Question

The solution of the differential equation
$\left(x \tan \left(\frac{y}{x}\right)-y \sec ^{2}\left(\frac{y}{x}\right)\right) d x+x \sec ^{2}\left(\frac{y}{x}\right) d y=0$
satisfying the initial condition $y(1)=\frac{\pi}{4}$ is

• A. $x \sec \frac{y}{x}=\sqrt{2}$
• B. $x \tan \frac{y}{x}=1$
• C. $x \tan ^{2} \frac{y}{x}=1$
• D. $x \sec ^{2} \frac{y}{x}=2$

Answer 1

Answer: (B)

Put $y=v x$
$\Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$
So, given equation reduces to
$\left(\tan \dot{v}-v \sec ^{2} v\right)+\sec ^{2} v\left(v+x \frac{d v}{d x}\right)=0$
$\Rightarrow \tan v-v \sec ^{2} v+v \sec ^{2} v+x \sec ^{2} v \frac{d v}{d x}=0 $
$\Rightarrow \int \frac{\sec ^{2} v}{\tan v} d v=-\int \frac{d x}{x}$
$\Rightarrow \log _{e}(\tan v)=-\log _{e} x+\log _{e} c $
$\Rightarrow x \tan v=c $
$\Rightarrow x \tan \frac{y}{x}=c $
Now, $ y(1)=\frac{\pi}{4} \Rightarrow \,\,\, c=1$
$\therefore \,\, x \tan \frac{y}{x}=1$