MHB Solve IVP: What is the solution to the given initial value problem?

[karush]

$\tiny{b.2.1.16}$
\begin{align*}\displaystyle
y^\prime +\frac{2}{x}y &=\frac{\cos x}{x^2} &&(1)\\
u(x)&=\exp\int\frac{2}{x} \, dx = e^{2\ln{x}}=2x&&(2)\\
(2x y)'&=\int\frac{\cos x}{x^2} \, dx &&(3)\\
y(x)&=\frac{1}{2x}\left[-\dfrac{\cos(x)}{x}-\int\dfrac{\sin(x)}{x}\, dx\right] &&(4)\\
y(\pi)&=0 &&(5)\\
&=\frac{1}{2(\pi)}
\left[-\dfrac{\cos(\pi)}{\pi}
-\int\dfrac{\sin(\pi)}{\pi}\, dx
\right]=0 &&(6)\\
&=\color{red}{\frac{\sin x}{x^2}}&&(7)\\
&=\frac{0}{\pi^2}=0 &&(8)
\end{align*}
ok (4)-(8) were ? red is bk answer

 

[MarkFL]

Your integrating factor is incorrect. Try finding that again. :)

 

[karush]

$\tiny{2.1.{8}}$
\begin{align*}\displaystyle
y^{\prime} +\frac{2}{x}y &=\frac{\cos x}{x^2} &&(1)\\
u(x)&=\exp\int\frac{2}{x} \, dx = e^{2\ln{x}}=x^2&&(2)\\
(x^2 y)'&=\int\frac{\cos x}{x^2} \, dx &&(3)\\
y(x)&=\frac{1}{x^2}\left[-\dfrac{\cos(x)}{x}-\int\dfrac{\sin(x)}{x}\, dx\right] &&(4)\\
y(\pi)&=0 &&(5)\\
&=\frac{1}{\pi^2}
\left[-\dfrac{\cos(\pi)}{\pi}
-\int\dfrac{\sin(\pi)}{\pi}\, dx
\right]=0 &&(6)\\
&=\color{red}{\frac{\sin x}{x^2}}&&(7)\\
\end{align*}ok $u(x)=x^2$ red is bk answer

 

[MarkFL]

You now have the correct integrating factor, but you didn't multiply the RHS by this factor. :)

You should eventually get:

$$\frac{d}{dx}\left(x^2y\right)=\cos(x)$$

Now, continue from there.

 

[karush]

You now have the correct integrating factor, but you didn't multiply the RHS by this factor. :)

You should eventually get:

$$\frac{d}{dx}\left(x^2y\right)=\cos(x)$$

Now, continue from there.

$\displaystyle x^2y=\int \cos x \, dx$
$\displaystyle x^2y=\sin x$
$\displaystyle y=\frac{\sin x}{x^2}$