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

The solution for x^2(dy)/(dx)-2xy=4y^4,y(1)= 1/3 is y=(\sqrt[3]{5}x^2)/(\sqrt[3]{3(-4x^5+49))}

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

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

Solution steps

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

Solve Bernoulli ODE:

x y^{^{'}} + y^{2} + y = 0

f (x) = 3 x^{2} + 2 x

f (x) = sin^{2} (x) + cos^{2} (x)

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

\frac{d}{d t} (s t^{2})

What is the solution for x^2(dy)/(dx)-2xy=4y^4,y(1)= 1/3 ?
The solution for x^2(dy)/(dx)-2xy=4y^4,y(1)= 1/3 is y=(\sqrt[3]{5}x^2)/(\sqrt[3]{3(-4x^5+49))}