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}}

Solution steps

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

Solve Bernoulli ODE:

\int_{1}^{2} \frac{5 ln ( x )}{x ^{2}} d x

\frac{\partial}{\partial x} (t)

\frac{d}{d x} (\frac{( 3 x + 1 )}{tan ( x )})

2 x

f (x) = \frac{x}{7 - 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}}