What is the solution for y^{'''}+3y^{''}+4y^'+12y=0 ?

The solution for y^{'''}+3y^{''}+4y^'+12y=0 is y=c_{1}e^{-3t}+c_{2}cos(2t)+c_{3}sin(2t)

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

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

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

Solution steps

y^{'''} (t) + 3 y^{''} (t) + 4 y^{'} (t) + 12 y = 0

Solve linear ODE:

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

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

y^{^{''}} + 4 y^{^{'}} + 5 y = 0

y^{^{''}} + 4 y = 0, y (\frac{π}{4}) = - 2

x^{^{''}} + \frac{1}{4} x^{^{'}} + x = 0

\frac{d ^{2} x}{d t ^{2}} = - k \frac{d x}{d t}

x^{^{''}} + \frac{32}{2} x = 0, x (0) = - 1, x^{^{'}} (0) = - 2

What is the solution for y^{'''}+3y^{''}+4y^'+12y=0 ?
The solution for y^{'''}+3y^{''}+4y^'+12y=0 is y=c_{1}e^{-3t}+c_{2}cos(2t)+c_{3}sin(2t)