What is the solution for -3y^'+(2y)/x =x^2y^4 ?

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

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-3y^'+(2y)/x =x^2y^4

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

y = 3 \frac{5 x ^{2}}{x ^{5} + c _{1}}

Solution steps

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

Solve Bernoulli ODE:

y = 3 \frac{5 x ^{2}}{x ^{5} + c _{1}}

y = 3 \frac{5 x ^{2}}{x ^{5} + c _{1}}

\frac{d}{d x} (3 2 x + 3)

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

(e^{2 x} - x^{2} - 1)^{^{'}}

\int_{- 1}^{2} x^{2} + 1 d x

\int u 1 - u^{2} d u

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