Let y = y(x) be the solution of the different | vedclass

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Question

(A) The given differential equation is $x \sin(\frac{y}{x}) dy = (y \sin(\frac{y}{x}) - x) dx$.
Substituting $y = vx$, we get $dy = v dx + x dv$.
Su

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(A) The given differential equation is $x \sin(\frac{y}{x}) dy = (y \sin(\frac{y}{x}) - x) dx$.
Substituting $y = vx$, we get $dy = v dx + x dv$.
Substituting into the equation: $x \sin v (v dx + x dv) = (vx \sin v - x) dx$.
$vx \sin v dx + x^2 \sin v dv = vx \sin v dx - x dx$.
$x^2 \sin v dv = -x dx \Rightarrow \sin v dv = -\frac{1}{x} dx$.
Integrating both sides: $-\cos v = -\ln|x| + C$.
Given $y(1) = \frac{\pi}{2}$, we have $v(1) = \frac{\pi}{2}$.
$-\cos(\frac{\pi}{2}) = -\ln(1) + C \Rightarrow 0 = 0 + C \Rightarrow C = 0$.
Thus, $\cos(\frac{y}{x}) = \ln x$.
Given $\alpha = \cos(\frac{e^{12}}{e^{12}}) = \cos(1)$.
The circle equation is $x^2 + y^2 - 2px + 2py + \alpha + 2 = 0$.
The radius $r$ is given by $\sqrt{p^2 + (-p)^2 - (\alpha + 2)} = \sqrt{2p^2 - \alpha - 2}$.
We require $r \leq 6$, so $2p^2 - \alpha - 2 \leq 36$.
$2p^2 \leq 38 + \alpha$.
Since $\alpha = \cos(1) \approx 0.54$, $2p^2 \leq 38.54 \Rightarrow p^2 \leq 19.27$.
The possible integral values for $p$ are $p \in \{-4, -3, -2, -1, 0, 1, 2, 3, 4\}$.
There are $9$ such values.

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