Question

The solution of the differential equation $x^{2} \frac{d y}{d x}-x y=1+\cos \frac{y}{x}$ is

• A. $\cos \frac{y}{x}=1+\frac{c}{x}$
• B. $x^{2}=\left(c+x^{2}\right) \tan \frac{y}{x}$
• C. $\tan \frac{y}{2 x}=c-\frac{1}{2 x^{2}}$
• D. $\tan \frac{y}{x}=c+\frac{1}{x}$

Answer 1

Answer: (C)

We have, $x^{2} \frac{d y}{d x}-x y=1+\cos \frac{y}{x}=2 \cos ^{2} \frac{y}{2 x} . $
$\Rightarrow \sec ^{2}\left(\frac{y}{2 x}\right) \cdot\left[x^{2} \frac{d y}{d x}-x y\right]=2$
$\Rightarrow \frac{1}{2} \sec ^{2}\left(\frac{y}{2 x}\right) \cdot \frac{x \frac{d y}{d x}-y}{x^{2}}=\frac{1}{x^{3}} $
$\Rightarrow \frac{d}{d x}\left(\tan \frac{y}{2 x}\right)=\frac{1}{x^{3}}$
Integrating, we get $\tan \frac{y}{2 x}=c-\frac{1}{2 x^{2}}$,
which is the required solution,