If sin(yx) = loge|x| + α/2 is the solution of the differential equation

Question

If \(\sin \left(\frac{y}{x}\right)=\log _e|x|+\frac{\alpha}{2}\) 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

(1) 3

(2) 12

(3) 4

(4) 9

Answer 1

Correct option is (1) 3

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 \(\frac{y}{x}=t\)

\( \cos \mathrm{t}\left(\frac{\mathrm{dt}}{\mathrm{dx}}\right)=\frac{1}{\mathrm{x}} \)

\(\cos \mathrm{dt}=\frac{1}{\mathrm{x}} \mathrm{dx}\)

Integrating both sides

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

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

Using \(\mathrm{y}(1)=\frac{\pi}{3}\), we get 

\(\mathrm{c}=\frac{\sqrt{3}}{2}\)

So, \(\alpha=\sqrt{3} \) 

\(\Rightarrow \alpha^2=3\)