What is the solution for x^2(dy)/(dx)-2xy=6y^4,y(1)= 1/3 ?

The solution for x^2(dy)/(dx)-2xy=6y^4,y(1)= 1/3 is y=(\sqrt[3]{5}x^2)/(\sqrt[3]{9(-2x^5+17))}

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x^2(dy)/(dx)-2xy=6y^4,y(1)= 1/3

x^{2} \frac{d y}{d x} - 2 x y = 6 y^{4}, y (1) = \frac{1}{3}

y = \frac{3 5 x ^{2}}{3 9 ( - 2 x ^{5} + 17 )}

Solution steps

x^{2} \frac{d y}{d x} - 2 x y = 6 y^{4}

Solve Bernoulli ODE:

y = \frac{3 5 x ^{2}}{3 9 ( - 2 x ^{5} + 17 )}

y = \frac{3 5 x ^{2}}{3 9 ( - 2 x ^{5} + 17 )}

x \to \infty lim (7 x)

\int \frac{1}{x + 7} d x

\int_{4}^{5} \frac{x}{8} d x

\int ln (a + b x) d x

\frac{1}{( s - 1 ) ^{2} - 4}

What is the solution for x^2(dy)/(dx)-2xy=6y^4,y(1)= 1/3 ?
The solution for x^2(dy)/(dx)-2xy=6y^4,y(1)= 1/3 is y=(\sqrt[3]{5}x^2)/(\sqrt[3]{9(-2x^5+17))}