y^'+ky=ka

Lösung

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

y = - e^{- k t - c_{1}} + a

Schritte zur Lösung

y^{'} (t) + k y = ka

Solve separable ODE:

y = - e^{- k t - c_{1}} + a

y = - e^{- k t - c_{1}} + a

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

y^{^{'}} - 3 t y = 5 t

x y^{^{'}} - 9 y^{2} = 0

x \cdot y^{^{'}} - 2 y \cdot ln (x) = 0

\frac{d y}{d x} = sin (x) e^{2 y}