y^{'''}+4y^{''}+13y^'=0

Lösung

y^{^{'''}} + 4 y^{^{''}} + 13 y^{^{'}} = 0

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

Schritte zur Lösung

y^{'''} (t) + 4 y^{''} (t) + 13 y^{'} (t) = 0

Solve linear ODE:

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

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

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

y^{^{''''}} + 3 y^{^{'''}} + 4 y^{^{''}} + 12 y^{^{'}} = 0

81 y^{^{''}} = - 16 y, y (0) = cos (k t)

25 y^{^{''}} + 10 y^{^{'}} + y = 0, y (0) = 2, y^{^{'}} (0) = 1

\frac{d ^{2} y}{d x ^{2}} + a^{2} y = 0