x^'=-2x+x

Lösung

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

x = e^{- t + c_{1}}

Schritte zur Lösung

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

Solve separable ODE:

x = e^{- t + c_{1}}

x = e^{- t + c_{1}}

\frac{d y}{d x} + x^{3} y = 4 x^{3}

\frac{d}{d x} (2 x^{3} + y^{2}) = 6

4 x - 2 y \frac{d y}{d x} = 0

y^{^{'}} + k y = ka

\frac{d y}{d t} = 2 y (1 - \frac{y}{5})