What is the solution for y^{'''}+8y^{''}+19y^'+12y=0 ?

The solution for y^{'''}+8y^{''}+19y^'+12y=0 is y=c_{1}e^{-t}+c_{2}e^{-3t}+c_{3}e^{-4t}

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y^{'''}+8y^{''}+19y^'+12y=0

y^{^{'''}} + 8 y^{^{''}} + 19 y^{^{'}} + 12 y = 0

y = c_{1} e^{- t} + c_{2} e^{- 3 t} + c_{3} e^{- 4 t}

Solution steps

y^{'''} (t) + 8 y^{''} (t) + 19 y^{'} (t) + 12 y = 0

Solve linear ODE:

y = c_{1} e^{- t} + c_{2} e^{- 3 t} + c_{3} e^{- 4 t}

y = c_{1} e^{- t} + c_{2} e^{- 3 t} + c_{3} e^{- 4 t}

y^{^{''}} + 25 y = 0, y (0) = 0, y^{^{'}} (0) = 4

81 y^{^{''}} + 108 y^{^{'}} - 13 y = 0, y (0) = 1, y^{^{'}} (0) = 0

x^{^{''}} + 0.5 x^{^{'}} + 4 x = 0

x^{^{''}} - c^{2} x = 0

y^{^{''}} - 8 y^{^{'}} + 65 y = 0, y (0) = 1, y^{^{'}} (0) = 2

What is the solution for y^{'''}+8y^{''}+19y^'+12y=0 ?
The solution for y^{'''}+8y^{''}+19y^'+12y=0 is y=c_{1}e^{-t}+c_{2}e^{-3t}+c_{3}e^{-4t}