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

The solution for x^2y^'-2xy=y^4 is y=(\sqrt[3]{5}x^2)/(\sqrt[3]{-3x^5+c_{1)}}

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x^2y^'-2xy=y^4

x^{2} y^{^{'}} - 2 x y = y^{4}

Solution steps

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

Solve Bernoulli ODE:

\frac{d}{d x} (cos^{2} (tan (x)))

x \to 0 lim (\frac{5 x}{x ^{3}})

\frac{d}{d x} (\frac{x - 1}{x ^{2} + 2 x + 6})

y^{^{'}} - 2 y = - 5 x^{2}

f (x) = 3 e^{- 4 x}, at x = 0

What is the solution for x^2y^'-2xy=y^4 ?
The solution for x^2y^'-2xy=y^4 is y=(\sqrt[3]{5}x^2)/(\sqrt[3]{-3x^5+c_{1)}}