What is the solution for y^{''}+4y^'+4y=e^{-t},y(0)=0,y^'(0)=1 ?

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 2 y^{^{'}} - 3 y = 3 t^{2} + 4 t - 5

y^{^{''}} - y = 7.1 cos (t), y (0) = 0, y^{^{'}} (0) = 7.1

y^{^{''}} - 4 y^{^{'}} + 3 y = 8 e^{x}

y^{^{''}} + 3 y = t^{3}, y (0) = 0, y^{^{'}} (0) = 0

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

What is the solution for y^{''}+4y^'+4y=e^{-t},y(0)=0,y^'(0)=1 ?
The solution for y^{''}+4y^'+4y=e^{-t},y(0)=0,y^'(0)=1 is y=-e^{-2t}+e^{-t}