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

The solution for v^{''}+2v^'+v=e^{-2t} is v(t)=c_{1}e^{-t}+c_{2}te^{-t}+e^{-2t}

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v^{''}+2v^'+v=e^{-2t}

v^{^{''}} + 2 v^{^{'}} + v = e^{- 2 t}

v (t) = c_{1} e^{- t} + c_{2} t e^{- t} + e^{- 2 t}

Solution steps

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

Solve linear ODE:

v (t) = c_{1} e^{- t} + c_{2} t e^{- t} + e^{- 2 t}

v (t) = c_{1} e^{- t} + c_{2} t e^{- t} + e^{- 2 t}

y^{^{''}} + y^{^{'}} - 2 y = 4

y^{^{''}} + y = 1, y (0) = 6, y^{^{'}} (0) = 0

y^{^{''}} + 2 y^{^{'}} + 4 y = e^{- 2 t}, y (0) = 1, y^{^{'}} (0) = 0

y^{^{''}} - 4 y = e^{3 x} + x sin (2 x)

y^{^{''}} - 2 y^{^{'}} + 1 = 2 x

What is the solution for v^{''}+2v^'+v=e^{-2t} ?
The solution for v^{''}+2v^'+v=e^{-2t} is v(t)=c_{1}e^{-t}+c_{2}te^{-t}+e^{-2t}