Soln of IVP: $y = \frac{e^x + 1 - e}{x}, \space x>0$

[karush]

$\textsf{ Find the solution of the given initial value problem.}$
$$xy^\prime+y=e^x, \qquad y(1)=1$$
$$\begin{array}{lrll}
\textit{Divide thru with x}\\
&\displaystyle y' +\frac{1}{x}y
&\displaystyle=\,\frac{e^x}{x} &_{(1)}\\
\textit {Find u(x)}\\
&\displaystyle u(x)
&\displaystyle=\exp\int\frac{1}{x}\,dx\\
&&=e^{\ln {x}}\\
&&=x &_{(2)}\\
\textit{Multiply thru with $x$} \\
&(xy)' +x'y&=e^x &_{(3)}\\
\textit{Rewrite:}\\
&(xy)'&=e^x &_{(4)}\\
\textit{Integrate }\\
&\displaystyle xy
&=\displaystyle\int e^x \, dx\\
&&=\displaystyle e^x+c &_{(5)}\\
\textit{Divide thru by $x$}\\
&\displaystyle y&=\displaystyle\frac{e^x}{x}+\frac{c}{e^x} &_{(6)}\\
\textit{So then if }\\
&\displaystyle y(1)&\displaystyle=e+\frac{c}{e}=1 &_{(7)}\\
\textit{with $c=?$ then }\\
&\displaystyle y
&=\color{red}{\displaystyle\frac{1}{x}(e^x + 1 - e)} &_{(8)}\\
\end{array}$$

ok (8) is the book answer but ? what wuld be c?
$\textit{State the interval in which the solution is valid. ?}\\$

 

[MarkFL]

Towards the end, when you divide through by $x$, you want:

$$y(x)=\frac{e^x}{x}+\frac{c}{x}$$

You mistakenly divided the constant by $e^x$.