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

The solution for y^{''}+2y^'+2y=e^{-t} is y=e^{-t}(c_{1}cos(t)+c_{2}sin(t))+e^{-t}

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

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

y = e^{- t} (c_{1} cos (t) + c_{2} sin (t)) + e^{- t}

Solution steps

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

Solve linear ODE:

y = e^{- t} (c_{1} cos (t) + c_{2} sin (t)) + e^{- t}

y = e^{- t} (c_{1} cos (t) + c_{2} sin (t)) + e^{- t}

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

y^{^{''}} - 4 y^{^{'}} + 4 y = (x + 1) e^{(2 x)}

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

y^{^{''}} - a^{2} y = - 2 sin (π \cdot x) - 4 sin (3 \cdot π \cdot x)

(d - 2) y = e^{2} x

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