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

Lösung

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}}

Schritte zur Lösung

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}

(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

(y + 1) d x + (4 x + 2)^{2} d y = 0