Brewer_MAT_275_ONLINE_B_Spring_2018.digamble.Section_6.2_Solution_of_I.V.P.pdf

David Gamble Brewer MAT 275 ONLINE B Spring 2018
Assignment Section 6.2 Solution of I.V.P due 04/14/2018 at 11:59pm MST
1. (1 point) Find the inverse Laplace transform f ( t ) =
L - 1
{ F ( s ) } of
F ( s ) =
2 s + 9
s 2 + 13
s > 0
f ( t ) = .
Solution:
L - 1
n
2 s + 9
s
2
+ 13
o
= 2
L - 1
n
s
s
2
+ 13
o
+
9
√
13
L - 1
n
√
13
s
2
+ 13
o
= 2cos (
√
13 t )+
9
√
13
sin (
√
13 t )
Correct Answers:
• 2*cos(3.60555127546399*t)+2.49615088301353*sin(3.60555127546399*t)
2. (1 point) Find the inverse Laplace transform f ( t ) =
L - 1
{ F ( s ) } of
F ( s ) =
9 s + 5
s 2 - 9
s > 3
f ( t ) = .
Substituting the initial conditions gives:
s
2
Y - 2 s + 2 + 3 sY - 6 - 10 Y = 0
(2) Solving for Y gives:
Y =
2 s + 4
s 2 + 3 s - 10
=
2 s + 4
( s - 2 )( s + 5 )
(3) Performing Partial Fraction Decomposition yields:
2 s + 4
( s - 2 )( s + 5 )
=
8
7
s - 2
+
6
7
s + 5
(4) Inverting the transform, we find
y ( t ) =
8
7
e
2 t
+
6
7
e
- 5 t
Correct Answers:
• sˆ2*Y-2*s--2+3*(s*Y-2)+-10*Y
• (2*s+4)/(sˆ2+3*s+-10)
• 1.14285714285714/(s-2)
• 0.857142857142857/(s+5)
• 1.14285714285714*exp(2*t)+0.857142857142857*exp(-5*t)
4. (1 point) Use the Laplace transform to solve the following
initial value problem:
y
00
+ 8 y
0
= 0 y ( 0 ) = 7 , y
0
( 0 ) = 7
1