Q120 / 160 · TS EAMCET 2020 (10 Sep Shift 2) · Mathematics

The general solution of the differential equation $x \cos \frac{y}{x}(y d x+x d y)=y \sin \frac{y}{x}(x d y-y d x)$ is

  • A
    $\log (x y)=\log \cos \frac{x}{y}+C$
  • Correct answer
    B
    $\cos \left(\frac{y}{x}\right)=\frac{C}{x y}$
  • C
    $\log (x y)=\log \sec \frac{x}{y}+C$
  • D
    $x+y+C=0$

Solution

We have, $x \cos \left(\frac{y}{x}\right)(y d x+x d y)$ $=y \sin \frac{y}{x}(x d y-y d x)$ $\Rightarrow \quad x y \sin \left(\frac{y}{x}\right) d y-y^2 \sin \left(\frac{y}{x}\right) d x$ $=x y \cos \left(\frac{y}{x}\right) d x+x^2 \cos \left(\frac{y}{x}\right) d y$ $\Rightarrow\left[x y \sin \left(\frac{y}{x}\right)-x^2 \cos \left(\frac{y}{x}\right)\right] d y$ $=\left[x y \cos \left(\frac{y}{x}\right)+y^2 \sin \left(\frac{y}{x}\right)\right] d x$ $\Rightarrow \frac{d y}{d x}=\frac{x y \cos \left(\frac{y}{x}\right)+y^2 \sin \left(\frac{y}{x}\right)}{x y \sin \left(\frac{y}{x}\right)-x^2 \cos \left(\frac{y}{x}\right)}$ $\Rightarrow \frac{d y}{d x}=\frac{\frac{y}{x} \cos \left(\frac{y}{x}\right)+\left(\frac{y}{x}\right)^2 \sin \left(\frac{y}{x}\right)}{\frac{y}{x} \sin \left(\frac{y}{x}\right)-\cos \left(\frac{y}{x}\right)}$ Put, $y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$ $\therefore \quad v+x \frac{d v}{d x}=\frac{v \cos v+v^2 \sin v}{v \sin v-\cos v}$ $\Rightarrow \quad x \frac{d v}{d x}=\frac{2 v \cos v}{v \sin v-\cos v}$ $\Rightarrow \quad \frac{v \sin v-\cos v}{v \cos v} d v=2 \frac{d x}{x}$ $\Rightarrow \quad \int\left(\tan v-\frac{1}{v}\right) d v=2 \int \frac{d x}{x}$ $\Rightarrow \quad \log |\sec v|-\log |v|=2 \log |x|+\log \left|C_1\right|$ $\Rightarrow \quad \log \left|\frac{\sec v}{v}\right|-\log |x|^2=\log \left|C_1\right|$ $\Rightarrow \quad \log \left|\frac{\sec v}{v x^2}\right|=\log \left|C_1\right|$ $\Rightarrow \quad \frac{\sec v}{v x^2}=C_1 \Rightarrow \frac{\sec \left(\frac{y}{x}\right)}{\frac{y}{x}\left(x^2\right)}=C_1$ $\Rightarrow \quad \frac{\sec \left(\frac{y}{x}\right)}{x y}=C_1 \Rightarrow \sec \left(\frac{y}{x}\right)=C_1 x y$ $\Rightarrow \quad \frac{1}{\cos \left(\frac{y}{x}\right)}=\frac{C}{x y} \quad\left(\because\right.$ where $\left.\frac{1}{C_1}=C\right)$

Hence, the correct answer is option: 'B'.

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