(t) = 0.

The characteristic roots: a22 + a1 + a0 = 0

The complementary solutions yc(t).

What is this note about? The Method ofUndetermined Coefficients: a method of finding yp(t), whenthe nonhomog term f(t) belongs a simple class.

Main Idea: Set up a trial function yp(t), by copying thefunction form of f(t).

Example 1: Solve 3y + y 2y = 10e4t, y(0) = 1, y(0) = 3.

Example 1: Solve 3y + y 2y = 10e4t, y(0) = 1, y(0) = 3.

General solutions y(t) = yc(t) + yp(t).

Example 1: Solve 3y + y 2y = 10e4t, y(0) = 1, y(0) = 3.

General solutions y(t) = yc(t) + yp(t).

Find complementary solutions yc(t):

Example 1: Solve 3y + y 2y = 10e4t, y(0) = 1, y(0) = 3.

General solutions y(t) = yc(t) + yp(t).

Find complementary solutions yc(t):

3yc + y

c 2yc = 0 (the corresponding homog eq)

32 + 2 = 0 1 = 1, 2 = 2/3 yc = C1et + C2e

2

3t

Example 1: Solve 3y + y 2y = 10e4t, y(0) = 1, y(0) = 3.

General solutions y(t) = yc(t) + yp(t).

Find complementary solutions yc(t):

3yc + y

c 2yc = 0 (the corresponding homog eq)

32 + 2 = 0 1 = 1, 2 = 2/3 yc = C1et + C2e

2

3t

To find yp(t), set the trial function

yp(t) = ae4t (form copied from f(t) = 10e4t)

where a is the undetermined coefficient.

Example 1: Solve 3y + y 2y = 10e4t, y(0) = 1, y(0) = 3.

General solutions y(t) = yc(t) + yp(t).

Find complementary solutions yc(t):

3yc + y

c 2yc = 0 (the corresponding homog eq)

32 + 2 = 0 1 = 1, 2 = 2/3 yc = C1et + C2e

2

3t

To find yp(t), set the trial function

yp(t) = ae4t (form copied from f(t) = 10e4t)

where a is the undetermined coefficient.

Substitute yp(t) in the nonhomog eq:

3(ae4t) + (ae4t) 2ae4t = 10e4t

= 3(16ae4t) + (4ae4t) 2ae4t

= 50ae4t

Example 1: Solve 3y + y 2y = 10e4t, y(0) = 1, y(0) = 3.

General solutions y(t) = yc(t) + yp(t).

Find complementary solutions yc(t):

3yc + y

c 2yc = 0 (the corresponding homog eq)

32 + 2 = 0 1 = 1, 2 = 2/3 yc = C1et + C2e

2

3t

To find yp(t), set the trial function

yp(t) = ae4t (form copied from f(t) = 10e4t)

where a is the undetermined coefficient.

Substitute yp(t) in the nonhomog eq:

3(ae4t) + (ae4t) 2ae4t = 10e4t

= 3(16ae4t) + (4ae4t) 2ae4t

= 50ae4t

Compare the coefficients of the two sides:

50a = 10 a =1

5

Example 1: Solve 3y + y 2y = 10e4t, y(0) = 1, y(0) = 3.

General solutions y(t) = yc(t) + yp(t).

Find complementary solutions yc(t):

3yc + y

c 2yc = 0 (the corresponding homog eq)

32 + 2 = 0 1 = 1, 2 = 2/3 yc = C1et + C2e

2

3t

To find yp(t), set the trial function

yp(t) = ae4t (form copied from f(t) = 10e4t)

where a is the undetermined coefficient.

Substitute yp(t) in the nonhomog eq:

3(ae4t) + (ae4t) 2ae4t = 10e4t

= 3(16ae4t) + (4ae4t) 2ae4t

= 50ae4t

Compare the coefficients of the two sides:

50a = 10 a =1

5 yp(t) =

1

5e4t

Example 1 (continued): Solve3y + y 2y = 10e4t, y(0) = 1, y(0) = 3.

Combine yc and yp to get

Gen Sols of Nonhomg Eq: y(t) =1

5e4t + C1e

t + C2e2

3t.

Example 1 (continued): Solve3y + y 2y = 10e4t, y(0) = 1, y(0) = 3.

Combine yc and yp to get

Gen Sols of Nonhomg Eq: y(t) =1

5e4t + C1e

t + C2e2

3t.

Use initial conditions:

y(0) = 1 15

+ C1 + C2 = 1

y(t) = 45e4t C1e

t + 23C2e

2

3t, y(0) = 3 4

5 C1 +

2

3C2 = 3

Example 1 (continued): Solve3y + y 2y = 10e4t, y(0) = 1, y(0) = 3.

Combine yc and yp to get

Gen Sols of Nonhomg Eq: y(t) =1

5e4t + C1e

t + C2e2

3t.

Use initial conditions:

y(0) = 1 15

+ C1 + C2 = 1

y(t) = 45e4t C1e

t + 23C2e

2

3t, y(0) = 3 4

5 C1 +

2

3C2 = 3

Solve this:

{

C1 = 9

5

C2 =3

5

Example 1 (continued): Solve3y + y 2y = 10e4t, y(0) = 1, y(0) = 3.

Combine yc and yp to get

Gen Sols of Nonhomg Eq: y(t) =1

5e4t + C1e

t + C2e2

3t.

Use initial conditions:

y(0) = 1 15

+ C1 + C2 = 1

y(t) = 45e4t C1e

t + 23C2e

2

3t, y(0) = 3 4

5 C1 +

2

3C2 = 3

Solve this:

{

C1 = 9

5

C2 =3

5

The solution of the initial value problem:

y(t) =1

5e4t

9

5et +

3

5e

2

3t.

Nonhomogeneous Linear Equations:

a2y(t) + a1y

(t) + a0y(t) = f(t),

Towards the Rules of Setting Up the Trial Function:

f(t) yp(t)

kert Aert

(to be continued)

Example 2: Solve 3y + y 2y = 8te2t.

Example 2: Solve 3y + y 2y = 8te2t.

Complementary solutions yc(t):

32 + 2 = 0 1 = 1, 2 = 2/3 yc = C1et + C2e

2

3t

Example 2: Solve 3y + y 2y = 8te2t.

Complementary solutions yc(t):

32 + 2 = 0 1 = 1, 2 = 2/3 yc = C1et + C2e

2

3t

To find yp(t), set the trial function

yp(t) = Ate2t.

Example 2: Solve 3y + y 2y = 8te2t.

Complementary solutions yc(t):

32 + 2 = 0 1 = 1, 2 = 2/3 yc = C1et + C2e

2

3t

To find yp(t), set the trial function

yp(t) = Ate2t.

Substitute yp(t) in the nonhomog eq:

8te2t = 3(Ate2t) + (Ate2t) 2Ate2t

= 3(4Ae2t + 4Ate2t) + (Ae2t 2Ate2t) 2Ate2t

= 11Ae2t + 8Ate2t

Example 2: Solve 3y + y 2y = 8te2t.

Complementary solutions yc(t):

32 + 2 = 0 1 = 1, 2 = 2/3 yc = C1et + C2e

2

3t

To find yp(t), set the trial function

yp(t) = Ate2t.

Substitute yp(t) in the nonhomog eq:

8te2t = 3(Ate2t) + (Ate2t) 2Ate2t

= 3(4Ae2t + 4Ate2t) + (Ae2t 2Ate2t) 2Ate2t

= 11Ae2t + 8Ate2t

Compare the coefficients of the two sides:{

11A = 08A = 8

Impossible!

Example 2: Solve 3y + y 2y = 8te2t.

Complementary solutions yc(t):

32 + 2 = 0 1 = 1, 2 = 2/3 yc = C1et + C2e

2

3t

To find yp(t), set the trial function

yp(t) = Ate2t.

Substitute yp(t) in the nonhomog eq:

8te2t = 3(Ate2t) + (Ate2t) 2Ate2t

= 3(4Ae2t + 4Ate2t) + (Ae2t 2Ate2t) 2Ate2t

= 11Ae2t + 8Ate2t

Compare the coefficients of the