What is the solution for y^'-1/x y=x^2y^4 ?

The solution for y^'-1/x y=x^2y^4 is y=(\sqrt[3]{2}x)/(\sqrt[3]{-x^6+c_{1)}}

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y^'-1/x y=x^2y^4

y^{^{'}} - \frac{1}{x} y = x^{2} y^{4}

Solution steps

y^{'} (x) - \frac{1}{x} y = x^{2} y^{4}

Solve Bernoulli ODE:

f (x) = 3 x^{2} - 3 x - 5

\frac{\partial}{\partial y} (x^{4} y^{- 2})

\int \frac{1}{( a ^{2} - x ^{2} ) ^{\frac{3}{2}}} d x

\int_{a}^{2 a} a x - \frac{a ^{3}}{x} d x

\int \frac{1}{( x ^{2} - 2 x + 5 ) ^{3}} d x

What is the solution for y^'-1/x y=x^2y^4 ?
The solution for y^'-1/x y=x^2y^4 is y=(\sqrt[3]{2}x)/(\sqrt[3]{-x^6+c_{1)}}