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

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

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

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

y = \frac{x ^{2}}{2} + 3 x + c_{1}, y = - \frac{x ^{2}}{2} + 3 x + c_{1}

Solution steps

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

Solve separable ODE:

y = \frac{x ^{2}}{2} + 3 x + c_{1}, y = - \frac{x ^{2}}{2} + 3 x + c_{1}

y = \frac{x ^{2}}{2} + 3 x + c_{1}, y = - \frac{x ^{2}}{2} + 3 x + c_{1}

(x y + 2 y + x + 2) d y - (x^{2} - 4) d x = 0

y^{^{'}} = \frac{3 x ^{2} + 4 x + 2}{2 y + 1}

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

y^{^{'}} = e^{t + y}, y (ln (2)) = 1

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

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