What is the solution for y^{''}+3y^'+2y=3t ?

The solution for y^{''}+3y^'+2y=3t is y=c_{1}e^{-t}+c_{2}e^{-2t}+(3t)/2-9/4

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y^{''}+3y^'+2y=3t

y^{^{''}} + 3 y^{^{'}} + 2 y = 3 t

y = c_{1} e^{- t} + c_{2} e^{- 2 t} + \frac{3 t}{2} - \frac{9}{4}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{- t} + c_{2} e^{- 2 t} + \frac{3 t}{2} - \frac{9}{4}

y = c_{1} e^{- t} + c_{2} e^{- 2 t} + \frac{3 t}{2} - \frac{9}{4}

y^{^{''}} - 5 y^{^{'}} + 6 y = 2 t e^{2 t}, y^{^{'}} (0) = - 3

y^{^{''''}} + y = cos (x)

x^{^{''}} + 2 x^{^{'}} - 8 x = 12 e^{2 t} + 8 t - 10

y^{^{''}} - 4 y^{^{'}} + 5 y = 4 e^{3 t}, y (0) = 2, y^{^{'}} (0) = 7

y^{^{'''}} - 2 y^{^{''}} - 5 y^{^{'}} + 6 y = e^{x} + x^{2}

What is the solution for y^{''}+3y^'+2y=3t ?
The solution for y^{''}+3y^'+2y=3t is y=c_{1}e^{-t}+c_{2}e^{-2t}+(3t)/2-9/4