What is the solution for (y^')/(y^{1/2)}=-u ?

The solution for (y^')/(y^{1/2)}=-u is y=(u^2t^2)/4-(c_{1}ut)/2+(c_{1}^2)/4

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(y^')/(y^{1/2)}=-u

\frac{y ^{^{'}}}{y ^{\frac{1}{2}}} = - u

y = \frac{u ^{2} t ^{2}}{4} - \frac{c _{1} u t}{2} + \frac{c _{1}^{2}}{4}

Solution steps

\frac{y ^{'} ( t )}{y ^{\frac{1}{2}}} = - u

Solve separable ODE:

y = \frac{u ^{2} t ^{2}}{4} - \frac{c _{1} u t}{2} + \frac{c _{1}^{2}}{4}

y = \frac{u ^{2} t ^{2}}{4} - \frac{c _{1} u t}{2} + \frac{c _{1}^{2}}{4}

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

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

y^{^{'}} = - 3 y^{\frac{4}{3}} sin (t)

\frac{d y}{d x} = \frac{x + 1}{e ^{y}}

\frac{d y}{d x} = - \frac{8}{9} x^{2} y^{6}

What is the solution for (y^')/(y^{1/2)}=-u ?
The solution for (y^')/(y^{1/2)}=-u is y=(u^2t^2)/4-(c_{1}ut)/2+(c_{1}^2)/4