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

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

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

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

Solution steps

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

Solve separable ODE:

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

\int \frac{1}{v ^{2} - 4} d v

\int (2 x + 5) d x

\frac{\partial}{\partial x} (cos (x z))

4 y^{^{''}} + y = - 4 cos (\frac{x}{2}) - 8 x sin (\frac{x}{2})

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