What is the solution for (y-9)^2=y^' ?

The solution for (y-9)^2=y^' is y=-1/(t+c_{1)}+(9t)/(t+c_{1)}+(9c_{1})/(t+c_{1)}

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(y-9)^2=y^'

(y - 9)^{2} = y^{^{'}}

y = - \frac{1}{t + c _{1}} + \frac{9 t}{t + c _{1}} + \frac{9 c _{1}}{t + c _{1}}

Solution steps

(y - 9)^{2} = y^{'} (t)

Solve separable ODE:

y = - \frac{1}{t + c _{1}} + \frac{9 t}{t + c _{1}} + \frac{9 c _{1}}{t + c _{1}}

y = - \frac{1}{t + c _{1}} + \frac{9 t}{t + c _{1}} + \frac{9 c _{1}}{t + c _{1}}

\frac{t}{t ^{2} + 1}

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x^{π}

What is the solution for (y-9)^2=y^' ?
The solution for (y-9)^2=y^' is y=-1/(t+c_{1)}+(9t)/(t+c_{1)}+(9c_{1})/(t+c_{1)}