9y^{''}+12y^'+29y=0

Lösung

9 y^{^{''}} + 12 y^{^{'}} + 29 y = 0

y = e^{- \frac{2 t}{3}} (c_{1} cos (\frac{5 t}{3}) + c_{2} sin (\frac{5 t}{3}))

Schritte zur Lösung

9 y^{''} (t) + 12 y^{'} (t) + 29 y = 0

Solve linear ODE:

y = e^{- \frac{2 t}{3}} (c_{1} cos (\frac{5 t}{3}) + c_{2} sin (\frac{5 t}{3}))

y = e^{- \frac{2 t}{3}} (c_{1} cos (\frac{5 t}{3}) + c_{2} sin (\frac{5 t}{3}))

16 y^{^{''}} - 40 y^{^{'}} + 25 y = 0, y (0) = 3, y^{^{'}} (0) = - \frac{9}{4}

y^{^{''}} + 8 y^{^{'}} + 20 y = 0, y (0) = 2, y^{^{'}} (0) = - 7

\frac{d ^{2} y}{d x ^{2}} - 5 y = 0

2 y^{^{''}} + 8 y^{^{'}} + 80 y = 0

y^{^{''}} + 6 y^{^{'}} + 9 y = 0, y (0) = 2, y^{^{'}} (0) = 3