If (y x )= _0|x|+ 2 is the solution of the differential equation … - Vidyadip

MCQ

If $\sin \left(\frac{y}{x}\right)=\log _0|x|+\frac{\alpha}{2}$ is the solution of the differential equation $x \cos \left(\frac{y}{x}\right) \frac{d y}{d x}=y \cos \left(\frac{y}{x}\right)+x$ and $y(1)=\frac{\pi}{3}$, then $\alpha^2$ is equal to

✓
$3$
B
$12$
C
$4$
D
$9$

✓

Answer

Correct option: A.

$3$

a
Differential equation :-

$ x \cos \frac{y}{x} \frac{d y}{d x}=y \cos \frac{y}{x}+x $

$ \cos \frac{y}{x}\left[x \frac{d y}{d x}-y\right]=x$

Divide both sides by $\mathrm{x}^2$

$\cos \frac{y}{x}\left(\frac{x \frac{d y}{d x}-y}{x^2}\right)=\frac{1}{x}$

Let $\frac{y}{x}=t$

$\cos t\left(\frac{d t}{d x}\right)=\frac{1}{x}$

$\cos \mathrm{t} \mathrm{dt}=\frac{1}{\mathrm{x}} \mathrm{dx}$

Integrating both sides

$ \sin \mathrm{t}=\ln |\mathrm{x}|+\mathrm{c} $

$ \sin \frac{\mathrm{y}}{\mathrm{x}}=\ln |\mathrm{x}|+\mathrm{c}$

Using $y(1)=\frac{\pi}{3}$, we get $c=\frac{\sqrt{3}}{2}$

So, $\alpha=\sqrt{3} \Rightarrow \alpha^2=3$

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