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Q2.6 - 12E

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Fundamentals Of Differential Equations And Boundary Value Problems
Found in: Page 76
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

Use the method discussed under “Homogeneous Equations” to solve problems 9 -16. (x2+y2)dx+2xy dy=0

Homogeneous equation for the given equation is x3+3xy2=C.

See the step by step solution

Step by Step Solution

General form of Homogeneous equation

If the right-hand side of the equation dydx=fx,y can be expressed as a function of the ratio yx alone, then we say the equation is homogeneous.

Evaluate the given equation

Given, x2+y2dx+2xy dy=0

Evaluate it.

x2+y2dx+2xy dy=02xydy=-x2+y2dxdydx=-x2+y22xy=-x2y-y2x=-12xy+yx

Substitution method

Let us take v=yx

Then y=vx

By Differentiating,

dydx=v+xdvdx-121v+v=v+xdvdx-121v+v-v=xdvdx-1+v22v-vdv=xdx

Integrate the equation

Now, integrate on both sides.

2v1+3v2dv=-1xdx2v1+3v2dv=-lnx+C

Integrate2v1+3v2dv separately.

Let us take w=1+3v2

Then,

dwdv=6vdv=16vdw

Now,

2vw16vdw=131wdw=13lnw

Substitute w=1+3v2

2v1+3v2dv=13lnw=13ln1+3v2

Then,

13ln1+3v2=-lnx+C113ln1+3v2+lnx=C1ln1+3v2+3lnx=3C1ln1+3v2+lnx3=C2

ln1+3v2×x3=C21+3v2×x3=eC21+3v2×x3=C

Substitute v=yx

x31+3yx2=Cx31+3y2x2=Cx3x2+3y2x2=Cxx2+3y2=Cx3+3xy2=C

Therefore, Homogeneous equation for the given equation is x3+3xy2=C.

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