MTH 352
Final Exam with Solutions
Prof. Townsend
Fall 2004
Exam:
Do problems A or B.
1) Find the Laplace Transform of
A) f ( t ) = 1 ! cos ( 2t )
B) f (t ) = 2 + sinh (3t )
2) Find the Laplace Transform of
A) f t = t 3 ! 3te!t
B) f t = t ! 3t 3e!t
()
3) Find the inverse Laplace Transform of
3
A) F s = 2
s !4
()
4) Find the inverse Laplace Transform of
s+3
A) F (s ) = 2
s + 4 s + 29
()
B) F (s ) =
()
B) F s =
s
s +9
2
s!3
s ! 2s + 26
2
()
()
5) Solve the following differential equation using Laplace Transforms. y 0 = 1 and y ' 0 = 0
A) y ''+ 2 y '+ y = sin (t )e ! t
B) y''+ 4y'+ 4y = cos ( t ) e!2t
Pattern matches from the Laplace Transform Tables:
#
1A
Rules Pattern Matches
f t
F s
()
Problem
f ( t ) = 1 ! cos ( 2t )
c
c
s
cos bt
1B
f (t ) = 2 + sinh (3t )
c
sinh bt
2A
()
f t = t 3 ! 3te!t
tn
e at t n
2B
f ( t ) = t ! 3t 3e!t
Fall 2004 Laplace Transform Final Solutions
t
()
n
s
s + b2
c
s
b
2
s ! b2
n!
s n +1
n!
2
( s ! a )n +1
n!
s n +1
Page 1 of 6
e at t n
n!
(s ! a )
n +1
3A
F (s) =
3
s !4
sinh bt
2
cosh bt
3B
F (s ) =
s
s +9
cos bt
2
sin bt
eat sinh bt
4A
()
F s =
b
s ! b2
s
2
s ! b2
s
2
s + b2
b
2
s + b2
b
2
(s ! a)
2
! b2
s!a
s+3
2
s + 4s + 29
eat cosh bt
( s ! a )2 ! b 2
s!a
eat cos bt
(s ! a)
eat sin bt
b
(s ! a)
2
2
+ b2
+ b2
b
eat sinh bt
(s ! a)
2
! b2
s!a
4B
F (s ) =
at
e cosh bt
s !3
s ! 2 s + 26
2
eat cos bt
( s ! a )2 ! b 2
s!a
(s ! a)
2
+ b2
b
()
5A
y ''+ 2 y '+ y = sin t e
5B
y ''+ 4 y '+ 4 y = cos (t )e
!t
eat sin bt
(s ! a)
f ' (t )
sL f t
f '' (t )
!2 t
Fall 2004 Laplace Transform Final Solutions
f ' (t )
f '' (t )
2
+ b2
{ ( )} ! f (0)
s L { f ( t )} ! sf ( 0 ) ! f ' ( 0 )
2
sL { f ( t )} ! f ( 0 )
{ ( )} ! sf (0) ! f '(0)
s2L f t
Page 2 of 6
Solutions :
1) Find the Laplace Transform of
A) f ( t ) = 1 ! cos ( 2t )
b = 2 F (s) =
B) f (t ) = 2 + sinh (3t )
b=3
()
F s =
1
s
! 2
s s + 22
2
3
+ 2 2
s s !3
2) Find the Laplace Transform of
A) f ( t ) = t 3 ! 3te!t
F s =
()
3!
1!
!3
4
2
s
s +1
B) f ( t ) = t ! 3t 3e!t
F (s) =
1!
3!
!3
2
s
( s + 1)4
(
)
3) Find the inverse Laplace Transform of
3
b=2
A) F s = 2
s !4
Assume solution based on denominator form f t = Asinh 2t + Bcosh 2t
()
()
Take the Laplace Transform of the solution
Combine fractions
Equate numerators
Try s = 0

Plug in 3 = 3 + sB
()
f t =
F (s) = A
2 A + sB
3
= 2
2
2
s !2
s ! 22
3 = 2 A + sB
3
3 = 2A 
A=
2
B=0

( )
( )
2
s
+B 2
2
s !2
s ! 22
2
()
F s =
3
sinh 2t
2
( )
s
b=3
s +9
Assume solution based on denominator form f t = Asin 3t + Bcos 3t
B) F (s ) =
2
Take the Laplace Transform of the solution
()
Combine fractions F s =
3A + Bs
s 2 + 32
Equate numerators
Fall 2004 Laplace Transform Final Solutions
()
( )
3
F ( s) = A
s +3
2
2
( )
+B
s
s + 32
2
s = 3A + sB
Page 3 of 6
Try s = 0
3A = 0 

Plug in s = sB
()
A=0
B =1

( )
f t = cos 3t
4) Find the inverse Laplace Transform of
s+3
A) F s = 2
s + 4s + 29
()
(
)
2
Complete the square s 2 + 4s + 29 = s 2 + 4s + 4 + 25 = s + 2 + 52
()
( )
( )
Assume solution based on denominator form f t = Ae!2t sin 5t + Be!2t cos 5t
Take the Laplace Transform of the solution
5
s+2
F s =A
+B
2
2
s + 2 + 52
s + 2 + 52
()
(
)
( )
5A + B ( s + 2 )
Combine fractions F ( s ) =
( s + 2) + 5
2
(
s + 3 = 5A + B s + 2
Equate numerators
Try s = !2
2
A=
1 = 5A 

(
Plug in s + 3 = 1 + B s + 2
)
)
1
5
(
)
s+2= B s+2 B =1

{ ( )
( )}
1
1
f t = e!2t sin 5t + e!2t cos 5t = e!2t sin 5t + 5cos 5t
5
5
()
B) F ( s ) =
( )
( )
s!3
s ! 2s + 26
2
(
)
f ( t ) = Ae sin (5t ) + Be cos (5t )
2
Complete the square s 2 ! 2s + 26 = s 2 ! 2s + 1 + 25 = s ! 1 + 52
Assume solution based on denominator form
t
t
Take the Laplace Transform of the solution
5
s !1
F s =A
+B
2
2
s ! 1 + 52
s ! 1 + 52
()
(
)
( )
5A + B ( s ! 1)
Combine fractions F ( s ) =
( s ! 1) + 5
2
Fall 2004 Laplace Transform Final Solutions
2
Page 4 of 6
s ! 3 = 5A + B ( s ! 1)
2
!2 = 5A 
A=!
5
Equate numerators
Try s = 1

(
)
Plug in s ! 3 = !2 + B s ! 1
B =1
So
s ! 1 = B ( s ! 1)

2
5
1
f ( t ) = ! et sin ( 5t ) + et cos ( 5t ) = et {!2 sin ( 5t ) + 5 cos ( 5t )}
5
5
5
5) Solve the following differential equation using Laplace Transforms
()
A) y ''+ 2 y '+ y = sin t e!t
Take the Laplace transform of the differential equation.
1
s 2Y s ! s "1 ! 0 + 2 sY s ! 1 + Y s =
2
s + 1 + 12
{ ()
} { () } ()
()
()
()
s 2Y s ! s + 2sY s ! 2 + Y s =
(
)
1
( s + 1)
2
+ 12
Note: you could put the above in solver and solve for Y.
1
s 2 + 2s + 1 Y s ! s ! 2 =
2
s + 1 + 12
{
} ()
s+2+
()
Y s =
{s
2
(
)
1
( s + 1)
2
+ 12
}
+ 2s + 1
F2/Expand gives the partial fraction expansion – Be sure to check Pretty Print to make
sure you entered the above equation correctly. If you have trouble, write it out on paper
first to get the parenthesis properly placed.
!1
1
2
Y s = 2
+
+
s + 2s + 2 s + 1 s + 1 2
()
(
)
Rewrite so the inverse transforms are obvious.
1
1
1!
Y (s) = !
+
+2
2
( s + 1) + 1 s + 1 ( s + 1)2
Take the inverse transforms
y ( t ) = !e!t sin ( t ) + e!t + 2te!t
Put in deSolve to check (radian mode) or solve using conventional techniques.
()
()
deSolve( y ''+ 2 y '+ y = sin (t )e ! t and y 0 = 1 and y ' 0 = 0 ,t,y)
The TI-89 gets
Fall 2004 Laplace Transform Final Solutions
Page 5 of 6
y ( t ) = ( ! sin ( t ) + 2t + 1) e!t
our solution, just rearranged.
()
B) y ''+ 4 y '+ 4 y = cos t e!2t
Take the Laplace transform of the differential equation.
{s Y ( s ) ! s "1 ! 0} + 4 {sY ( s ) ! 1} + 4Y ( s ) = ( s +s2+) 2+ 1
2
2
s 2Y ( s ) ! s + 4sY ( s ) ! 4 + 4Y ( s ) =
2
s+2
( s + 2 )2 + 12
Note: you could put the above in solver and solve for Y.
s+2
s 2 + 4s + 4 Y ( s ) ! s ! 4 =
( s + 2 )2 + 12
{
}
s+4+
()
Y s =
{s
2
s+2
( s + 2)
2
+ 12
}
+ 4s + 4
F2/Expand gives the partial fraction expansion – Be sure to check Pretty Print to make
sure you entered the above equation correctly. If you have trouble, write it out on paper
first to get the parenthesis properly placed.
!s
2
2
2
Y s = 2
! 2
+
+
s + 4s + 5 s + 4s + 5 s + 2 s + 2 2
()
(
Rewrite so the inverse transforms are obvious.
s+2
1
1!
Y s =!
+2
+2
2
s+2
s + 2 + 12
s+2
()
(
)
(
)
)
2
Take the inverse transforms
y ( t ) = !e!2t cos ( t ) + 2e!2t + 2te!2t
Put in deSolve to check (radian mode) or solve using conventional techniques.
()
()
()
deSolve( y ''+ 4 y '+ 4 y = cos t e!2t and y 0 = 1 and y ' 0 = 0 ,t,y)
The TI-89 gets
y ( t ) = ( ! cos ( t ) + 2t + 2 ) e!2t
our solution, just rearranged.
Fall 2004 Laplace Transform Final Solutions
Page 6 of 6