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

The solution for y^'=(2x^4+y^4)/(xy^3) is y=x\sqrt[4]{8ln(x)+c_{1}},y=-x\sqrt[4]{8ln(x)+c_{1}}

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

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

y = x 4 8 ln (x) + c_{1}, y = - x 4 8 ln (x) + c_{1}

Solution steps

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

Solve Bernoulli ODE:

y = x 4 8 ln (x) + c_{1}, y = - x 4 8 ln (x) + c_{1}

y = x 4 8 ln (x) + c_{1}, y = - x 4 8 ln (x) + c_{1}

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

y^{^{'}} = 9 x + y - 9

\int \frac{x}{x - 6} d x

\frac{\partial}{\partial x} (2 y - sin (x))

\frac{d}{d x} (x^{3} e^{4 x})

What is the solution for y^'=(2x^4+y^4)/(xy^3) ?
The solution for y^'=(2x^4+y^4)/(xy^3) is y=x\sqrt[4]{8ln(x)+c_{1}},y=-x\sqrt[4]{8ln(x)+c_{1}}