What is the solution for y^'+1/x y= 2/3 x^2y^2 ?

The solution for y^'+1/x y= 2/3 x^2y^2 is y= 3/(x(-x^2+c_{1))}

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y^'+1/x y= 2/3 x^2y^2

y^{^{'}} + \frac{1}{x} y = \frac{2}{3} x^{2} y^{2}

y = \frac{3}{x ( - x ^{2} + c _{1} )}

Solution steps

y^{'} (x) + \frac{1}{x} y = \frac{2}{3} x^{2} y^{2}

Solve Bernoulli ODE:

y = \frac{3}{x ( - x ^{2} + c _{1} )}

y = \frac{3}{x ( - x ^{2} + c _{1} )}

\int sec^{6} (x) d x

x \to π + lim (cot (x))

\frac{d ^{2}}{d x ^{2}} (5 x^{\frac{3}{2}})

\frac{r}{r ^{2} + 6}

\frac{d}{d x} (e^{\frac{9}{x}})

What is the solution for y^'+1/x y= 2/3 x^2y^2 ?
The solution for y^'+1/x y= 2/3 x^2y^2 is y= 3/(x(-x^2+c_{1))}