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

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

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

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

Solution steps

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

Solve separable ODE:

x y^{^{'}} - y = x y

d y = (y^{2} - 4) d x

h^{^{'}} = \frac{2 h}{x ln ( x )}

(y^{2} - 1) \frac{d x}{d y} = x

\frac{d y}{d x} + 12 x^{2} y = 5 x^{2}, y (0) = 4

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