2y^'+4y=-8y^3

Lösung

2 y^{^{'}} + 4 y = - 8 y^{3}

y = \frac{1}{e ^{4 t - c_{1}} - 2}, y = - \frac{1}{e ^{4 t - c_{1}} - 2}

Schritte zur Lösung

2 y^{'} (t) + 4 y = - 8 y^{3}

Solve Bernoulli ODE:

y = \frac{1}{e ^{4 t - c_{1}} - 2}, y = - \frac{1}{e ^{4 t - c_{1}} - 2}

y = \frac{1}{e ^{4 t - c_{1}} - 2}, y = - \frac{1}{e ^{4 t - c_{1}} - 2}

d er i v a t i v e y = f (y, x) (g (x))

\frac{d}{d x} (x y z)

x \to - 0.999 lim (\frac{1}{x ^{2}})

\frac{\partial}{\partial z} (y e^{z} sin (x z))

x \to - \infty lim (5 f (x))