JEE Main 2022 (Online) 30th July Evening ShiftQuestion 1
If $y=y(x)$ satisfies the differential equation $x\,dy+\left(x\tan\left(\frac yx\right)-y\right)dx=0$, $x>0$, and $y(1)=\frac\pi2$, then a value of $y(\sqrt2)$ is
Differential Equation
Solution
Concept. Put $y=vx$; the equation separates into $\cot v\,dv=-\frac{dx}{x}$. Formulas used. (1) $y=vx\Rightarrow dy=v\,dx+x\,dv$(2) $\int\cot v\,dv=\ln|\sin v|$Step 1 - Substitute. By (1), $x(v\,dx+x\,dv)+(x\tan v-vx)\,dx=0$, so $x^2\,dv+x\tan v\,dx=0$ and $$\cot v\,dv=-\frac{dx}{x}.$$ Step 2 - Integrate. By (2), $x\sin v=C$. $y(1)=\frac\pi2$: $C=1$. So $\sin\frac yx=\frac1x$. Step 3 - At $x=\sqrt2$. $\sin v=\frac{1}{\sqrt2}$, so $v=\frac\pi4$ or $\frac{3\pi}{4}$, and $y=\sqrt2\,v=\frac{\pi}{2\sqrt2}$ or $\frac{3\pi}{2\sqrt2}$. Step 4 - Choose. The value among the options is $\frac{3\pi}{2\sqrt2}$. Answer: (D)