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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 6 y^{^{'}} + 9 y = e^{3 x} (x^{2} - 4 x + 5)

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

y^{^{''}} - 5 y^{^{'}} + 9 y = x e^{x}

y^{^{''}} - y^{^{'}} - 30 y = 2 x + 4

x^{^{''}} + 4 x = 1

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