What is the solution for 1/(sqrt(y))y^'=sqrt(b) ?

The solution for 1/(sqrt(y))y^'=sqrt(b) is y=(bt^2)/4+(c_{1}sqrt(b)t)/2+(c_{1}^2)/4

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1/(sqrt(y))y^'=sqrt(b)

\frac{1}{y} y^{^{'}} = b

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

Solution steps

\frac{1}{y} y^{'} (t) = b

Solve separable ODE:

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

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

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

\frac{( t ^{2} )}{x} d t = d x

\frac{d y}{d x} y = \frac{4 sin ( x )}{9 - 8 sin ( x )}

\frac{d y}{d x} y = - 3 x^{7} + \frac{3}{5} x^{4} + 568 x + 12

y y^{^{'}} = x e^{x^{2} + y}

What is the solution for 1/(sqrt(y))y^'=sqrt(b) ?
The solution for 1/(sqrt(y))y^'=sqrt(b) is y=(bt^2)/4+(c_{1}sqrt(b)t)/2+(c_{1}^2)/4