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Q15E

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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. dydx=x2-y23xy

Homogeneous equation for the given equation is (x2-4y2)3x2=C.

See the step by step solution

Step by Step Solution

Step 1: 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.

Step 2: Evaluate the given equation

Given, dydx=x2-y23xy.

Evaluate it.

dydx=x2-y23xydydx=x23xy-y23xy=x3y-y3x=13xy-yx

Step 3: Substitution method

Let us take v=yx.

Then y=vx.

By Differentiating,

dydx=v+xdvdx131v-v=v+xdvdx13v-v3-v=xdvdx1-v2-3v23v=xdvdx1-4v23v=xdvdxv1-4v2dv=13xdx

Step 4: Integrate the equation

Now, integrate on both sides.

v1-4v2dv=13xdx v1-4v2dv=lnx+C

Find the value of v1-4v2dv separately.

Let w=1-4v2dwdv=-8vdv=-18vdw

v1-4v2dv=-vw18vdw=-181wdw=-18lnw

Then, v1-4v2dv=-18ln1-4v2.

And role="math" localid="1655139371679" -18ln1-4v2=13lnx+C1

Step 5: Substitution method

Substitute v=yx.

-18ln1-4yx2=13lnx+C1-3ln1-4yx2=8lnx+C2-3lnx2-4y2x2=8lnx+C2

Where lnMN=ln M-ln N

Then,

-3lnx2-4y2+3lnx2=8lnx+C2-3lnx2-4y2+3lnx2-8lnx=C2lnx6-lnx2-4y23-lnx8=C2lnx6x2-4y23-lnx8=C2

ln1x2-4y23x2=C21x2-4y23x2=eC21x2-4y23x2=C3x2-4y23x2=1C3x2-4y23x2=C

Therefore, Homogeneous equation for the given equation is (x2-4y2)3x2=C

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