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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{'''}} - 2 y^{^{''}} - 7 y^{^{'}} - 4 y = 0

y^{^{''}} - 5 y + 2 y = 0

y^{^{''}} - y = 0, y (0) = 5, y^{^{'}} (0) = 1

y^{^{''}} - 10 y^{^{'}} + 25 y = 0, y (0) = 1

y^{^{''}} + 3 y^{^{'}} - 5 y = 0

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