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 $\alpha=\cos \left(\frac{y\left(e^{12}\right)}{e^{12}}\right)$. Then the number of integral value of $p$, for which the equation $x^{2}+y^{2}-2px+2py+\alpha+2=0$ represents a circle of radius $r \leq 6$, is ______.

Differential Equations — JEE Main · 2026 · 05 Apr 2026 · Shift 1 Question with Solution

Let

y = y (x)

be the solution of the differential equation

x sin (\frac{y}{x}) d y = (y sin (\frac{y}{x}) - x) d x, y (1) = \frac{π}{2}

and let

α = cos (\frac{y ( e ^{12} )}{e ^{12}})

. Then the number of integral value of

p

, for which the equation

x^{2} + y^{2} - 2 p x + 2 p y + α + 2 = 0

represents a circle of radius

r \leq 6

, is ______.

Answer: 11

The given differential equation is

x sin (\frac{y}{x}) d y = (y sin (\frac{y}{x}) - x) d x

.

Dividing by

d x

and rearranging:

sin \frac{y}{x} \frac{d y}{d x} = \frac{y}{x} sin \frac{y}{x} - 1

.

Substitute

y = t x

where

t = \frac{y}{x}

, so

\frac{d y}{d x} = t + x \frac{d t}{d x}

.

This gives:

sin t (t + x \frac{d t}{d x}) = t sin t - 1

, which simplifies to

x sin t \frac{d t}{d x} = - 1

.

Separating variables:

sin t d t + \frac{d x}{x} = 0

.

Integrating:

- cos t + ln x = C

, or

- cos (\frac{y}{x}) + ln x = C

.

Using the initial condition

y (1) = \frac{π}{2}

- cos (\frac{π}{2}) + ln 1 = C \Rightarrow C = 0

.

Thus

cos (\frac{y}{x}) = ln x

.

At

x = e^{12}

cos (\frac{y ( e ^{12} )}{e ^{12}}) = ln (e^{12}) = 12

, so

α = 12

.

The circle equation becomes

x^{2} + y^{2} - 2 p x + 2 p y + 14 = 0

.

The radius is

r = p^{2} + p^{2} - 14 = 2 p^{2} - 14

.

For

r \leq 6

2 p^{2} - 14 \leq 36 \Rightarrow p^{2} \leq 25 \Rightarrow - 5 \leq p \leq 5

.

The integral values are

p \in {- 5, - 4, - 3, - 2, - 1, 0, 1, 2, 3, 4, 5}

, giving 11 values.

JEE Main · 2026 · 05 Apr 2026 · Shift 1 · Integer answer