Given Differential Equation

$x\sin\left(\dfrac{y}{x}\right)dy=\left(y\sin\left(\dfrac{y}{x}\right)-x\right)dx$

with

$y(1)=\dfrac{\pi}{2}$

and

$\alpha=\cos\left(\dfrac{y\left(e^{1/2}\right)}{e^{1/2}}\right)$

Step 1 : Convert into differential form

Divide by $dx$:

$x\sin\left(\dfrac{y}{x}\right)\dfrac{dy}{dx}=y\sin\left(\dfrac{y}{x}\right)-x$

$\dfrac{dy}{dx}=\dfrac{y}{x}-\dfrac{1}{\sin(y/x)}$

Let

$v=\dfrac{y}{x}$

Then

$y=vx$

So,

$\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}$

Substitute:

$v+x\dfrac{dv}{dx}=v-\csc v$

$x\dfrac{dv}{dx}=-\csc v$

$\sin v; dv=-\dfrac{dx}{x}$

Integrating,

$-\cos v=-\ln x+C$

$\cos v=\ln x+C$

Since $v=\dfrac{y}{x}$,

$\cos\left(\dfrac{y}{x}\right)=\ln x+C$

Step 2 : Use initial condition

Given

$y(1)=\dfrac{\pi}{2}$

So,

$\cos\left(\dfrac{\pi}{2}\right)=\ln 1+C$

$0=0+C$

Hence,

$C=0$

Therefore,

$\cos\left(\dfrac{y}{x}\right)=\ln x$

Step 3 : Find $\alpha$

At $x=e^{1/2}$,

$\cos\left(\dfrac{y\left(e^{1/2}\right)}{e^{1/2}}\right)=\ln\left(e^{1/2}\right)=\dfrac12$

Thus,

$\alpha=\dfrac12$

Step 4 : Equation of circle

Given equation:

$x^2+y^2-2px+2py+\alpha+2=0$

Substitute $\alpha=\dfrac12$:

$x^2+y^2-2px+2py+\dfrac52=0$

Compare with standard form

$x^2+y^2+2gx+2fy+c=0$

Radius:

$r=\sqrt{g^2+f^2-c}$

Here,

$g=-p,\quad f=p,\quad c=\dfrac52$

So,

$r^2=p^2+p^2-\dfrac52$

$r^2=2p^2-\dfrac52$

Given

$r\leq 6$

Therefore,

$2p^2-\dfrac52\leq 36$

$2p^2\leq \dfrac{77}{2}$

$p^2\leq \dfrac{77}{4}$

$p^2\leq 19.25$

Hence,

$|p|\leq 4$

Integral values of $p$ are

$-4,-3,-2,-1,0,1,2,3,4$
It can be any non-negative integer