(x+1)y^'-y=0

Lösung

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

y = e^{c_{1}} x + e^{c_{1}}

Schritte zur Lösung

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

Solve separable ODE:

y = e^{c_{1}} x + e^{c_{1}}

y = e^{c_{1}} x + e^{c_{1}}

\frac{d V}{d t} = 0.5 (50 - V), V < 50

y = a y^{^{'}}

\frac{d y}{d x} + 2 y x = x

y^{^{'}} = y \cdot t

\frac{d y}{d x} y = (x^{3} - x^{2})^{4}