Question

The solution of the differential equation
$\left(x\tan \left(\frac{y}{x}\right)-y{\sec }^2\left(\frac{y}{x}\right)\right)dx+x{\sec }^2\left(\frac{y}{x}\right)dy=0$
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 $y=vx$
$\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$
So, given equation reduces to
$\left(\tan \dot{v}-v{\sec }^{2}v\right)+{\sec }^{2}v\left(v+x\frac{dv}{dx}\right)=0$
$\Rightarrow \tan v-v{\sec }^{2}v+v{\sec }^{2}v+x{\sec }^{2}v\frac{dv}{dx}=0$
$\Rightarrow \int \frac{{\sec }^{2}v}{\tan v}dv=-\int \frac{dx}{x}$
$\Rightarrow {\log }_e(\tan v)=-{\log }_{e}x+{\log }_{e}c$
$\Rightarrow x\tan v=c$
$\Rightarrow x\tan \frac{y}{x}=c$
Now, $y(1)=\frac{\pi }{4}\Rightarrow c=1$
$\therefore x\tan \frac{y}{x}=1$