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

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

ln (x) y^{^{'}} - (\frac{2 \cdot y}{x}) = 0

\frac{d x}{d y} = sin (5 x)

\frac{d y}{d x} y = ln (ln (ln (x)))

\frac{d y}{d t} = 2 t cos^{2} (y), y (0) = \frac{π}{4}

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}