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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 7 y^{^{'}} + 10 y = 9 cos (t) + 7 sin (t)

y^{^{''}} - y = cosh (x), y (0) = 2, y^{^{'}} (0) = 12

y^{^{''}} - y^{^{'}} - 6 y = 2 e^{4 t}

y^{^{''}} + 2 y^{^{'}} + 2 y = 3, y (0) = 1, y^{^{'}} (0) = 0

y^{^{''}} + 5 y^{^{'}} + 6 y = e^{- x}

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