What is the solution for y^'=(y-50)^2 ?

The solution for y^'=(y-50)^2 is y=-1/(t+c_{1)}+(50t)/(t+c_{1)}+(50c_{1})/(t+c_{1)}

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y^'=(y-50)^2

y^{^{'}} = (y - 50)^{2}

y = - \frac{1}{t + c _{1}} + \frac{50 t}{t + c _{1}} + \frac{50 c _{1}}{t + c _{1}}

Solution steps

y^{'} (t) = (y - 50)^{2}

Solve separable ODE:

y = - \frac{1}{t + c _{1}} + \frac{50 t}{t + c _{1}} + \frac{50 c _{1}}{t + c _{1}}

y = - \frac{1}{t + c _{1}} + \frac{50 t}{t + c _{1}} + \frac{50 c _{1}}{t + c _{1}}

\frac{d v}{d t} = 60 - \frac{7}{100} v, v = 0, t = 0

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

y^{^{'}} = (y - 1)^{2}, y (0) = 1.01

(x + 4) (2 y + 6) d y = x y^{2} d x

y^{3} (x^{4} + 1) \frac{d y}{d x} - x^{3} (y^{4} + 1) = 0

What is the solution for y^'=(y-50)^2 ?
The solution for y^'=(y-50)^2 is y=-1/(t+c_{1)}+(50t)/(t+c_{1)}+(50c_{1})/(t+c_{1)}