Question

Let $y = y(x)$ be the solution of the differential equation $x \sin \left( \frac{y}{x} \right) dy = \left( y \sin \left( \frac{y}{x} \right) - x \right) dx, y(1) = \frac{\pi}{2}$ and let $\alpha = \cos \left( \frac{y(e^{12})}{e^{12}} \right)$. Then the number of integral values of $p$, for which the equation $x^2 + y^2 - 2px + 2py + \alpha + 2 = 0$ represents a circle of radius $r \le 6$, is _________.

Answer 1

Correct Answer: 6

Step 1: Understanding the Question:
First, solve the homogeneous differential equation to find $y(x)$. Then determine $\alpha$. Finally, use circle properties to find the range of $p$. 
Step 2: Key Formula or Approach: 
Use substitution $y = vx$ for homogeneous differential equations. Radius of a circle $x^2 + y^2 + 2gx + 2fy + c = 0$ is $\sqrt{g^2 + f^2 - c}$. 
Step 3: Detailed Explanation: 
Differential equation: $\frac{dy}{dx} = \frac{y \sin(y/x) - x}{x \sin(y/x)} = \frac{y}{x} - \frac{1}{\sin(y/x)}$. 
Put $y = vx$, so $\frac{dy}{dx} = v + x \frac{dv}{dx}$. 
\[ v + x \frac{dv}{dx} = v - \frac{1}{\sin v} \implies x \frac{dv}{dx} = -\frac{1}{\sin v} \] 
Separating variables: $\int \sin v dv = -\int \frac{1}{x} dx \implies -\cos v = -\ln x + C$. 
$\cos(y/x) = \ln x + C'$. 
Using $y(1) = \pi/2$: $\cos(\pi/2) = \ln 1 + C' \implies 0 = 0 + C' \implies C' = 0$. 
So, $\cos(y/x) = \ln x$. 
Now, $\alpha = \cos \left( \frac{y(e^{12})}{e^{12}} \right) = \ln(e^{12}) = 12$. 
Equation of circle: $x^2 + y^2 - 2px + 2py + 12 + 2 = 0 \implies x^2 + y^2 - 2px + 2py + 14 = 0$. 
Center $(p, -p)$. Radius $r = \sqrt{p^2 + (-p)^2 - 14} = \sqrt{2p^2 - 14}$. 
Conditions: 
1. For circle to exist: $2p^2 - 14>0 \implies p^2>7$. 
2. Radius $r \le 6 \implies \sqrt{2p^2 - 14} \le 6 \implies 2p^2 - 14 \le 36 \implies 2p^2 \le 50 \implies p^2 \le 25$. 
Combining: $7 < p^2 \le 25$. 
Possible integers for $p$: $\pm 3, \pm 4, \pm 5$. 
Set of integral values: $\{-5, -4, -3, 3, 4, 5\}$. Count = 6. 
Step 4: Final Answer: 
The number of such integral values of $p$ is 6.