Quiz 6 Solutions

1. Use the definition of the Laplace Transform to compute L(f(t)), where f(t) is given as:

f(t) =

{t2 if 0 ≤ t < 1

2 + t if t > 1

Be sure to include the details about the convergence of the improper integral (L’Hospital’srule might come in handy).

L(f(t)) =∫ ∞0

e−stf(t) dt =∫ 1

0t2e−st dt +

∫ ∞1

(2 + t)e−st dt

To integrate each of these, we need to integrate by parts:

+ t2 e−st

− 2t −1se−st

+ 2 1s2 e

−st

− 0 − 1s3 e

−st

+ 2 + t e−st

− 1 −1se−st

+ 0 1s2 e

−st

For the first integral,

−e−st

(t2

s+

2t

s2+

2

s3

)∣∣∣∣∣1

0

= −e−s(

1

s+

2

s2+

2

s3

)+

2

s3

For the second integral,

−e−st(

2 + t

s+

1

s2

)∣∣∣∣∞1

= limt→∞

[−e−st

(2 + t

s+

1

s2

)]+ e−s

(3

s+

1

s2

)For this limit, we see that, by L’Hospital’s rule:

limt→∞

[−e−st

(2 + t

s+

1

s2

)]= − lim

t→∞

2+ts

+ 1s2

est= lim

t→∞

1s

sest= 0, s > 0

Overall, the solution is:

−e−s(

1

s+

2

s2+

2

s3

)+

2

s3+ e−s

(3

s+

1

s2

)

Note: You could check your answer by rewriting f(t) in terms of the Heaviside function,then use the table (not necessary, but shown here for completeness). If t ≥ 0,

f(t) = (1− u(t− 1))t2 + u(t− 1)(2 + t) = t2 − u1(t)t2 + u1(t)(2 + t)

For the term:

u1(t)t2 ⇒ f(t− 1) = t2 ⇒ f(t) = (t + 1)2 = t2 + 2t + 1

So the Laplace transform is:

L(u1(t)t2) = e−s

(2

s3+

2

s2+

1

s

)

1

For the third term,

u1(t)(2 + t) ⇒ f(t− 1) = 2 + t ⇒ f(t) = 2 + (t + 1) = 3 + t

so the Laplace transform is:

L(u1(t)(2 + t) = e−s(

3

s+

1

s2

)

2. Problem 14, page 313. You may either use the formula given in the text, or the formula:

e(a+ib)t = eat cos(bt) + ieat sin(bt)

and you may use Table Entry 2. Of course, you may NOT use Table Entry 10.

Given Euler’s formula,e(a+ib)t = eat cos(bt) + ieat sin(bt)

The desired Laplace transform can be found via:

L(e(a+ib)t) = L(eat cos(bt)) + iL(eat sin(bt))

Using Table Entry 2,

1

s− (a + ib)= L(eat cos(bt)) + iL(eat sin(bt))

1

(s− a)− ib= L(eat cos(bt)) + iL(eat sin(bt))

(s− a) + i b

(s− a)2 + b2= L(eat cos(bt)) + iL(eat sin(bt))

(s− a)

(s− a)2 + b2+ i

b

(s− a)2 + b2= L(eat cos(bt)) + iL(eat sin(bt))

Therefore, the Laplace transform of eat cos(bt) is

(s− a)

(s− a)2 + b2

3. Find, using the table: L−1(

s

s2 + 2s− 3

)We first perform Partial Fraction Decomposition:

s

s2 + 2s− 3=

A

s + 3+

B

s− 1=

3

4

1

s + 3+

1

4

1

s− 1

so that the Laplace transform is:3

4e−3t +

1

4et

2

4. Find an expression for Y (s), do not solve for y(t):

y′′ − 4y′ + 4y = et cos(t), y(0) = 0, y′(0) = 1

Taking the Laplace transform of both sides,

(s2Y − 0− 1)− 4(sY − 0) + 4Y =s− 1

(s− 1)2 + 1

Y (s) =s− 1

((s− 1)2 + 1)(s2 − 4s + 4)+

1

s2 − 4s + 4

5. Solve the following IVP using the method of Laplace Transforms:

y′′ + 3y′ + 2y = e−t y(0) = 1 y′(0) = 0

s2Y − s + 3(sY − 1) + 2Y =1

s + 1⇒ (s2 + 3s + 2)Y =

1

s + 1+ s + 3

Therefore,

Y =1

(s + 1)(s2 + 3s + 2)+

s + 3

s2 + 3s + 2

We could keep these separate, or combine them (the following solution will combinethem). Factor the denominator:

Y =1

(s + 1)2(s + 2)+

s + 3

(s + 1)(s + 2)=

s2 + 4s + 4

(s + 1)2(s + 2)=

s + 2

(s + 1)2

Use partial fractions, or see that:

Y =s + 1 + 1

(s + 1)2=

1

s + 1+

1

(s + 1)2

so that y(t) = e−t + te−t.

3