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:
This is a multi-step problem.
1. Solve the given homogeneous differential equation with the initial condition to find the function $y(x)$.
2. Use this solution to calculate the value of the constant $\alpha$.
3. Substitute $\alpha$ into the given equation of a circle and find the range of integer values for the parameter $p$ that satisfy the given conditions on the circle's radius.
Step 2: Key Formula or Approach:
- For the homogeneous DE, use the substitution $y = vx$. - For the circle $x^2 + y^2 + 2gx + 2fy + c = 0$, the radius is $r = \sqrt{g^2 + f^2 - c}$. For a real circle, we need the condition $g^2+f^2-c>0$.
Step 3: Detailed Explanation:
Part 1: Solve the Differential Equation
First, rearrange the DE to the form $\frac{dy}{dx}$: \[ \frac{dy}{dx} = \frac{y \sin(y/x) - x}{x \sin(y/x)} = \frac{y}{x} - \frac{1}{\sin(y/x)} \] This is a homogeneous equation. Let $y=vx$, so $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
\[ v + x \frac{dv}{dx} = v - \frac{1}{\sin v} \implies x \frac{dv}{dx} = -\frac{1}{\sin v} \] Separate the variables:
\[ \sin v \, dv = -\frac{1}{x} \, dx \] Integrate both sides:
\[ \int \sin v \, dv = -\int \frac{1}{x} \, dx \implies -\cos v = -\ln|x| + C \] Substitute back $v = y/x$: $\cos(y/x) = \ln|x| - C$.
Use the initial condition $y(1) = \pi/2$:
$\cos(\frac{\pi/2}{1}) = \ln(1) - C \implies 0 = 0 - C \implies C=0$.
The solution is $\cos(y/x) = \ln x$ (for $x>0$).
Part 2: Find the value of $\alpha$
We need to evaluate $\alpha = \cos \left( \frac{y(e^{12})}{e^{12}} \right)$.
From our solution, $\cos(y(x)/x) = \ln x$.
Let $x = e^{12}$. Then $\cos(y(e^{12})/e^{12}) = \ln(e^{12}) = 12$. So, $\alpha = 12$.
Part 3: Analyze the Circle
Substitute $\alpha=12$ into the circle's equation:
$x^2 + y^2 - 2px + 2py + 12 + 2 = 0 \implies x^2 + y^2 - 2px + 2py + 14 = 0$.
The radius is $r = \sqrt{(-p)^2 + p^2 - 14} = \sqrt{2p^2 - 14}$.
We have two conditions:
1. The equation must represent a real circle, so $r^2>0$:
$2p^2 - 14>0 \implies 2p^2>14 \implies p^2>7$.
2. The radius must be less than or equal to 6, so $r \le 6$:
$\sqrt{2p^2 - 14} \le 6 \implies 2p^2 - 14 \le 36 \implies 2p^2 \le 50 \implies p^2 \le 25$.
Combining the two conditions, we get $7<p^2 \le 25$.
We need to find the number of integers $p$ that satisfy this. The integers whose squares are in this range are:
$p = \pm 3$ (since $3^2=9$)
$p = \pm 4$ (since $4^2=16$)
$p = \pm 5$ (since $5^2=25$)
The set of integer values is $\{-5, -4, -3, 3, 4, 5\}$. There are 6 such values.
Step 4: Final Answer:
The number of integral values of $p$ is 6.