What is the solution for 1-1/(y^2-4y+4)y^'=0 ?

The solution for 1-1/(y^2-4y+4)y^'=0 is y=-1/(t+c_{1)}+(2t)/(t+c_{1)}+(2c_{1})/(t+c_{1)}

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1-1/(y^2-4y+4)y^'=0

1 - \frac{1}{y ^{2} - 4 y + 4} y^{^{'}} = 0

y = - \frac{1}{t + c _{1}} + \frac{2 t}{t + c _{1}} + \frac{2 c _{1}}{t + c _{1}}

Solution steps

1 - \frac{1}{y ^{2} - 4 y + 4} y^{'} (t) = 0

Solve separable ODE:

y = - \frac{1}{t + c _{1}} + \frac{2 t}{t + c _{1}} + \frac{2 c _{1}}{t + c _{1}}

y = - \frac{1}{t + c _{1}} + \frac{2 t}{t + c _{1}} + \frac{2 c _{1}}{t + c _{1}}

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

\frac{d x}{d t} = 4 (x^{2} + 1), x (\frac{π}{4}) = 1

(4 x - 8) d x - (6 y - 10) d y = 0

\frac{d x}{d t} = 80 - \frac{x}{100}

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

What is the solution for 1-1/(y^2-4y+4)y^'=0 ?
The solution for 1-1/(y^2-4y+4)y^'=0 is y=-1/(t+c_{1)}+(2t)/(t+c_{1)}+(2c_{1})/(t+c_{1)}