What is the solution for x^2y^'+2xy=5y^4 ?

The solution for x^2y^'+2xy=5y^4 is y=\sqrt[3]{(7x)/(15+7c_{1)x^7}}

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x^2y^'+2xy=5y^4

x^{2} y^{^{'}} + 2 x y = 5 y^{4}

Solution steps

x^{2} y^{'} (x) + 2 x y = 5 y^{4}

Solve Bernoulli ODE:

\frac{\partial}{\partial x} (x^{3} + 2^{y} + x^{y})

\frac{\partial}{\partial x} (5 x y^{5} z)

\frac{d}{d x} (x e^{6 x})

- \frac{x ^{2}}{4} + \frac{22}{4}

f (x) = x^{2} + 5, (1, 6)

What is the solution for x^2y^'+2xy=5y^4 ?
The solution for x^2y^'+2xy=5y^4 is y=\sqrt[3]{(7x)/(15+7c_{1)x^7}}