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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\int \frac{x ^{2}}{( x ^{2} + a ^{2} ) ^{3}} d x

(sin (x) sin (2 x) sin (3 x))^{^{'}}

y = e^{x} sin (2 x)

x \to 0 + lim (x ln (sin (x)))

\int \frac{1}{5 x - 2} d x

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