Hallo,
Lösung durch Trennung der Variablen:
y‘(t) = 3 * (1-2y(t)) * y(t)
\( \frac{dy}{dt} \) = 3 * (1-2y(t)) * y(t) |:(1-2y(t) *y(t) *dt
\( \frac{dy}{ y(t)(1-2y(t))} \) = 3 dt --->Lösung durch Partialbruchzerlegung (linkes Integral)
\( \frac{1}{y}-\frac{2}{2 y-1} \) = 3 dt
ln|y| - ln| 2y-1| = 3t+C
ln|\( \frac{y}{2y-1} \)| =3t+C | e hoch
|\( \frac{y}{2y-1} \)| =e^(3t+C) = e^(3t) *e^C
\( \frac{y}{2y-1} \) = e^(3t) * ± e^C ---------->± e^C = C1
\( \frac{y}{2y-1} \) = C1 e^(3t) | *(2y-1)
y= C1 e^(3t) *(2y-1)
y= C1 e^(3t) 2y-C1 e^(3t)
y - C1 e^(3t) 2y = - C1 e^(3t)
y(1 -C1 e^(3t)* 2) = - C1 e^(3t)
y=\( \frac{- C1 e^(3t)}{1 -2C1 e^(3t) } \)
\( y(t)=\frac{C_{1} e^{3 t}}{2 C_{1} e^{3 t}-1} \)