y+y^'-2y=0

Lösung

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

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

Schritte zur Lösung

y + y^{'} (t) - 2 y = 0

Solve separable ODE:

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

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

(2 x y) d x = (3 - x^{2}) d y

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

\frac{d y}{d x} = 8 y (- 3 - y)

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

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