What is the solution for y^{'''}+3y^{''}+y^'+3y=0 ?

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

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

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

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

Solution steps

y^{'''} (t) + 3 y^{''} (t) + y^{'} (t) + 3 y = 0

Solve linear ODE:

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

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

\frac{d ^{2} y}{d x ^{2}} + 3 \frac{d y}{d x} - 4 y = 0

y^{^{''}} + (e + 1) y^{^{'}} + ey = 0

3 y^{^{''}} - y^{^{'}} = 5 y

y^{^{''}} - y^{^{'}} - 42 y = 0, y (0) = 1, y^{^{'}} (0) = - 1

y^{^{''}} - y = 0, y (0) = 0, y^{^{'}} (0) = 5

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