What is the solution for y^{''}+2y^'+y=e^{-t},y(0)=1,y^'(0)=3 ?

The solution for y^{''}+2y^'+y=e^{-t},y(0)=1,y^'(0)=3 is y=e^{-t}+4e^{-t}t+(e^{-t}t^2)/2

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y^{''}+2y^'+y=e^{-t},y(0)=1,y^'(0)=3

y^{^{''}} + 2 y^{^{'}} + y = e^{- t}, y (0) = 1, y^{^{'}} (0) = 3

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

Solution steps

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

Solve linear ODE:

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

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

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

x^{^{''}} + 25 x = 2 cos (5 t)

x^{^{''}} + 9 x = sin (3 t)

4 y^{^{''}} - y = e^{\frac{x}{2}} + 2

y^{^{''}} - 4 y = 6 cos (3 t + 3 0^{\circ})

What is the solution for y^{''}+2y^'+y=e^{-t},y(0)=1,y^'(0)=3 ?
The solution for y^{''}+2y^'+y=e^{-t},y(0)=1,y^'(0)=3 is y=e^{-t}+4e^{-t}t+(e^{-t}t^2)/2