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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''''}} - 4 y^{^{'''}} + 4 y^{^{''}} = sin (x)

y^{^{'''}} - 25 y^{^{'}} = t e^{- 5 t} + 6 cos (5 t)

y^{^{''}} - 2 y^{^{'}} + y = \frac{e ^{t}}{t ^{2} + 1}

y^{^{'''}} + y^{^{''}} + y^{^{'}} + y = e^{- t} + 13 t

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

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