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

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

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y^'= x/(2y)

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

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

Solution steps

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

Solve separable ODE:

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

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

x \frac{d y}{d x} = e^{9 y + x}, x > 0

(1 + y^{2}) cos (x) d x - \frac{1}{x} d y = 0, y (0) = \frac{π}{4}

y^{^{'}} = x^{3} y^{3}

d x - (8 x y + 3 y) d y = 0

(y^{2} + 1) sin (x) d x - y cos (x) d y = 0, y (0) = 0

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