What is the solution for y^'=2(t+1)sqrt(y-1) ?

The solution for y^'=2(t+1)sqrt(y-1) is y=(t^4+c_{1}^2)/4+(c_{1}t^2)/2+t^3+c_{1}t+t^2+1

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y^'=2(t+1)sqrt(y-1)

y^{^{'}} = 2 (t + 1) y - 1

y = \frac{t ^{4} + c _{1}^{2}}{4} + \frac{c _{1} t ^{2}}{2} + t^{3} + c_{1} t + t^{2} + 1

Solution steps

y^{'} (t) = 2 (t + 1) y - 1

Solve separable ODE:

y = \frac{t ^{4} + c _{1}^{2}}{4} + \frac{c _{1} t ^{2}}{2} + t^{3} + c_{1} t + t^{2} + 1

y = \frac{t ^{4} + c _{1}^{2}}{4} + \frac{c _{1} t ^{2}}{2} + t^{3} + c_{1} t + t^{2} + 1

\frac{d v}{d t} = 32.2 - \frac{1}{40} v

\frac{d P}{d t} = (0.1) P - (0.05) P (t)^{2}

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

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

\frac{d y}{d x} = \frac{3 - y}{x ^{2} - 3 x + 2}

What is the solution for y^'=2(t+1)sqrt(y-1) ?
The solution for y^'=2(t+1)sqrt(y-1) is y=(t^4+c_{1}^2)/4+(c_{1}t^2)/2+t^3+c_{1}t+t^2+1