What is the solution for (y^')/(y^{1/2)}=-10 ?

The solution for (y^')/(y^{1/2)}=-10 is y=25t^2-5c_{1}t+(c_{1}^2)/4

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(y^')/(y^{1/2)}=-10

\frac{y ^{^{'}}}{y ^{\frac{1}{2}}} = - 10

y = 25 t^{2} - 5 c_{1} t + \frac{c _{1}^{2}}{4}

Solution steps

\frac{y ^{'} ( t )}{y ^{\frac{1}{2}}} = - 10

Solve separable ODE:

y = 25 t^{2} - 5 c_{1} t + \frac{c _{1}^{2}}{4}

y = 25 t^{2} - 5 c_{1} t + \frac{c _{1}^{2}}{4}

(3 x^{2} y) d x + (2 x^{3}) d y = 0

(9 + x) y^{^{'}} = 2 y

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

\frac{d y}{d t} = \frac{t}{y}

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

What is the solution for (y^')/(y^{1/2)}=-10 ?
The solution for (y^')/(y^{1/2)}=-10 is y=25t^2-5c_{1}t+(c_{1}^2)/4