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

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

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

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

y = 4 e^{- 2 t} t + 2 e^{- t}

Solution steps

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

Solve linear ODE:

y = 4 e^{- 2 t} t + 2 e^{- t}

y = 4 e^{- 2 t} t + 2 e^{- t}

y^{^{''}} + 3 y^{^{'}} + 2 y = \frac{1}{6 + e ^{x}}

y^{^{''}} + 3 y^{^{'}} + 2 y = \frac{1}{5 + e ^{x}}

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

y^{^{''}} + 2 y^{^{'}} - 6 x - 3 = 0

y^{^{''}} + 3 y^{^{'}} = 2

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