+y^'+y=0

Lösung

+ y^{^{'}} + y = 0

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

Schritte zur Lösung

+ y^{'} (t) + y = 0

Solve separable ODE:

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

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

(x y + x) d x + (y x + y) d y = 0

(e^{2 x} + 1) \frac{d y}{d x} = y

y^{^{'}} = e^{3 x} y^{2}

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

\frac{d x}{d t} = e^{x} t cos (t)