Solve DE for y: $\displaystyle y^\prime +y = xe^{-x}+1$

[karush]

$\textsf{Given:}$
$$\displaystyle y^\prime +y = xe^{-x}+1$$
$\textit{Solve the given differential equation}$
$\textit{From:$\displaystyle\frac{dy}{dx}+Py=Q$}$
$\textit{then:}$
$$\displaystyle
e^x y=\int x+e^{-2x} \, dx
+ c \\
\displaystyle e^x y=\frac{1}{2}(x^2-e^{-2x})+c$$
$\textit{divide every term by $e^x$}$
$$\displaystyle y=\frac{1}{2(e^x)}(x^2)
-\frac{e^{-2x}}{2(e^x)}+\frac{c}{(e^x)}$$
$\textit{simplify and reorder terms}$
$$\displaystyle y=c_1e^{-x}+\frac{1}{2}e^{-x}x^2+1$$
$\textit{Answer by W|A}$
$$y(x)=\color{red}
{\displaystyle c_1e^{-x}+\frac{1}{2}e^{-x}x^2+1}$$

any bugs any suggest?

 

[MarkFL]

The integrating factor is:

\(\displaystyle \mu(x)=e^x\)

And so we get:

\(\displaystyle y'e^x+ye^x=x+e^x\)

\(\displaystyle \frac{d}{dx}\left(e^xy\right)=x+e^x\)

Integrate:

\(\displaystyle e^xy=\frac{x^2}{2}+e^x+c_1\)

Hence:

\(\displaystyle y(x)=\frac{x^2}{2}e^{-x}+1+c_1e^{-x}\)

Somehow you arrived at the correct answer, but your work doesn't show how.

 

[karush]

Hence:

\(\displaystyle y(x)=\frac{x^2}{2}e^{-x}+1+c_1e^{-x}\)

Somehow you arrived at the correct answer, but your work doesn't show how.

ok I hit and missed with some examples
but see your steps make a lot more sense