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

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

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y^'=(y^2-3)/((2xy))

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

y = c_{1} x + 3, y = - c_{1} x + 3

Solution steps

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

Solve separable ODE:

y = c_{1} x + 3, y = - c_{1} x + 3

y = c_{1} x + 3, y = - c_{1} x + 3

y^{^{'}} + y \frac{1 + 6 t ^{2}}{2 t + 3 t ^{3}} = 0

\frac{d y}{d t} + 2 y = 7

\frac{d x}{d t} = 2 t 1 - x^{2}

3 x [\frac{d y}{d x}] = 4 x y

\frac{d ^{2} y}{d x ^{2}} - 2 \frac{d y}{d x} + y = 3 x^{2} - 5

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