5y^{''}+40y^'+80y=0

Lösung

5 y^{^{''}} + 40 y^{^{'}} + 80 y = 0

y = c_{1} e^{- 4 t} + c_{2} t e^{- 4 t}

Schritte zur Lösung

5 y^{''} (t) + 40 y^{'} (t) + 80 y = 0

Solve linear ODE:

y = c_{1} e^{- 4 t} + c_{2} t e^{- 4 t}

y = c_{1} e^{- 4 t} + c_{2} t e^{- 4 t}

x \to 9 lim (5 x + 3)

\int x d x

y^{^{'}} + 2 y = 4 t

\frac{\partial}{\partial y} (4 x^{3 y})

3 y^{^{'}} + 4 y = e^{- x}