What is the solution for (x^2)/y-4xe^{y^2}y^'=0 ?

The solution for (x^2)/y-4xe^{y^2}y^'=0 is y=sqrt(ln((x^2)/4+c_{1))},y=-sqrt(ln((x^2)/4+c_{1))}

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(x^2)/y-4xe^{y^2}y^'=0

\frac{x ^{2}}{y} - 4 x e^{y^{2}} y^{^{'}} = 0

y = ln (\frac{x ^{2}}{4} + c_{1}), y = - ln (\frac{x ^{2}}{4} + c_{1})

Solution steps

\frac{x ^{2}}{y} - 4 x e^{y^{2}} y^{'} (x) = 0

Solve separable ODE:

y = ln (\frac{x ^{2}}{4} + c_{1}), y = - ln (\frac{x ^{2}}{4} + c_{1})

y = ln (\frac{x ^{2}}{4} + c_{1}), y = - ln (\frac{x ^{2}}{4} + c_{1})

What is the solution for (x^2)/y-4xe^{y^2}y^'=0 ?
The solution for (x^2)/y-4xe^{y^2}y^'=0 is y=sqrt(ln((x^2)/4+c_{1))},y=-sqrt(ln((x^2)/4+c_{1))}