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

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

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y^{''}+2y^'+y=(2+1)te^{(-t)}

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

y = c_{1} e^{- t} + c_{2} t e^{- t} + \frac{e ^{- t} t ^{3}}{2}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{- t} + c_{2} t e^{- t} + \frac{e ^{- t} t ^{3}}{2}

y = c_{1} e^{- t} + c_{2} t e^{- t} + \frac{e ^{- t} t ^{3}}{2}

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

y^{^{'''}} - 2 y^{^{''}} + y^{^{'}} = 5 + 4 x e^{x}

1.2 x^{^{''}} + 1600 x^{^{'}} + 100000 = 18

\frac{d ^{2} x}{d t ^{2}} + x = 2 sin (t)

y^{^{''}} + 11 y = cos (3 t)

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