y=1+y^'

Lösung

y = 1 + y^{^{'}}

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

Schritte zur Lösung

y = 1 + y^{'} (t)

Solve separable ODE:

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

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

\frac{d P}{d t} = P (1 - \frac{P}{1800})

sec (x) y^{^{'}} + y = 2

\frac{d y}{d x} = (1 - y) \cdot cos (x)

(2 x + 3) y^{^{'}} = x y (2 x + 3)^{\frac{1}{2}}

1 + y^{2} - x y^{^{'}} = 0