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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 4 y^{^{'}} + 3 y = 6 t e^{3 t}

y^{^{''}} - 9 y^{^{'}} + 14 y = 30 t - 34

y^{^{''}} + y = 6 sin (2 t), y (0) = 3, y^{^{'}} (0) = 1

\frac{d ^{2} y}{d x ^{2}} + 4 y = x

y^{^{''}} + 6 y^{^{'}} + 8 y = e sin (x)

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