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

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

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

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

y = 4 c_{1} x^{2} - 1, y = - 4 c_{1} x^{2} - 1

Solution steps

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

Solve separable ODE:

y = 4 c_{1} x^{2} - 1, y = - 4 c_{1} x^{2} - 1

y = 4 c_{1} x^{2} - 1, y = - 4 c_{1} x^{2} - 1

y^{^{'}} = y + 5, y (0) = 11 e^{x} - 5

\frac{d y}{d x} = e^{(6 y + 7 x)}, y (0) = 15

50 y^{^{'}} + 20 y^{^{'}} + y = 5000

\frac{d y}{d x} x^{2} - y + y^{2} = - 3

y cos^{2} (x) \frac{d y}{d x} = 2 + tan (x)

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