What is the solution for y^{''}+3y^'+2y=20e^{-t} ?

The solution for y^{''}+3y^'+2y=20e^{-t} is y=c_{1}e^{-t}+c_{2}e^{-2t}+20te^{-t}

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y^{''}+3y^'+2y=20e^{-t}

y^{^{''}} + 3 y^{^{'}} + 2 y = 20 e^{- t}

y = c_{1} e^{- t} + c_{2} e^{- 2 t} + 20 t e^{- t}

Solution steps

y^{''} (t) + 3 y^{'} (t) + 2 y = 20 e^{- t}

Solve linear ODE:

y = c_{1} e^{- t} + c_{2} e^{- 2 t} + 20 t e^{- t}

y = c_{1} e^{- t} + c_{2} e^{- 2 t} + 20 t e^{- t}

y^{^{''}} + 9 y = 2 x^{2}

\frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} + 8 y = cos (x)

2 \frac{d ^{2} y}{d x ^{2}} + 1 \frac{d y}{d x} + 1 y = 2

\frac{d ^{2} y}{d t ^{2}} + 3 \frac{d y}{d t} + 6 y = 2 t

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

What is the solution for y^{''}+3y^'+2y=20e^{-t} ?
The solution for y^{''}+3y^'+2y=20e^{-t} is y=c_{1}e^{-t}+c_{2}e^{-2t}+20te^{-t}