What is the solution for x^3yy^'+x^2y^2=y^5 ?

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

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x^3yy^'+x^2y^2=y^5

x^{3} y y^{^{'}} + x^{2} y^{2} = y^{5}

Solution steps

x^{3} y y^{'} (x) + x^{2} y^{2} = y^{5}

Solve Bernoulli ODE:

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

\int 13 e^{u} d u

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

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

\int \frac{x ^{2} - 12}{x ^{3} + 4 x} d x

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