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

The solution for y^{'''}+3y^{''}+2y^'=u is y=c_{1}+c_{2}e^{-t}+c_{3}e^{-2t}+(ut)/2

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y^{'''}+3y^{''}+2y^'=u

y^{^{'''}} + 3 y^{^{''}} + 2 y^{^{'}} = u

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 2 y^{^{'}} + 2 y = 2 x^{2} - 2 x + 1 + sin (x)

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

y^{^{''}} + 4 y^{^{'}} + 4 y = 3 x + 6

y^{^{''}} - 2 y^{^{'}} - 8 y = 3 e^{3 x}

y^{^{''}} + 3 y^{^{'}} - 5 y = (4 x^{2} - 7 x + 0) e^{4 x}

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