y^{''}+2y^'+9y=0

Lösung

y^{^{''}} + 2 y^{^{'}} + 9 y = 0

y = e^{- t} (c_{1} cos (2^{1 + \frac{1}{2}} t) + c_{2} sin (2^{1 + \frac{1}{2}} t))

Schritte zur Lösung

y^{''} (t) + 2 y^{'} (t) + 9 y = 0

Solve linear ODE:

y = e^{- t} (c_{1} cos (2^{1 + \frac{1}{2}} t) + c_{2} sin (2^{1 + \frac{1}{2}} t))

y = e^{- t} (c_{1} cos (2^{1 + \frac{1}{2}} t) + c_{2} sin (2^{1 + \frac{1}{2}} t))

y^{^{''}} - 2 y^{^{'}} - 2 y = 0, y (0) = 0, y^{^{'}} (0) = 3

2 y^{^{''}} - 3 y + y = 0

y^{^{''}} - cy = 0

y^{^{''}} - 13 y^{^{'}} + 40 y = 0

4 y^{^{''}} + 3 y^{^{'}} = 0, y (0) = - 2, y^{^{'}} (0) = 3