What is the solution for y^{'''}+7y^{''}+19y^'+13y=0 ?

The solution for y^{'''}+7y^{''}+19y^'+13y=0 is y=c_{1}e^{-t}+e^{-3t}(c_{2}cos(2t)+c_{3}sin(2t))

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y^{'''}+7y^{''}+19y^'+13y=0

y^{^{'''}} + 7 y^{^{''}} + 19 y^{^{'}} + 13 y = 0

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

Solution steps

y^{'''} (t) + 7 y^{''} (t) + 19 y^{'} (t) + 13 y = 0

Solve linear ODE:

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

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

y^{^{''}} + 6 y^{^{'}} + 9 y = 0, y (0) = 4, y^{^{'}} (0) = - 9

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

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

\frac{d ^{2} y}{d x ^{2}} + 6 \frac{d y}{d x} + 13 y = 0

y^{^{''}} - 4 y^{^{'}} - y = 0

What is the solution for y^{'''}+7y^{''}+19y^'+13y=0 ?
The solution for y^{'''}+7y^{''}+19y^'+13y=0 is y=c_{1}e^{-t}+e^{-3t}(c_{2}cos(2t)+c_{3}sin(2t))