21 Care must be taken when the solution of a DE involves inverse functions. Solutions may be inadvertently lost or added, as in the following example. It demonstrates that any solution of a DE should be thoroughly checked before it is accepted as correct. dy -V1 - y', with y (0) dx Consider the initial value problem = 1. a Find the implicit general solution of the DE. b Evaluate the unknown constant by applying the initial condition. c Making y the subject without taking care, it would seem that the solution of the IVP is y = cos x. Explain why this solution is not valid for all values of x. It may help to substitute this solution into each side of the DE. d What is the correct solution of the IVP in which y is the subject?
21 Care must be taken when the solution of a DE involves inverse functions. Solutions may be inadvertently lost or added, as in the following example. It demonstrates that any solution of a DE should be thoroughly checked before it is accepted as correct. dy -V1 - y', with y (0) dx Consider the initial value problem = 1. a Find the implicit general solution of the DE. b Evaluate the unknown constant by applying the initial condition. c Making y the subject without taking care, it would seem that the solution of the IVP is y = cos x. Explain why this solution is not valid for all values of x. It may help to substitute this solution into each side of the DE. d What is the correct solution of the IVP in which y is the subject?
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
A and D on
Transcribed Image Text:21 Care must be taken when the solution of a DE involves inverse functions. Solutions may be inadvertently
lost or added, as in the following example. It demonstrates that any solution of a DE should be
thoroughly checked before it is accepted as correct.
dy
Consider the initial value problem
-V1 - y?, with y (0) = 1.
dx
a Find the implicit general solution of the DE.
b Evaluate the unknown constant by applying the initial condition.
c Making y the subject without taking care, it would seem that the solution of the IVP is y
= cos x.
Explain why this solution is not valid for all values of x. It may help to substitute this solution into
each side of the DE.
d What is the correct solution of the IVP in which y is the subject?
Transcribed Image Text:21 a cos- y = x + C
b C = 0 so cos -ly = x
c LHS
- sin x, RHS
-V1 - cos?x = - sin x
X = – sin x
|
|
These are unequal when sin x < 0.
dy = cos x with domain 0 < x < n.
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