Let y=y(x) be the solution of the differential equation x ((y)/(x))dy=(y ((y)/(x))-x)dx,y(…

Question

Let $y=y(x)$ be the solution of the differential equation $x\sin (\frac{y}{x})dy=(y\sin (\frac{y}{x})-x)dx,y(1)=\frac{\pi }{2}$ and let $\alpha =\cos (\frac{y({e}^{12})}{{e}^{12}})$. 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 $r\le 6$, is $____$ .

Answer: 11

Accepted Answer

We are given the differential equation

$x\sin (\frac{y}{x})dy=(y\sin (\frac{y}{x})-x)dx$

with initial condition $y(1)=\frac{\pi }{2}$.

Express the differential equation in standard form: $x\sin (\frac{y}{x})\frac{dy}{dx}=y\sin (\frac{y}{x})-x$

$\Rightarrow \frac{dy}{dx}=\frac{y\sin (\frac{y}{x})-x}{x\sin (\frac{y}{x})}$

$\Rightarrow \frac{dy}{dx}=\frac{y}{x}-\frac{1}{\sin (\frac{y}{x})}$

Substitution $y=vx$: Let $y=vx$, where $v$ is a function of $x$.

Then

$\frac{dy}{dx}=v+x\frac{dv}{dx}$

Substitute in the differential equation:

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

Simplify:

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

$\Rightarrow \sin v\, dv=-\frac{dx}{x}$

Integrate both sides: $\int \sin v\, dv=-\int \frac{dx}{x}$

$-\cos v=-\ln |x|+C$

$\Rightarrow \cos v=\ln |x|+C$

Now recall $v=\frac{y}{x}$, so:

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

Apply the initial condition $y(1)=\frac{\pi }{2}$: $x=1,\, y=\frac{\pi }{2}\Rightarrow \frac{y}{x}=\frac{\pi }{2}$

$\cos (\frac{\pi }{2})=0=\ln 1+C\Rightarrow C=0$

So the particular solution is:

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

Find $\alpha $: We are told that

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

From the equation above:

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

Thus at $x={e}^{12}$:

$\alpha =\ln ({e}^{12})=12$

$\alpha =12$

Equation of circle: Given:

${x}^{2}+{y}^{2}-2px+2py+\alpha +2=0$

Substitute $\alpha =12$:

${x}^{2}+{y}^{2}-2px+2py+14=0$

Write in standard form of circle: We know:

${x}^{2}+{y}^{2}-2gx-2fy+c=0$

represents a circle with center $(g,f)$ and radius $r=\sqrt{{g}^{2}+{f}^{2}-c}$.

Compare with the given equation:

${x}^{2}+{y}^{2}-2px+2py+14=0$

We can write $+2py=-2fy\Rightarrow f=-p$.

Hence:

$g=p,\, f=-p,\, c=14$

Radius:

$r=\sqrt{{g}^{2}+{f}^{2}-c}=\sqrt{{p}^{2}+(-p{)}^{2}-14}=\sqrt{2{p}^{2}-14}$

Condition for $r\le 6$: $\sqrt{2{p}^{2}-14}\le 6$

$\Rightarrow 2{p}^{2}-14\le 36$

$\Rightarrow 2{p}^{2}\le 50$

$\Rightarrow {p}^{2}\le 25$

$\Rightarrow -5\le p\le 5$

Count the integral values: Integral values of $p$ are:

$-5,-4,-3,-2,-1,0,1,2,3,4,5$

Hence, total number of integral values $=11$.

$11$

Solutions · 1