y^{''}-8y^'+9y=0,y(0)=0,y^'(0)=9

Lösung

y^{^{''}} - 8 y^{^{'}} + 9 y = 0, y (0) = 0, y^{^{'}} (0) = 9

y = \frac{9}{2 7} e^{(4 + 7) t} - \frac{9}{2 7} e^{(4 - 7) t}

Schritte zur Lösung

y^{''} (t) - 8 y^{'} (t) + 9 y = 0

Solve linear ODE:

y = \frac{9}{2 7} e^{(4 + 7) t} - \frac{9}{2 7} e^{(4 - 7) t}

y = \frac{9}{2 7} e^{(4 + 7) t} - \frac{9}{2 7} e^{(4 - 7) t}

\frac{\partial}{\partial x} (x^{- 3})

x \to - 1 lim (4 x^{2} - 8 x + 3)

\int e^{- x} cos (3 x) d x

7 t \cdot \frac{d y}{d t} - y = t

x \to 4 + lim (e^{4 - x})