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

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

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y^'+y=te^{-t}+1

y^{^{'}} + y = t e^{- t} + 1

y = \frac{t ^{2}}{2 e ^{t}} + 1 + \frac{c _{1}}{e ^{t}}

Solution steps

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

Solve linear ODE:

y = \frac{t ^{2}}{2 e ^{t}} + 1 + \frac{c _{1}}{e ^{t}}

y = \frac{t ^{2}}{2 e ^{t}} + 1 + \frac{c _{1}}{e ^{t}}

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x \to + 0 + lim (\frac{1}{x} - \frac{1}{x ^{2}})

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