2x^'+8x=0

Lösung

2 x^{^{'}} + 8 x = 0

x = e^{- 4 t + c_{1}}

Schritte zur Lösung

2 x^{'} (t) + 8 x = 0

Solve separable ODE:

x = e^{- 4 t + c_{1}}

x = e^{- 4 t + c_{1}}

y = x \cdot \frac{d y}{d x}

\frac{d y}{d x} = 3 y, y = e^{3 x}

y^{^{'}} = e^{- y} (\frac{x ^{5}}{7})

y^{^{'}} + y = \frac{1}{e}

(x + 1) (y - 1) d x = (x - 1) (y + 1) d y