y^'+y=2,y(-2)=1

Lösung

y^{^{'}} + y = 2, y (- 2) = 1

y = - e^{- t - 2} + 2

Schritte zur Lösung

y^{'} (t) + y = 2

Solve separable ODE:

y = - e^{- t - 2} + 2

y = - e^{- t - 2} + 2

\frac{df}{d x} = - 4 x f (x) (x)^{2}

y^{^{'}} = 1 + \frac{( y ) ^{2}}{4}, y (0) = 0.5

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

y^{^{'}} = - 8 y + 16

\frac{d y}{d x} = 0.5 (y e^{x})^{\frac{1}{3}}