What is the solution for y^{'''}+6y^{''}+10y^'+8y=0 ?

The solution for y^{'''}+6y^{''}+10y^'+8y=0 is y=c_{1}e^{-4t}+e^{-t}(c_{2}cos(t)+c_{3}sin(t))

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y^{'''}+6y^{''}+10y^'+8y=0

y^{^{'''}} + 6 y^{^{''}} + 10 y^{^{'}} + 8 y = 0

y = c_{1} e^{- 4 t} + e^{- t} (c_{2} cos (t) + c_{3} sin (t))

Solution steps

y^{'''} (t) + 6 y^{''} (t) + 10 y^{'} (t) + 8 y = 0

Solve linear ODE:

y = c_{1} e^{- 4 t} + e^{- t} (c_{2} cos (t) + c_{3} sin (t))

y = c_{1} e^{- 4 t} + e^{- t} (c_{2} cos (t) + c_{3} sin (t))

y^{^{''}} + 2 y^{^{'}} + 26 y = 0, y (0) = 1, y^{^{'}} (0) = 4

4 y^{^{''}} - 9 y = 0, y (0) = 0, y^{^{'}} (0) = 3

y^{^{'''}} - 2 y^{^{''}} - 21 y^{^{'}} - 18 y = 0

4 x^{^{''}} + 8 x^{^{'}} + 4 x = 0

\frac{d ^{4} y}{d x ^{4}} - 16 y = 0

What is the solution for y^{'''}+6y^{''}+10y^'+8y=0 ?
The solution for y^{'''}+6y^{''}+10y^'+8y=0 is y=c_{1}e^{-4t}+e^{-t}(c_{2}cos(t)+c_{3}sin(t))