y^'+2y+6=0

Lösung

y^{^{'}} + 2 y + 6 = 0

y = e^{- 2 t + c_{1}} - 3

Schritte zur Lösung

y^{'} (t) + 2 y + 6 = 0

Solve separable ODE:

y = e^{- 2 t + c_{1}} - 3

y = e^{- 2 t + c_{1}} - 3

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

\frac{d y}{d x} = \frac{4 x}{y - 1}

\frac{d y}{d x} = cos (x) tan (y)

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

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