y^'=1-x-y+xy

Lösung

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

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

Schritte zur Lösung

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

Solve separable ODE:

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

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

\frac{d y}{d x} = 3 x^{5} y

\frac{d y}{d x} = - \frac{3}{4} x^{7} y^{8}

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

y^{^{'}} = 5 y^{- 4}, y (2) = 5

- y^{^{'}} + y = 10