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Question: In Problems 1-30, solve the equation.

 x2+y2dx+3xy dy=0

Short Answer

Expert verified

The solution of the given equation is x2+4y23x2=C

Step by step solution

01

Given information and simplification

Given that, x2+y2dx+3xy dy=0⋅⋅⋅⋅⋅⋅1

Evaluate the equation (1).

x2+y2dx+3xy dy=03xy dy=-x2+y2dxdydx=-x2+y23xy

Let us take y=tx. Differentiate with respect to x.

dydx=xdtdx+t

Then,

xdtdx+t=-1+t23txdtdx=-1+t23t-txdtdx=-1+t2-3t23t

xdtdx=-1-4t23t3t1+4t2dt=-dxx

Let us takep=1+4t2 and dp=8t dt.

38pdp=-dxx

Now integrate on both sides.

∫38pdp=-∫dxx38lnp=-lnx+Cp=C'x-83

02

Evaluation method

Substitute the value of p.

1+4t2=C'x-831+4t23=C''x-8

Substitute the value of t.

x2+4y2x23=C''x-8x2+4y23x2=C

So, the solution is (x2+4y2)3x2=C

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