2y^{''}-8y=0

Lösung

2 y^{^{''}} - 8 y = 0

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

Schritte zur Lösung

2 y^{''} (t) - 8 y = 0

Solve linear ODE:

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

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

\frac{d}{d x} (- e^{x} (x - 7))

\frac{\partial}{\partial y} (- 10 y sin (x y))

d er i v a t i v e y = x^{x^{2}}

\frac{\partial}{\partial x} (- x cos (y))

\int 3 cos (x) - \frac{1}{x} d x