7/30/2019 331spr10hw19sol

1/3

Math 331.5: Homework 19

Solutions

1. Compute the Laplace transform of the following functions.

(i) f(t) = sin(bt), where b is a constant.

L[sin(bt)] =b

s2 + b2, s > 0

(ii) f(t) = t

L[t] =1

s2, s > 0

(iii) f(t) = t2

L[t2] =2

s3, s > 0

(iv) f(t) = tn, where n is a positive integer.

L[tn] =n!

sn+1, s > 0

2. Recall that

cosh t =et + et

2sinh t =

et et

2

Compute the Laplace transform of the following functions where a and b are constant.

(i) f(t) = sinh(bt)

L[sinh(bt)] =b

s2 b2

(ii) f(t) = cosh(bt)

L[cosh(bt)] =s

s2 b2

(iii) f(t) = eat cosh(bt)

L[cosh(bt)] =s a

(s a)2 b2

3. Follow the steps to compute the Laplace transform of eat cos(bt) and eat sin(bt).

(i) Use Eulers formula

eix = cos x + i sin x

to rewrite e(a+ib)t in the form u(t) + iv(t) where u(t) and v(t) are real valued functions oft.

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

7/30/2019 331spr10hw19sol

2/3

2

(ii) Show that the Laplace transform of f(t) = e(a+ib)t is F(s) =1

s a ib.

L[e(a+ib)t] =

0

e(a+ib)test dt

= limR

R

0

e(as+ib)t dt

= limR

1

a s + ibe(as+ib)t

R

0

= limR

1

a s + ib

e(as+ib)R 1

=1

s a ibif s > a

(iii) Rewrite F(s) in the form U(s) + iV(s) where U(s) and V(s) are real valued functions ofs. (Hint: Multiply in the numerator and denominator by the conjugate of s a ib).

1

s a ib =1

s a ib s a + ib

s a + ib

=s a + ib

(s a)2 + b2

=s a

(s a)2 + b2+ i

b

(s a)2 + b2

(iv) Equate real and imaginary parts to conclude that

L[u(t)] = U(s) and L[v(t)] = V(s)

0

e(a+ib)test dt =

0

(eat cos(bt) + ieat sin(bt))est dt

=0 e

(as)t

cos(bt) dt + i0 e

(as)t

sin(bt) dt

=s a

(s a)2 + b2+ i

b

(s a)2 + b2

Equating real and imaginary parts,

L[eat cos(bt)] =s a

(s a)2 + b2L[eat sin(bt)] =

b

(s a)2 + b2

4. Find the inverse Laplace transform of the following functions.

(i) F(s) =1

s 3f(t) = e3t

(ii) F(s) = 3s2 + 4

f(t) =3

2sin(2t)

7/30/2019 331spr10hw19sol

3/3

3

(iii) F(s) =2

s2 + 3s 42

s2 + 3s 4=

2/5

s + 4+

2/5

s 1

f(t) = 2

5 e4t

+2

5 et

(iv) F(s) =2s + 2

s2 + 2s + 5(Hint: Complete the square in the denominator and use problem 3).

2s + 2

s2 + 2s + 5=

2(s + 1)

(s + 1)2 + 4

f(t) = 2et cos(2t)

(v) F(s) =2s 3

s2 42s 3

s2 4=

1/4

s 2+

7/4

s + 2

f(t) =

1

4 e2t

+

7

4 e2t

(vi) F(s) =8s2 4s + 12

s(s2 + 4)

8s2 4s + 12

s(s2 + 4)=

3

s+

5s 4

s2 + 4

= 31

s+ 5

s

s2 + 4 2

2

s2 + 4

f(t) = 3 + 5 cos(2t) 2 sin(2t)

(vii) F(s) =1 2s

s2 + 4s + 5

1

2ss2 + 4s + 5 = 1

2s(s + 2)2 + 1

= 2s 12

(s + 2)2 + 1

= 2s + 2 52

(s + 2)2 + 1

= 2s + 2

(s + 2)2 + 1+

5

(s + 2)2 + 1

f(t) = 2e2t cos t + 5e2t sin t