Solve differential equation using homogeneous equation method. Solve differentia

Question

Solve differential equation using homogeneous equation method. Solve differential equation using homogeneous equation method. (2xy+x^2)dx + y^2dy=0 I started off the problem like this: x^2(2(y/x)+1) +y^2dy=0 dy/dx=[-2(y/x)+1]/[(y/x)^2] g(v)=[-2v+1]/[v^2] Please help me finish it I have to go back in terms of y.

Explanation / Answer

This is a homogeneous DE, since it can be written in the form dy/dx = (y/x)^2 - 2(y/x). Let v = y/x. ==> y = xv and dy/dx = v + x dv/dx. Thus, the DE transforms to v + x dv/dx = v^2 - 2v Rewrite this in separable form dv / (v^2 - 3v) = dx/x By partial fractions, 1/(v^2 - 3v) = (1/3) [1/(v - 3) - 1/v] So, we now have (1/3) [1/(v - 3) - 1/v] dv = dx/x ==> [1/(v - 3) - 1/v] dv = 3 dx/x Integrate both sides: ln(v - 3) - ln v = 3 ln(x) + A ==> ln[(v - 3) / v] = 3 ln(x^3) + A ==> 1 - 3/v = Cx^3, where C = e^A. Since v = y/x, we have 1 - 3x/y = Cx^3.

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