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Q 2.6-41E

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Fundamentals Of Differential Equations And Boundary Value Problems
Found in: Page 77
Fundamentals Of Differential Equations And Boundary Value Problems

Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069

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Short Answer

Question: Use the substitution v=x-y+2 to solve equation (8).

dydx=y-x-1+x-y+2-1

x-y+22=Ce2x+1

See the step by step solution

Step by Step Solution

Step 1: Use the given substitution v=x-y+2  to solve the given equation,

The given substitution,

v=x-y+2

Simplify the above equation,

1-v=1-x-y+21-v=y-x-1

And

role="math" localid="1663933729097" 1-v=y-x-1y=1-v+x+1y=x-v+2dydx=1-dvdx

Given equation,

dydx=y-x-1+x-y+2-1

Substitute v=x-y+2,1-v=y-x-1 and dydx=1-dvdx in the above equation,

1-dvdx=1-v+v-1

Simplify the above equation,

-dvdx=-v+v-1dvdx=v-1vdvdx=v2-1v

Cross multiplication on both sides in the above equation,

vv2-1dv=dx2vv2-1dv=2dx

Step 2: Integrating

Taking integration both sides in the above equation

2vv2-1dv=2dx

Solve the above equation,

2vv2-1dv=2dxlogv2-1=2x+logCv2-1C=e2xv2-1=Ce2xv2=Ce2x+1

Substitute the value of v=x-y+2 in the above equation,

x-y+22=Ce2x+1

Hence the solution is x-y+22=Ce2x+1

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