Consider the following fifth-order linear homogeneous initial value problem with constant coefficients: y ‚ (5) − y (4) − y ' + y =0, y (0) = 0, y '(0) =0, y" (0) =0, y (3) (0) = 1, y (4) (0)=0 Hide answer choices A B The solution is given by: y=(1/4) e*- (1/4) e*- (1/2) sin(x) The solution is given by: y=(1/4) e*- (1/4) e¯*+ (1/2) sin(x) The solution is given by: y=(1/4) e* + (1/4) e*- (1/2) cos(x)
Consider the following fifth-order linear homogeneous initial value problem with constant coefficients: y ‚ (5) − y (4) − y ' + y =0, y (0) = 0, y '(0) =0, y" (0) =0, y (3) (0) = 1, y (4) (0)=0 Hide answer choices A B The solution is given by: y=(1/4) e*- (1/4) e*- (1/2) sin(x) The solution is given by: y=(1/4) e*- (1/4) e¯*+ (1/2) sin(x) The solution is given by: y=(1/4) e* + (1/4) e*- (1/2) cos(x)
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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Question
![Consider the following fifth-order linear homogeneous initial value problem with constant coefficients:
\( y^{(5)} - y^{(4)} - y' + y = 0, \)
\( y(0) = 0, \, y'(0) = 0, \, y''(0) = 0, \, y^{(3)}(0) = 1, \, y^{(4)}(0) = 0 \)
The possible solutions are:
A) The solution is given by:
\[
y = \left(\frac{1}{4}\right) e^x - \left(\frac{1}{4}\right) e^{-x} - \left(\frac{1}{2}\right) \sin(x)
\]
B) The solution is given by:
\[
y = \left(\frac{1}{4}\right) e^x - \left(\frac{1}{4}\right) e^{-x} + \left(\frac{1}{2}\right) \sin(x)
\]
C) The solution is given by:
\[
y = \left(\frac{1}{4}\right) e^x + \left(\frac{1}{4}\right) e^{-x} - \left(\frac{1}{2}\right) \cos(x)
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6d11a2cb-f5e2-441d-8b2a-04ce207dc9df%2Fe7da6785-05ef-4d53-bb10-fbd2ccfa183d%2Fl90b0se_processed.png&w=3840&q=75)
Transcribed Image Text:Consider the following fifth-order linear homogeneous initial value problem with constant coefficients:
\( y^{(5)} - y^{(4)} - y' + y = 0, \)
\( y(0) = 0, \, y'(0) = 0, \, y''(0) = 0, \, y^{(3)}(0) = 1, \, y^{(4)}(0) = 0 \)
The possible solutions are:
A) The solution is given by:
\[
y = \left(\frac{1}{4}\right) e^x - \left(\frac{1}{4}\right) e^{-x} - \left(\frac{1}{2}\right) \sin(x)
\]
B) The solution is given by:
\[
y = \left(\frac{1}{4}\right) e^x - \left(\frac{1}{4}\right) e^{-x} + \left(\frac{1}{2}\right) \sin(x)
\]
C) The solution is given by:
\[
y = \left(\frac{1}{4}\right) e^x + \left(\frac{1}{4}\right) e^{-x} - \left(\frac{1}{2}\right) \cos(x)
\]
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