What is the solution for y^{''}+2y^'+y= t/(e^t) ?

The solution for y^{''}+2y^'+y= t/(e^t) is y=c_{1}e^{-t}+c_{2}te^{-t}+(e^{-t}t^3)/6

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

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

y = c_{1} e^{- t} + c_{2} t e^{- t} + \frac{e ^{- t} t ^{3}}{6}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{- t} + c_{2} t e^{- t} + \frac{e ^{- t} t ^{3}}{6}

y = c_{1} e^{- t} + c_{2} t e^{- t} + \frac{e ^{- t} t ^{3}}{6}

y^{^{''}} + y = 2 sin (2 x)

y^{^{'''}} + 49 y^{^{'}} = 5 x + 6

y^{^{''}} - 6 y^{^{'}} + 9 y = x^{- 4} \cdot e^{3} x

3 y^{^{''}} + 4 y = 3 e^{3} x

y^{^{''}} - 2 y^{^{'}} + 2 y = e x, y (\frac{π}{2}) = 0, y (0) = 0

What is the solution for y^{''}+2y^'+y= t/(e^t) ?
The solution for y^{''}+2y^'+y= t/(e^t) is y=c_{1}e^{-t}+c_{2}te^{-t}+(e^{-t}t^3)/6