xyy' = x^2 + y^2
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1 Answer

Let y=v(x)x, then y'=v+xdv/dx and we have:


v(v+xdv/dx)=1+v², assuming x≠0.

So v²+vxdv/dx=1+v², vxdv/dx=1, vdv=dx/x which integrates to v²/2=ln|x|+c, where c is the constant of integration.

Replacing v=y/x, we get (y/x)²=ln(x²)+2c, y²=x²ln(x²)+Cx² where C=2c (replacing one constant with another).

by Top Rated User (763k points)

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