MHB 2.2.8 de xy'+2y=sin x,y(pi)=1/pi find solution

[karush]

$\textsf{ Find the solution of the given initial value problem. State the interval in which the solution is valid.}$
$$xy^\prime+2y=\sin x, \qquad y(\pi)=\frac{1}{\pi}$$
$$\begin{array}{lrll}
\textit{rewrite as}\\
&\displaystyle y' +P(x)y
&\displaystyle=g(x) &_{(2)}\\
\textit{Divide thru by x}\\
&\displaystyle y'+\frac{2}{x}y
&\displaystyle=\frac{\sin x}{x}\\
\textit {Find u(x) }\\
&\displaystyle u(x)
&\displaystyle =\exp\int \frac{2}{x} \, dx &_{(3)}\\
&\displaystyle&=e^{\ln x^2}\\
&\displaystyle&=x^2\\
\textit{multiply thru w/ $u(x)$} \\
&\displaystyle x^2y' +2xy
&\displaystyle =x^2\sin x &_{(4)}\\
\textsf{rewrite:}\\
&(x^2 y)'&=u(x)g(x)&_{(5)}\\
\textit{Integrate }\\
&\displaystyle x^2 y
&\displaystyle=\int x^2\sin x \, dx&_{(6)}\\
&\displaystyle &=2x \sin (x)-x^2\cos(x)-2\cos(x)+c_1 &_{(5)}\\
\textit{divide thru by $x^2$}\\
&\displaystyle y
&\displaystyle=\frac{2\sin (x)}{x}
-\cos(x)-\frac{2\cos(x)}{x^2}
+\frac{c_1}{x^2} &_{(6)}\\
\textit{W|A }\\
&y&=\color{red}{\displaystyle\frac{\sin x - \cos x}{x^2}}, \qquad x>0
\end{array}$$

ok I stopped here to see if so far is ok
but the interval is kinda ? to me

 

[MarkFL]

You wound up with $x^2\sin(x)$ on the RHS before integrating, but it should be $x\sin(x)$. :)

 

[karush]

$$\begin{array}{lrll}
\textit{multiply thru w/ $x$} \\
&\displaystyle x^2y' +2xy
&\displaystyle =x\sin x\\
\textit{Integrate }\\
&\displaystyle x^2 y
&\displaystyle=\int x\sin x \, dx\\
&\displaystyle &=\sin{x}-x\cos{x}+c_1 \\
\textit{divide thru by $x^2$}\\
&\displaystyle y
&\displaystyle=\frac{\sin{x}}{x^2}-\frac{\cos{x}}{x}+\frac{c_1}{x^2}
\end{array}$$
$\textit{eMahthHelp returned }$
$$y{\left(x \right)} = \frac{1}{x} \left(\frac{C_{1}}{x} - \cos{\left(x \right)} + \frac{1}{x} \sin{\left(x \right)}\right)$$
ok now dealing with
$$\displaystyle y(\pi)=\frac{\sin{\pi}}{\pi^2}-\frac{\cos{\pi}}{\pi}+\frac{c_1}{\pi^2}=\frac{1}{\pi}$$
$$\displaystyle y(\pi)=\frac{0}{\pi^2}+\frac{1}{\pi}+\frac{c_1}{\pi^2}=\frac{1}{\pi}$$
$\textit{so}$
$\displaystyle y(\pi)
=\frac{\sin{\pi}}{\pi^2}
-\frac{\cos{\pi}}{\pi}
+\frac{c_1}{\pi^2}=\frac{1}{\pi}\\$
$\textit{then $c_1=0$ so}$
$y=\color{red}{\displaystyle\frac{\sin x - x\cos x}{x^2}}, \qquad x>0 $