What is the solution for y^{''}+3y^'+2y=12e^t ?

The solution for y^{''}+3y^'+2y=12e^t is y=c_{1}e^{-t}+c_{2}e^{-2t}+2e^t

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y^{''}+3y^'+2y=12e^t

y^{^{''}} + 3 y^{^{'}} + 2 y = 12 e^{t}

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

Solution steps

y^{''} (t) + 3 y^{'} (t) + 2 y = 12 e^{t}

Solve linear ODE:

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

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

D^{2} y = lo g_{10} (x)

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

(D^{2} - 2 D + 3) y = x^{3} + sin (x)

\frac{d ^{2} y}{d x ^{2}} + 9 y = sec (3 x)

y^{^{''}} + 2 y^{^{'}} + 5 y = e^{x} cos (2 x)

What is the solution for y^{''}+3y^'+2y=12e^t ?
The solution for y^{''}+3y^'+2y=12e^t is y=c_{1}e^{-t}+c_{2}e^{-2t}+2e^t