How did they get Y(s)? I tried, but I don't get how to convert the y' notation.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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How did they get Y(s)?
I tried, but I don't get how to convert the y' notation.
Transcribed Image Text:Solve the initial value problem below using the method of Laplace transforms.
y'' - y' - 2y = 0, y(0) = -1, y'(0) = 7
Click here to view the table of Laplace transforms.
Click here to view the table of properties of Laplace transforms.
What is the Laplace transform Y(s) of the solution y(t)?
2
3
²2
S-2 s +1
Y(s) =
Solve the initial value problem.
y(t) = 2e²t - 3e
(Type an exact answer in terms of e.)
![De
2
y"-y!-2y = 0; y(@) = -1; y'(0) = 7
↓
ⒸL{y} = Y
[₁²Y-(-1) -(7)]-[₂Y-(+)] = Y = n² Y-3 + ₂Y
[₂2² +2=7]-[~Y +1]=Y=²Y+₂Y÷3
-27 +2-8-Y = 27-3
-1=227
227+1=0
DL{y} = 2²Y-₁4(0)-7² (0) =
2² Y-3
Ⓒ-2y = 0
-2Y=0
@L_{y} =nY=y(0) = Y
2- Y
=227 +5.
-Y = 2₂7+5-s
·+ = -2₂7 + (-5) +2
Y(2₂+1) = 2-5
Y = 2-5
22+1
A-3
дахт
2²4 +2Y-7 = 224-3
0 Y₁ =4 = 0 → Y₂-4 +2₂4 +1-2Y=0
[x+1] =
+
22+1
324-3-24=0
3₂Y - 2Y = 3
Y(32-2) = 3
Y = (³) 3
-(1) 322
Y =](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F48ae96ba-8296-4f48-84d9-a65ba8f4727e%2F98bb1d2b-fe7b-4222-9524-bfc928a8c84c%2Fguwown8_processed.jpeg&w=3840&q=75)
Transcribed Image Text:De
2
y"-y!-2y = 0; y(@) = -1; y'(0) = 7
↓
ⒸL{y} = Y
[₁²Y-(-1) -(7)]-[₂Y-(+)] = Y = n² Y-3 + ₂Y
[₂2² +2=7]-[~Y +1]=Y=²Y+₂Y÷3
-27 +2-8-Y = 27-3
-1=227
227+1=0
DL{y} = 2²Y-₁4(0)-7² (0) =
2² Y-3
Ⓒ-2y = 0
-2Y=0
@L_{y} =nY=y(0) = Y
2- Y
=227 +5.
-Y = 2₂7+5-s
·+ = -2₂7 + (-5) +2
Y(2₂+1) = 2-5
Y = 2-5
22+1
A-3
дахт
2²4 +2Y-7 = 224-3
0 Y₁ =4 = 0 → Y₂-4 +2₂4 +1-2Y=0
[x+1] =
+
22+1
324-3-24=0
3₂Y - 2Y = 3
Y(32-2) = 3
Y = (³) 3
-(1) 322
Y =
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