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) = 1, y (3) (0)=0, y (4) (0) =0 A E B The solution is given by: y=(1/4) e + (1/4) e-(1/2) cos(x) The solution is given by: y=(1/4).xe*+ (1/8) e*- (3/8) e*+ (1/4) cos(x) + (1/4) sin(x) The solution is given by: Ⓒy=(1/8) e*- (3/8)e~*— (1/4) xe¨*+ (1/4) cos(x) − (1/4) sin(x) The solution is given by: y=(1/4).xe*+ (1/8) e*- (3/8) e*+ (1/2) cos(x) + (1/2) sin(x) The solution is given by: y=(1/4) xe¹ + (1/8) e*- (3/8)e+ (1/2) cos(x) − (1/2) sin(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) = 1, y (3) (0)=0, y (4) (0) =0 A E B The solution is given by: y=(1/4) e + (1/4) e-(1/2) cos(x) The solution is given by: y=(1/4).xe*+ (1/8) e*- (3/8) e*+ (1/4) cos(x) + (1/4) sin(x) The solution is given by: Ⓒy=(1/8) e*- (3/8)e~*— (1/4) xe¨*+ (1/4) cos(x) − (1/4) sin(x) The solution is given by: y=(1/4).xe*+ (1/8) e*- (3/8) e*+ (1/2) cos(x) + (1/2) sin(x) The solution is given by: y=(1/4) xe¹ + (1/8) e*- (3/8)e+ (1/2) cos(x) − (1/2) sin(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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![**Consider the following fifth-order linear homogeneous initial value problem with constant coefficients:**
\[ y^{(5)} + y^{(4)} - y' = y = 0, \]
\[ y(0) = 0, \, y'(0) = 0, \, y''(0) = 1, \, y^{(3)}(0) = 0, \, y^{(4)}(0) = 0 \]
---
The solution is given by:
**A**
\[ y = (1/4)e^{x} + (1/4)e^{-x} - (1/2)\cos(x) \]
---
The solution is given by:
**B**
\[ y = (1/4)xe^{x} + (1/8)e^{-x} - (3/8)e^{x} + (1/4)\cos(x) + (1/4)\sin(x) \]
---
The solution is given by:
**C**
\[ y = (1/8)e^{x} - (3/8)e^{-x} - (1/4)xe^{-x} + (1/4)\cos(x) - (1/4)\sin(x) \]
---
The solution is given by:
**D**
\[ y = (1/4)xe^{x} + (1/8)e^{-x} - (3/8)e^{x} + (1/2)\cos(x) + (1/2)\sin(x) \]
---
The solution is given by:
**E**
\[ y = (1/4)e^{x} + (1/8)e^{-x} - (3/8)e^{x} + (1/2)\cos(x) - (1/2)\sin(x) \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa6a5d441-691e-4a20-a4ea-52e6f9df3d14%2Fcbe9d968-5389-40ce-afc9-b0e15d045377%2Fb9sjj66_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Consider the following fifth-order linear homogeneous initial value problem with constant coefficients:**
\[ y^{(5)} + y^{(4)} - y' = y = 0, \]
\[ y(0) = 0, \, y'(0) = 0, \, y''(0) = 1, \, y^{(3)}(0) = 0, \, y^{(4)}(0) = 0 \]
---
The solution is given by:
**A**
\[ y = (1/4)e^{x} + (1/4)e^{-x} - (1/2)\cos(x) \]
---
The solution is given by:
**B**
\[ y = (1/4)xe^{x} + (1/8)e^{-x} - (3/8)e^{x} + (1/4)\cos(x) + (1/4)\sin(x) \]
---
The solution is given by:
**C**
\[ y = (1/8)e^{x} - (3/8)e^{-x} - (1/4)xe^{-x} + (1/4)\cos(x) - (1/4)\sin(x) \]
---
The solution is given by:
**D**
\[ y = (1/4)xe^{x} + (1/8)e^{-x} - (3/8)e^{x} + (1/2)\cos(x) + (1/2)\sin(x) \]
---
The solution is given by:
**E**
\[ y = (1/4)e^{x} + (1/8)e^{-x} - (3/8)e^{x} + (1/2)\cos(x) - (1/2)\sin(x) \]
Transcribed Image Text:The image contains a series of mathematical expressions labeled from E to J. Each expression is presented as a solution to a given problem. Here are the transcriptions of each expression:
---
**E**
The solution is given by:
\( y = (1/4)xe^x + (1/8)e^{-x} - (3/8)e^x + (1/2)\cos(x) - (1/2)\sin(x) \)
---
**F**
The solution is given by:
\( y = (1/8)e^x + (5/8)e^{-x} + (1/4)xe^{-x} - (1/4)\cos(x) - (1/4)\sin(x) \)
---
**G**
The solution is given by:
\( y = (1/4)e^{-x} - (1/4)e^{-x} + (1/2)\sin(x) \)
---
**H**
The solution is given by:
\( y = (5/8)e^x + (1/8)e^{-x} - (1/4)xe^{-x} + (1/4)\cos(x) - (1/4)\sin(x) \)
---
**I**
The solution is given by:
\( y = (1/8)e^x + (5/8)e^{-x} + (1/4)xe^{-x} + (1/4)\cos(x) + (1/4)\sin(x) \)
---
**J**
The solution is given by:
\( y = (1/4)e^{-x} - (1/4)e^{-x} - (1/2)\sin(x) \)
---
These solutions seem to pertain to problems involving exponential, trigonometric, and polynomial functions of \(x\).
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