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

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

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

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

y = x 4 - 1 + c_{1} x^{4}, y = - x 4 - 1 + c_{1} x^{4}

Solution steps

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

Solve Bernoulli ODE:

y = x 4 - 1 + c_{1} x^{4}, y = - x 4 - 1 + c_{1} x^{4}

y = x 4 - 1 + c_{1} x^{4}, y = - x 4 - 1 + c_{1} x^{4}

\frac{\partial}{\partial x} (2 k^{3} s^{2})

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

x \to 0 + lim (\frac{x ^{2} - 15}{x})

x \to - 0 lim (\frac{1}{x ^{2}})

3 (x^{3} + 1)^{2}

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