y^'=-y^3,y(0)=1

Lösung

y^{^{'}} = - y^{3}, y (0) = 1

y = \frac{1}{2 t + 1}

Schritte zur Lösung

y^{'} (t) = - y^{3}

Solve separable ODE:

y = \frac{1}{2 t + 1}

y = \frac{1}{2 t + 1}

\int \frac{1}{x ^{2}} + x d x

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

x \to 2 - lim (3 x + 6)

\frac{\partial}{\partial y} (x cos (x y))

\frac{\partial}{\partial y} (x^{2} + x y - 25)