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

The solution for y^'=(3x^4+y^4)/(xy^3) is y=x\sqrt[4]{12ln(x)+c_{1}},y=-x\sqrt[4]{12ln(x)+c_{1}}

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

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

y = x 4 12 ln (x) + c_{1}, y = - x 4 12 ln (x) + c_{1}

Solution steps

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

Solve Bernoulli ODE:

y = x 4 12 ln (x) + c_{1}, y = - x 4 12 ln (x) + c_{1}

y = x 4 12 ln (x) + c_{1}, y = - x 4 12 ln (x) + c_{1}

f (x) = \frac{1}{5 + 3 x}, (- 1, \frac{1}{2})

\frac{d}{d x} (π 5 x cos (x))

\frac{d}{d x} (C e^{x})

\frac{\partial}{\partial x} (y ln (x^{2} + y^{2} + 4))

\frac{d}{d x} (x^{3} - \frac{3}{x} + e^{2})

What is the solution for y^'=(3x^4+y^4)/(xy^3) ?
The solution for y^'=(3x^4+y^4)/(xy^3) is y=x\sqrt[4]{12ln(x)+c_{1}},y=-x\sqrt[4]{12ln(x)+c_{1}}