x^'+cx=a

Lösung

x^{^{'}} + c x = a

x = - \frac{e ^{- c t - c_{1}}}{c} + \frac{a}{c}

Schritte zur Lösung

x^{'} (t) + c x = a

Solve separable ODE:

x = - \frac{e ^{- c t - c_{1}}}{c} + \frac{a}{c}

x = - \frac{e ^{- c t - c_{1}}}{c} + \frac{a}{c}

\frac{d p}{d t} = c (1 - p)

\frac{d p}{d t} = t^{2} p - 5 p + t^{2} - 5

y^{^{'}} = \frac{2}{x} y

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

5 \frac{d Q}{d t} + \frac{1}{0.05} Q = 80