5 LAPLACE TRANSFORMS

CHAPTER

5

LAPLACE TRANSFORMS

5.1 Introduction and Definition

In this section we introduce the notion of the Laplace transform. We will usethis idea to solve differential equations, but the method also can be used tosum series or compute integrals. We begin with the definition:

Laplace TransformLet f(t) be a function whose domain includes (0,∞) then the Laplace trans-form of f(t) is:

L(f(t))(s) =

∫

∞

0

f(t)e−st dt

Note that the Laplace transform of a function is another function whosevariable is usually denoted as s. Also note that the Laplace transform involvesan improper integral. We compute the Laplace transform of several functions:

Example 5.1 Compute the Laplace transform of f(t) = 1

127

128 CHAPTER 5. LAPLACE TRANSFORMS

Solution:

L(1)(s) =

∫

∞

0

1e−st dt = limc→∞

−1

se−st

∣

∣

∣

∣

c

0

In order for this limit to exist, we must insist that s 6= 0 and that s > 0 sothat e−sc has a limit (of zero). When s > 0, we obtain

−1

slimc→∞

(e−sc − 1) =1

s

So

L(1)(s) =1

s; s > 0.

¤

Example 5.2 Compute the Laplace transform of f(t) = t

Solution:

L(t)(s) =

∫

∞

0

te−st dt

We integrate by Parts (letting u = t and dv = e−st dt) to obtain:∫

te−st dt = −1

ste−st −

1

s2e−st,

so∫

∞

0

te−st dt = limc→∞

(

−1

ste−st −

1

s2e−st

)∣

∣

∣

∣

c

0

In order for this limit to exist, we again must insist that s 6= 0 and that s > 0so that e−sc has a limit (of zero). We obtain

−1

slimc→∞

(ce−sc − 0) −1

s2limc→∞

(e−sc − 1)

which exists for s > 0 and after L’Hopital’s rule yields

L(t)(s) =1

s2; s > 0.

¤

The previous example can be upgraded to find the Laplace transform off(t) = tn for any positive integer n.

5.1. INTRODUCTION AND DEFINITION 129

Example 5.3 Show that if f(t) = tn, for any positive integer n then

L(tn)(s) =n

sL(tn−1)(s)

Solution:

L(tn)(s) =

∫

∞

0

tne−st dt

We integrate by Parts (letting u = tn and dv = e−st dt) to obtain:

∫

tne−st dt = −1

stne−st +

1

s

∫

ntn−1e−st dt,

so∫

∞

0

tne−st dt = limc→∞

(

−1

stne−st

)∣

∣

∣

∣

c

0

+n

sL(tn−1)(s).

In order for this limit to exist, we again must insist that s 6= 0 and that s > 0so that e−sc has a limit (of zero). Using L’Hopital’s rule n times, we see that

limc→∞

(

−1

scne−sc

)

= 0.

¤

Using this result inductively, we compute:

L(t2)(s) =2

sL(t)(s) =

2

s·

1

s2=

2

s3

L(t3)(s) =3

sL(t2)(s) =

3

s·2

s·

1

s2=

6

s4

L(t4)(s) =4

sL(t3)(s) =

4

s·3

s·2

s·

1

s2=

24

s5

Note that in all cases above, we must have s > 0. In summary,

Laplace Transform of single term monic polynomials

L(tn)(s) =n!

sn+1; s > 0


Linear Combinations and Laplace TransformTheorem 5.4 Let f(t) and g(t) be functions with Laplace transforms F (s)and G(s) respectively then for any constants a and b

L(af(t) + bg(t))(s) = aF (s) + bG(s)

Proof:

L(af(t) + bg(t))(s) =

∫

∞

0

[af(t) + bg(t)]e−st dt

=

∫

∞

0

af(t)e−st dt +

∫

∞

0

bg(t)e−st dt = aF (s) + bG(s)

¤

From this result we can take the Laplace transform of any arbitrary poly-nomial in the next example,

Example 5.5 Find L(4t3 + 8t2 − 7)(s)

Solution:

L(4t3 + 8t2 − 7)(s) = 4L(t3)(s) + 8L(t2)(s) − 7L(1)(s)

= 4

(

6

s4

)

+ 8

(

2

s3

)

− 71

s

=24

s4+

16

s3−

7

s,

where s > 0. ¤

Laplace Transform of exponential functions

L(eat)(s) =1

s − a; s > a


Proof:

L(eat)(s) =

∫

∞

0

eate−st dt

=

∫

∞

0

e(a−s)t dt,

which can be integrated with respect to t (use u = (a − s)t and du = (a −s) dt). We obtain

= limb→∞

1

a − se(a−s)b −

1

a − s

The above limit exists exactly when s > a, so

L(eat)(s) =1

s − a, s > a

¤

Example 5.6 Find

L(2t)(s)

Solution: Rewriting 2t = eln2t

= etln2 and taking a = ln 2,

L(e(ln2)t)(s) =1

s − ln 2, s > ln 2

¤

Lastly,

Laplace Transform of sine and cosine

L(cos(βt))(s) =s

s2 + β2, s > 0

L(sin(βt))(s) =β

s2 + β2, s > 0


We derive the second formula and leave the derivation of the first formulaas an exercise. By definition:

L(sin(βt))(s) =

∫

∞

0

e−st sin(βt) dt.

Using integration by parts with u = sin(βt) and dv = e−st dt we obtain∫

∞

0

e−st sin(βt) dt = −1

ssin(βt)e−st

∣

∣

∣

∣

∞

0

−

∫

∞

0

(−β

s)e−st cos(βt) dt.

Using parts again with u = cos(βt) and dv = e−st dt, we obtain

= −1

ssin(βt)e−st

∣

∣

∣

∣

∞

0

+ (β

s)

[

(−1

s)e−st cos(βt)

∣

∣

∣

∣

∞

0

−

∫

∞

0

(β

s)e−st sin(βt) dt

]

From the squeeze theorem we see that for s > 0 limc→∞

e−sc cos(βc) = 0 and

limc→∞

e−sc sin(βc) = 0, so the above reduces to

= (β

s2) −

(

β2

s2

)∫

∞

0

(e−st sin(βt) dt

Now recall that the left-hand side of this equation (what we were solvingfor) is

∫

∞

0

e−st sin(βt) dt

so we have∫

∞

0

e−st sin(βt) dt =

(

β

s2

)

−

(

β2

s2

)∫

∞

0

e−st sin(βt) dt

So moving the term with the integral on the right side to the left sidegives

∫

∞

0

e−st sin(βt) dt +

(

β2

s2

)∫

∞

0

e−st sin(βt) dt = (β

s2)

Factoring,(

1 +β2

s2

)∫

∞

0

e−st sin(βt) dt = (β

s2)

So∫

∞

0

e−st sin(βt) dt =( β

s2 )(

1 + β2

s2

) =β

s2 + β2


NOTE: It is customary to use the independent variable s for a functionthat is an output of a Laplace transform and the independent variable t fora function that is an output of a Laplace transform. This convention will behandy in the later sections.

The below result gives a condition that guarantees the existence of theLaplace transform.

A condition that guarantees the existence of L[f(t)](s)Suppose that there are numbers α and M so that

|f(t)| ≤ Meαt

for all t > 0. Then the Laplace transform of f(t) exists with a domain of atleast s > α.

Proof: Suppose that there are numbers α and M so that

|f(t)| ≤ Meαt

for all t > 0. Then

L(|f(t)|)(s) = limb→∞

∫ b

0

e−st|f(t)| dt

≤

∫

∞

0

e−stMeαt dt = ML[eαt](s) =M

s − α,

for s > α.Therefore, since

∫ b

0

e−st|f(t)| dt

is increasing in b and bounded for all b,

limb→∞

∫ b

0

e−st|f(t)| dt

exists. This implies that

limb→∞

∫ b

0

e−stf(t) dt


also must exist. ¤

Note that most all exponential functions, polynomials, and the trig func-tions sine and cosine satisfy this condition but ln x, tan x and et2 do not.These functions do not have Laplace transforms.

Exercises

1. Compute L[f(t)](s) for f(t) = 0.

2. Compute L[f(t)](s) for

f(t) =

{

1 if t < 80 if t ≥ 8.

3. Compute L[f(t)](s) for

f(t) =

t if t < 12 − t if 1 ≤ t < 20 if t ≥ 2.

4. Compute the formula for L[cosh(βt)](s), where β is a constant.

5. Compute L[sinh(βt)](s),

6. Find the Laplace transform of

f(t) =

t − 8 if t < 8et+6 if 8 < t < 10t2 if 10 < t < 110 if t > 11.

In 7-12, take the Laplace transform of the function f(t) using the resultsin this section (do not derive using the definition)

7. f(t) = t3 − 7t2 + 8

8. f(t) = 4 sin t − 3 cos t

9. f(t) = cos(2t) − e9t

10. f(t) = 1 + 7 sin(5t)


11. f(t) = cosh(βt), where β is a constant. (Recall cosh(z) = ez+e−z

2).

12. f(t) = sinh(βt), where β is a constant. (Recall sinh(z) = ez−e−z

2).

In 13-15, use the definition to take the Laplace transform of the functionf(t) using integration by parts and formulas given in this section

13. f(t) = tet

14. f(t) = t sin(2t)

15. f(t) = t2 cos(t)

16. Recall that L[f(t) + g(t)] = L[f(t)] + L[g(t)]. Is it true in general thatL[f(t) · g(t)] = L[f(t)] · L[g(t)]?

(a) Try f(t) = t and g(t) = 1 to see if L[f(t) · g(t)] = L[f(t)] · L[g(t)].

(b) Is it ever true that L[f(t) · g(t)] = L[f(t)] · L[g(t)]?


5.2 Laplace Transforms, The Inverse Laplace

Transform, and ODEs

In this section we will see how the Laplace transform can be used tosolve differential equations. The key result that allows us to do this isthe following:

Laplace Transform of y′(t)Suppose that L[y(t)](s) exists and that y(t) is differentiable (0,∞) withderivative y′(t) then

L[y′(t)](s) = sL[y(t)](s) − y(0)

Proof: By definition,

L[y′(t)](s) =

∫

∞

0

e−sty′(t) dt

Setting u = e−st and dv = y′(t) dt and using integration by parts, weobtain

L[y′(t)](s) = e−sty(t)∣

∣

∞

0−

∫

∞

0

(−s)e−sty(t) dt

For s big enough, limc→∞ e−scy(c) = 0, since the Laplace transform ofy(t) exists (this follows since the continuous integrand of a convergentimproper integral must tend to zero – we will not prove this fact here.)

So we obtainL[y′(t)](s) = −y(0) + sL[y(t)](s)

¤

From this result, we derive:

Laplace Transform of y′′(t)Suppose that L[y(t)](s) exists and that y(t) is twice differentiable on (0,∞)with second derivative y′′(t) then

L[y′′(t)](s) = s2L[y(t)](s) − sy(0) − y′(0)

5.2. LAPLACE TRANSFORMS, THE INVERSE LAPLACE TRANSFORM, AND ODES137

Proof: We simply us the previous result twice

L[y′′(t)](s) = sL[y′(t)](s) − y′(0) = s [sL[y(t)](s) − y(0)] − y′(0).

¤

Example 5.7 Solve y′′(t) = t using the Laplace Transform

Solution: Taking the Laplace transform of both sides we obtain

L[y′′(t)](s) = L[t]

L[y′′(t)](s) =1

s2

so

s2L[y(t)](s) − sy(0) − y′(0) =1

s2

Solving for L[y(t)](s) we obtain

L(y(t))(s) =1

s4+

y(0)

s+

y′(0)

s2

Now if we could reverse engineer which function y(t) has Laplace trans-form equal to the right hand side, then we would be done.

By playing a bit, we can see that y(t) = 16t3 + y(0)

2+ y′(0)t has this

Laplace transform. So voila! We have solved the differential equation.

(Note that the general solution obtained from undetermined coefficientswould be y(t) = 1

6t3 + C1 + C2t which agrees with our solution since

y(0) and y′(0) are unspecified constants.) ¤

The above example illustrates a common checklist for solving DEs usingthe Laplace transform:

1. Transform the original problem to one involving L(y(t))(s),


2. Solve for L[y(t)](s),

3. Undo the Laplace transform to recover y(t).

Definition 5.8 A continuous function f(t) is the Inverse Laplace

Transform of a function F (s) if L[f(t)](s) = F (s). In this case, wewrite f(t) = L−1[F (s)]

It is customary shorthand notation to denote F (s) to be the Laplacetransform of f(t) and G(s) to be the Laplace transform of g(t). Forconstants a, b, since

L[af(t) + bg(t)] = aL[f(t)] + bL[g(t] = aF (s) + bG(s)

we see thatL−1[aF (s) + bG(s)] = af(t) + bg(t)

orL−1[aF (s) + bG(s)] = aL−1[F (s)] + bL−1[G(s)],

so the Laplace transform of a linear combination is the linear combi-nation of the Laplace transforms.

The following table lists several inverses which come from the previoussection:

Some Basic Inverse Laplace Transform Facts

F (s) L−1[F (s)]1s

1n!

sn+1 tn

1s−a

eat

ss2+β2 cos βt

β

s2+β2 sin βt

aF (s) + bG(s) af(t) + bg(t)

Example 5.9 Find L−1[

7s3

]


Solution:

L−1

[

7

s3

]

= 7L−1

[

1

s3

]

We multiply the inner fraction by 22

to obtain the form in the previoustable.

= 7L−1

[

2

2s3

]

=7

2L−1

[

2

s3

]

=7

2t2

¤


2s−1s2+4

]

Solution:

L−1

[

2s − 1

s2 + 4

]

= 2L−1

[

s

s2 + 4

]

− L−1

[

1

s2 + 4

]

Again, the inner fraction of the second must be multiplied by 22

to putit in the form in the table (β = 2 since β2 = 4).

= 2 cos 2t − L−1

[

2

2(s2 + 4)

]

= 2 cos 2t −1

2L−1

[

2

s2 + 4

]

= 2 cos 2t −1

2sin 2t

¤

Example 5.11 Solve y′′(t) + 9y(t) = e4t with y(0) = 1 and y′(0) = 0


Solution: We first take the Laplace transform of both sides to obtain

L[y′′(t) + 9y(t)] = L[e4t]

s2L[y(t)] − sy(0) − y′(0) + 9L[y(t)] =1

s − 4

(s2 + 9)L[y(t)] =1

s − 4+ s

L[y(t)] =1

(s − 4)(s2 + 9)+

s

s2 + 9

We use partial fractions to decompose the first term into a sum of termsthat are recognizable inverses.

Noting that by partial fractions:

1

(s − 4)(s2 + 9)=

A

s − 4+

Bs + C

s2 + 9,

where A = 125

B = − 125

C = − 425

. Substituting in, we obtain

L[y(t)] =1

25

(

1

s − 4

)

−1

25

(

s

s2 + 9

)

−4

25

(

1

s2 + 9

)

+s

s2 + 9

L[y(t)] =1

25

(

1

s − 4

)

+24

25

(

s

s2 + 9

)

−4

25

(

1

s2 + 9

)

So

y(t) =1

25e4t +

24

25cos 3t −

4

75sin 3t

(note that the ’3’ appeared in the demoninator since β = 3 and in order

to find L−1

[

1

s2 + 9

]

we multiply the inner fraction by 33.) ¤

We end this section by noting the following extension


Laplace Transform of y(n)(t)Suppose that L[y(t)](s) exists and that y(t) is differentiable n times on (0,∞)with nth derivative y(n)(t) then

L[y(n)(t)](s) = snL[y(t)](s)−sn−1y(0)−sn−2y′(0)− ...−sy(n−2)(0)−y(n−1)(0)

Exercises

In 1-7, find the Inverse Laplace transform of the function F (s).You may need partial fractions

1. F (s) = 1s+2

2. F (s) = 1s4

3. F (s) = s+5s2+16

4. F (s) = 4s+6

5. F (s) = 1(s+1)(s2+1)

6. F (s) = s(s+1)(s2+1)

7. F (s) = 3s3+s

8. Use the Laplace transform to find the general solution of

y′′ + 6y′ + 5y = t


y′′ − y = et


y′′ + y = et

11. Use the Laplace transform to find the solution of the IVP

y′′ + y = 2, y(0) = 1, y′(0) = −1

12. Use the Laplace transform to find the solution of the IVP

y′′ + 2y′ + y = 2, y(2) = 1, y′(2) = −1

(Hint: find the general solution first).


5.3 Advanced Properties of the Laplace

Transform

In this section, we look at several theorems which can be used to solveDEs using the Laplace transform method. We start with:

Laplace Transform Exponential Shift Theorem (Forward)

L[eatf(t)](s) = L[f(t)](s − a) = F (s − a)

Effectively, this says L[eatf(t)](s) is equal to L[f(t)](w), (using w for thetraditional s in the definition) and replacing w = s − a.

Proof: By definition:

L[eatf(t)] =

∫

∞

0

eate−stf(t) dt

=

∫

∞

0

e−(s−a)tf(t) dt.

Also by definition,

L[f(t)](s − a) =

∫

∞

0

e−(s−a)tf(t) dt.

¤

We can use this result to compute Laplace transforms of exponentialsmultiplied by functions whose Laplace transforms we already know.

Example 5.12 Find L[e−2t cos(3t)](s)

Solution: By the shifting theorem

L[e−2t cos(3t)](s) = L[cos(3t)](s − (−2)) = L[cos(3t)](s + 2) =

5.3. ADVANCED PROPERTIES OF THE LAPLACE TRANSFORM 143

The Laplace transform of cos(3t) with (different) variable w is

L[cos(3t)](w) =w

w2 + 32,

so, replacing w with s + 2,

= L[cos(3t)](s + 2) =s + 2

(s + 2)2 + 9.

=s + 2

s2 + 4s + 13.

Further note that in order for L[cos(3t)](w) to exist, w > 0 so thedomain is s + 2 > 0 or s > −2. ¤

Example 5.13 Find L[t3et](s)

Solution: By the shifting theorem

L[t3et](s) = L[t3](w) =6

w4,

where w = s − 1, so

L[t3et](s) =6

(s − 1)4

¤

Similarly, we obtain the following result:

Laplace Transform Exponential Shift Theorem (Backward)

L−1[F (s − a)]) = eatf(t),

where F (w) is the Laplace transform of f(t) (with variable w).


To apply this result, we look for familiar outputs of Laplace transformswith the variable s replaced by a shift.


24(s−8)5

]

Solution: This is just a shift of

24

w5

which is the Laplace transform of t4 (with variable w0. So 24(s−8)5

=

F (s − 8) where F is the Laplace transform of f(t) = t4. Therefore, bythe Exponential Shifting Theorem

L−1[F (s − 8)] = e8tt4.

¤


ss2+2s+5

]

Solution: The trick here is to complete the square in the denominatorand then use the backward shift theorem.

s

s2 + 2s + 5=

s

s2 + 2s + 1 + 4=

s

(s + 1)2 + 4

We also need an s+1 in the numerator to use the backwards shift withcosine, so we write the numerator as:

=s + 1 − 1

(s + 1)2 + 4=

s + 1

(s + 1)2 + 4−

1

(s + 1)2 + 4

Now we can take the inverse of each term using the backward shifttheorem to get

L−1

[

s

s2 + 2s + 5

]

= L−1

[

s + 1

(s + 1)2 + 4

]

− L−1

[

1

(s + 1)2 + 4

]


= e−t cos 2t −1

2e−t sin 2t

¤

The next result equates differentiation of the Laplace transform F withrespect to s to the Laplace transform of the product tf(t).

Differentiation TheoremLet F (s) be the Laplace transform of f(t). Then

d

ds[F (s)] = −L[tf(t)](s)

or

−d

ds[F (s)] = L[tf(t)](s)

Proof: By definition:

F ′(s) = lim∆s→0

F (s + ∆s) − F (s)

∆s

Note that

F (s+∆s)−F (s) =

∫

∞

0

e−(s+∆s)tf(t) dt−

∫

∞

0

e−stf(t) dt =

∫

∞

0

[e−(s+∆s)t−e−st]f(t) dt

SoF (s + ∆s) − F (s)

∆s=

∫

∞

0

e−(s+∆s)t − e−st

∆sf(t) dt

Therefore

lim∆s→0

F (s + ∆s) − F (s)

∆s=

∫

∞

0

lim∆s→0

e−(s+δs)t − e−st

∆sf(t) dt


Noting that lim∆s→0e−(s+∆s)t

−e−st

∆sis the definition of d

ds(e−st), we know

this limit is −te−st.

Substituting, we obtain

F ′(s) =

∫

∞

0

(−te−st)f(t) dt.

By definition, the right hand side is L[−tf(t)](s) ¤

We provide several examples which illustrate how to use this result.

Example 5.16 Find L [te2t]

By the Theorem,L[tf(t)](s) = −F ′(s).

Here f(t) = e2t, so F (s) = 1s−2

.

NowF ′(s) = −(s − 2)−2

so

L[te2t](s) = −(−(s − 2)−2) =1

(s − 2)2

¤

Note that we could have also obtained this answer by using the ForwardExponential Shift Theorem.

Example 5.17 Find L [t cos(3t)]

By the Theorem,L[tf(t)](s) = −F ′(s).

Here f(t) = cos(3t), so F (s) = ss2+9

.

Now

F ′(s) =(s2 + 9) − s(2s)

(s2 + 9)2=

9 − s2

(s2 + 9)2


so

L[t cos(3t)](s) = −9 − s2

(s2 + 9)2=

s2 − 9

(s2 + 9)2

¤

Next we solve several differential equations using these new results.

Example 5.18 Find the general solution to y′′ + 6y′ + 6y = sin 2t

Taking the Laplace transform, we obtain

L[y′′ + 6y′ + 10y] = L[sin 2t]

so

s2L − sy(0) − y′(0) + 6sL − 6y(0) + 10L =2

s2 + 4

(here L is shorthand for L[y(t)](s)), so

(s2 + 6s + 10)L = sy(0) + y′(0) + 6y(0) +2

s2 + 4

Solving for L,

L =sy(0) + y′(0) + 6y(0) + 2

s2+4

s2 + 6s + 10

=sy(0)

s2 + 6s + 6+

y′(0) + 6y(0)

s2 + 6s + 10+

2

(s2 + 4)(s2 + 6s + 10)(5.1)

We now have to use the inverse Laplace transform. We start with thefirst term, which we will have to write as a shift by completing thesquare of the denominator. In particular the first term is

sy(0)

(s + 3)2 + 1

which we can rewrite as

sy(0)

(s + 3)2 + 1=

(s + 3 − 3)y(0)

(s + 3)2 + 1=

(s + 3)y(0)

(s + 3)2 + 1−

3y(0)

(s + 3)2 + 1.


We can invert this using the shift theorem, so the inverse of the firstterm is:

y(0)e−3t cos t − 3y(0)e−3t sin t

Similarly, the second term in ?? can be realized as a shift, It is

y′(0) + 6y(0)

(s + 3)2 + 1

which has inverse transform

(y′(0) + 6y(0))e−3t sin t.

Lastly, we use partial fractions on the final term to get

2

(s2 + 4)(s2 + 6s + 10)=

As + B

s2 + 4+

Cs + D

s2 + 6s + 10

where A = − 115

, B = 115

, C = 115

, D = 13. (The reader should be familiar

with the method of Partial Fractions, normally covered in a secondsemester calculus class).

We again can easily invert the first term

As + B

s2 + 4= A

s

s2 + 4+

B

2

2

s2 + 4

which inverts to

A cos 2t +B

2sin 2t.

The second term, after again setting up for the shifting theorem bycompleting the square, becomes

Cs + D

s2 + 6s + 10=

C(s + 3 − 3)

(s + 3)2 + 1+

D

(s + 3)2 + 1

= C(s + 3)

(s + 3)2 + 1+ (D − 3C)

1

(s + 3)2 + 1

Taking the inverse, we obtain:


= Ce−3t cos t + (D − 3C)e−3t sin t

Putting it altogether,

y(t) = y(0)e−3t cos t − 3y(0)e−3t sin t + (y′(0) + 6y(0))e−3t sin t

+Ce−3t cos t + (D − 3C)e−3t sin t

−1

15cos 2t +

1

30sin 2t,

where C = 115

, D = 13.

Note that this simplifies to

y(t) = (y(0) + C)e−3t cos t + ((D − 3C) + y′(0) + 3y(0))e−3t sin t

−1

15cos 2t +

1

30sin 2t.

If we had used the method of undetermined coefficients, we would haveobtained (possibly in a lot less time)

y(t) = C1e−3t cos t + C2e

−3t sin t −1

15cos 2t +

1

30sin 2t.

¤

The previous example could (and probably should) have been solvedusing other methods. The next example is one where these other meth-ods would not apply.

Example 5.19 Find the solution to

y′′ + ty′ − 2y = 2, y(0) = 0, y′(0) = 0


We take the Laplace transform of both sides of the differential equation.

L[y′′] + L[ty′] − 2L[y] =2

s,

s2L[y] −d

ds(L[y′]) − 2L[y] =

2

s,

s2L[y] −d

ds(sL[y]) − 2L[y] =

2

s,

Using the product rule, note dds

(sL[y]) = sL′[y] + L[y] (recall: L[y] isalso a function of s). So,

s2L[y] − sL′[y] − L[y] − 2L[y] =2

s,

or

(s2 − 3)L[y] − sL′[y] =2

s,

This is a first order linear differential equation in L with variable s!

L′[y] +

(

−s +3

s

)

L[y] = −2

s2,

This has integrating factor e−s2

2+3 ln s which simplifies to s3e−

s2

2 , so

L[y] =

∫

s3e−s2

2 (− 2s2 ) ds + C

s3e−s2

2

This resolves to

2e−s2

2 + C

s3e−s2

2

We take C = 0 and obtain L[y] = 2s3 (all other choices do not give valid

outputs of Laplace transforms).


Which implies that y(t) = t2 solves the DE. (One may easily checkthat, indeed y(t) = t2 does solve the DE/IVP. ¤

Exercises

In 1-8, solve the ODE/IVP using Laplace Transform

1. y′′ + 4y′ + 3y = 0, y(0) = 1, y′(0) = 0

2. y′′ + 4y′ + 3y = t2, y(0) = 1, y′(0) = 0

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

4. y′′ − 3y′ + 2y = et, y(0) = 1, y′(0) = 0

5. y′′ − 2y = t2, y(0) = 1, y′(0) = 0

6. y′′ − 4y = e2t, y(0) = 0, y′(0) = −1

7. y′′ + 3ty′ − y = 6t, y(0) = 0, y′(0) = 0

8. y′′+ty′−3y = −2t, y(0) = 0, y′(0) = 1 (You will need integrationby parts or use technology)

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5 LAPLACE TRANSFORMS