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}}

Solution steps

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

Solve Bernoulli ODE:

\frac{d}{d x} (x^{3} - cos (x))

\int - 7 sin (x) d x

- \frac{7}{x ^{3}}

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

(2 x + 2 x^{- 2} y) d x + (x y - x^{- 1}) d y = 0

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}}