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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

x \to 0 lim (ln (\frac{x}{x - 1}))

\frac{e ^{- s}}{s - 8}

\int 4 - 2 x d x

2 x - y \frac{d y}{d x} = 3

\int \frac{x ^{2} + 1}{x ^{4}} d x

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