What is the solution for x^2y^'-2xy=5y^4,y(1)= 1/4 ?

The solution for x^2y^'-2xy=5y^4,y(1)= 1/4 is y=(x^2)/(\sqrt[3]{-3x^5+67)}

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x^2y^'-2xy=5y^4,y(1)= 1/4

x^{2} y^{^{'}} - 2 x y = 5 y^{4}, y (1) = \frac{1}{4}

y = \frac{x ^{2}}{3 - 3 x ^{5} + 67}

Solution steps

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

Solve Bernoulli ODE:

y = \frac{x ^{2}}{3 - 3 x ^{5} + 67}

y = \frac{x ^{2}}{3 - 3 x ^{5} + 67}

\int_{1}^{2} \frac{y ^{3}}{2} + \frac{1}{2 y ^{3}} d y

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

\int_{0}^{\frac{π}{4}} cos^{2} (\frac{x}{3}) d x

\int_{1}^{2} (x^{2} + 2 x) d x

f (x) = x^{2} + x, at x = - 1, 2

What is the solution for x^2y^'-2xy=5y^4,y(1)= 1/4 ?
The solution for x^2y^'-2xy=5y^4,y(1)= 1/4 is y=(x^2)/(\sqrt[3]{-3x^5+67)}