Integral Values of Parameter in Circle Equation from Differential Equation Solution

To find the number of integral values of $p$, we first solve the given differential equation, determine the value of $\alpha$, and then analyze the condition for the equation to represent a circle with the specified radius constraint.

Step 1: Solving the Differential Equation

The given differential equation is: $x \sin\left(\frac{y}{x}\right) dy = \left(y \sin\left(\frac{y}{x}\right) - x\right) dx$

Rewriting in standard form: $\frac{dy}{dx} = \frac{y}{x} - \frac{1}{\sin\left(\frac{y}{x}\right)}$

This is a homogeneous differential equation. Let $y = v x$, so $\frac{dy}{dx} = v + x \frac{dv}{dx}$. Substituting these into the equation: $v + x \frac{dv}{dx} = v - \frac{1}{\sin v}$

Simplifying: $x \frac{dv}{dx} = -\frac{1}{\sin v}$

Separating the variables: $\sin v \, dv = -\frac{dx}{x}$

Integrating both sides: $\int \sin v \, dv = -\int \frac{dx}{x}$ $-\cos v = -\ln|x| + C \implies \cos\left(\frac{y}{x}\right) = \ln|x| + C'$

Using the initial condition $y(1) = \frac{\pi}{2}$: $\cos\left(\frac{\pi/2}{1}\right) = \ln(1) + C' \implies 0 = 0 + C' \implies C' = 0$

Thus, the solution to the differential equation for $x > 0$ is: $\cos\left(\frac{y}{x}\right) = \ln x$

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Step 2: Evaluating $\alpha$

We are given $\alpha = \cos\left(\frac{y(e^{1/2})}{e^{1/2}}\right)$.

Substituting $x = e^{1/2}$ into our solution: $\alpha = \cos\left(\frac{y(e^{1/2})}{e^{1/2}}\right) = \ln\left(e^{1/2}\right) = \frac{1}{2}$

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Step 3: Finding the Radius of the Circle

The equation of the circle is given as: $x^2 + y^2 - 2px + 2py + \alpha + 2 = 0$

Substituting $\alpha = \frac{1}{2}$: $x^2 + y^2 - 2px + 2py + \frac{5}{2} = 0$

Comparing this with the general equation of a circle $x^2 + y^2 + 2gx + 2fy + c = 0$, we have: $g = -p, \quad f = p, \quad c = \frac{5}{2}$

The radius $r$ of the circle is given by: $r = \sqrt{g^2 + f^2 - c} = \sqrt{(-p)^2 + p^2 - \frac{5}{2}} = \sqrt{2p^2 - \frac{5}{2}}$

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Step 4: Finding the Number of Integral Values of $p$

For the equation to represent a real circle, the radius must be strictly positive: $r > 0 \implies 2p^2 - \frac{5}{2} > 0 \implies p^2 > \frac{5}{4} = 1.25$

Under the given condition $r^2 \le 6$: $2p^2 - \frac{5}{2} \le 6 \implies 2p^2 \le \frac{17}{2} \implies p^2 \le \frac{17}{4} = 4.25$

Combining the inequalities: $1.25 < p^2 \le 4.25$

Since $p$ is an integer parameter, $p^2$ must be an integer square in the range $(1.25, 4.25]$: $p^2 = 4 \implies p = 2$

Thus, there is only $1$ integral value of $p$ that satisfies the given condition.