y^{''}+20y^'+36y=0,y^'(0)=0,y(0)=5

Lösung

y^{^{''}} + 20 y^{^{'}} + 36 y = 0, y^{^{'}} (0) = 0, y (0) = 5

y = \frac{45}{8} e^{- 2 t} - \frac{5}{8} e^{- 18 t}

Schritte zur Lösung

y^{''} (t) + 20 y^{'} (t) + 36 y = 0

Solve linear ODE:

y = \frac{45}{8} e^{- 2 t} - \frac{5}{8} e^{- 18 t}

y = \frac{45}{8} e^{- 2 t} - \frac{5}{8} e^{- 18 t}

- 2 y^{^{''}} + 8 y^{^{'}} = 0, y (- 1) = 2, y^{^{'}} (2) = 2

y^{^{''}} + 7 y^{^{'}} + 12 y = 0, y (0) = 1, y^{^{'}} (0) = 9

y^{^{''}} = - y^{^{'}} - y

y^{^{''}} - 3 y^{^{'}} - 4 y = 0, y (0) = 0, y^{^{'}} (0) = 1

y^{^{''}} + y^{^{'}} = 0, y (\frac{π}{4}) = 0, y^{^{'}} (\frac{π}{4}) = - 3