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

Lösung

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

y = e^{- \frac{x ^{2}}{2} + x} - 1

Schritte zur Lösung

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

Solve separable ODE:

y = e^{- \frac{x ^{2}}{2} + x} - 1

y = e^{- \frac{x ^{2}}{2} + x} - 1

5 y^{^{'}} = y cos (x), y (0) = 2

\frac{d v}{d t} = 9.81 - 3 v

y^{^{'}} = x y^{2} sin (x^{2})

\frac{d v}{d t} = 2 - 4 v

\frac{d r}{d s} = e^{r + s}