P^'=kP(t)

Lösung

P^{^{'}} = k P

P (t) = e^{k t + c_{1}}

Schritte zur Lösung

P^{'} (t) = k P (t)

Solve separable ODE:

P (t) = e^{k t + c_{1}}

P (t) = e^{k t + c_{1}}

x y^{^{'}} y + 4 = 0

y^{^{'}} = \frac{cos ( x )}{sin ( 2 y )}

\frac{d y}{d x} = 12 x y

y^{^{'}} - a \cdot y = 0

y \frac{d y}{d x} = 9 x (y^{2} + 5)^{3}