What is the solution for -y^{'''}+2y^{''}+3y^'+3=0 ?

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

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-y^{'''}+2y^{''}+3y^'+3=0

- y^{^{'''}} + 2 y^{^{''}} + 3 y^{^{'}} + 3 = 0

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

Solution steps

- y^{'''} (t) + 2 y^{''} (t) + 3 y^{'} (t) + 3 = 0

Solve linear ODE:

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

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

y^{^{''}} - 6 y^{^{'}} + 9 y = 2 e^{4 x}

y^{^{''}} + 2 y^{^{'}} = 1 + (t - 2), y (0) = 0, y^{^{'}} (0) = 1

4 x^{^{''}} + 85 x = cos (6 t)

y^{^{''}} - 4 y = 4 e^{x}

y^{^{''}} + 4 y = g (x), y (0) = 1, y^{^{'}} (0) = 3

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