y^'=(2+y)(2-y),y(0)=4

Lösung

y^{^{'}} = (2 + y) (2 - y), y (0) = 4

y = - \frac{6 e ^{4 t}}{1 - 3 e ^{4 t}} - \frac{2}{1 - 3 e ^{4 t}}

Schritte zur Lösung

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

Solve separable ODE:

y = - \frac{6 e ^{4 t}}{1 - 3 e ^{4 t}} - \frac{2}{1 - 3 e ^{4 t}}

y = - \frac{6 e ^{4 t}}{1 - 3 e ^{4 t}} - \frac{2}{1 - 3 e ^{4 t}}

y^{^{'}} + x y = 2 x

(e^{y} - 1) y^{^{'}} = 2 + cos (x)

\frac{d y}{d x} = 8 x^{7} y

(x + 3) d y - \frac{1}{y ^{2}} d x = 0

y^{^{'}} = 4 y + 3