What is the solution for y^{''}+4y^'+3y=2e^{-2t} ?

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

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y^{''}+4y^'+3y=2e^{-2t}

y^{^{''}} + 4 y^{^{'}} + 3 y = 2 e^{- 2 t}

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

Solution steps

y^{''} (t) + 4 y^{'} (t) + 3 y = 2 e^{- 2 t}

Solve linear ODE:

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

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

y^{^{''}} - 10 y^{^{'}} + 16 y = 17 sin (2 x)

y^{^{''}} + 25 y = 12 cos (5 x) + 18 sin (5 x)

y^{^{''}} + 8 y^{^{'}} + 7 y = 16 x^{2}

y^{^{''}} - 6 y^{^{'}} + 9 y = x e^{3 x}

y^{^{''}} + 3 y = - 48 x^{2} e^{3} x

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