What is the solution for y^{'''}+6y^{''}+9y^'+4y=0 ?

The solution for y^{'''}+6y^{''}+9y^'+4y=0 is y=c_{1}e^{-4t}+c_{2}e^{-t}+c_{3}te^{-t}

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y^{'''}+6y^{''}+9y^'+4y=0

y^{^{'''}} + 6 y^{^{''}} + 9 y^{^{'}} + 4 y = 0

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

Solution steps

y^{'''} (t) + 6 y^{''} (t) + 9 y^{'} (t) + 4 y = 0

Solve linear ODE:

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

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

y^{^{''}} - 5 y^{^{'}} + 4 y = 0, y (0) = 6, y^{^{'}} (0) = 2

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

y^{^{''}} + 0.8 y^{^{'}} + 100 y = 0, y (0) = 0, y^{^{'}} (0) = 4

18 y^{^{''}} + 72 y = 0

10 x^{^{''}} + \frac{1}{90} x = 0, x (0) = 100, x^{^{'}} (0) = 25

What is the solution for y^{'''}+6y^{''}+9y^'+4y=0 ?
The solution for y^{'''}+6y^{''}+9y^'+4y=0 is y=c_{1}e^{-4t}+c_{2}e^{-t}+c_{3}te^{-t}