What is the solution for x^2y^'=2y^2-x^2 ?

The solution for x^2y^'=2y^2-x^2 is y=-(e^{-3c_{1}}x)/(x^3-e^{-3c_{1)}}-(x^4)/(2(x^3-e^{-3c_{1)})}

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x^2y^'=2y^2-x^2

x^{2} y^{^{'}} = 2 y^{2} - x^{2}

y = - \frac{e ^{- 3 c_{1}} x}{x ^{3} - e ^{- 3 c_{1}}} - \frac{x ^{4}}{2 ( x ^{3} - e ^{- 3 c_{1}} )}

Solution steps

x^{2} y^{'} (x) = 2 y^{2} - x^{2}

Solve using substitution:

y = - \frac{e ^{- 3 c_{1}} x}{x ^{3} - e ^{- 3 c_{1}}} - \frac{x ^{4}}{2 ( x ^{3} - e ^{- 3 c_{1}} )}

y = - \frac{e ^{- 3 c_{1}} x}{x ^{3} - e ^{- 3 c_{1}}} - \frac{x ^{4}}{2 ( x ^{3} - e ^{- 3 c_{1}} )}

What is the solution for x^2y^'=2y^2-x^2 ?
The solution for x^2y^'=2y^2-x^2 is y=-(e^{-3c_{1}}x)/(x^3-e^{-3c_{1)}}-(x^4)/(2(x^3-e^{-3c_{1)})}