JEE Main 2021 (Online) 17th March Evening ShiftQuestion 1
Let $y=y(x)$ be the solution of the differential equation $\cos x(3\sin x+\cos x+3)dy=(1+y\sin x(3\sin x+\cos x+3))dx$, $0\le x\le\frac{\pi}{2}$, $y(0)=0$. Then, $y\left(\frac{\pi}{3}\right)$ is equal to :
Differential Equation
Solution
Concept. With $g=3\sin x+\cos x+3$ the equation is linear with IF $\cos x$; the integral $\int\frac{dx}{g}$ needs $t=\tan\frac x2$. Formulas used. (1) $\sin x=\dfrac{2t}{1+t^2}$, $\cos x=\dfrac{1-t^2}{1+t^2}$, $dx=\dfrac{2\,dt}{1+t^2}$(2) $\dfrac{1}{(t+1)(t+2)}=\dfrac{1}{t+1}-\dfrac{1}{t+2}$Step 1 - Linear form. Dividing by $g\cos x\,dx$: $$y'-y\tan x=\frac{1}{g\cos x}\ \Longrightarrow\ \big(y\cos x\big)'=\frac1g.$$ Step 2 - The integral. By (1), $g=\dfrac{2(t^2+3t+2)}{1+t^2}$, so $\dfrac{dx}{g}=\dfrac{dt}{(t+1)(t+2)}$. By (2), $$\int\frac{dx}{g}=\ln\frac{t+1}{t+2}.$$ Step 3 - From $x=0$ to $\frac\pi3$. $t$ goes from $0$ to $\frac1{\sqrt3}$, and $y(0)=0$: $$\frac12\,y\left(\frac\pi3\right)=\ln\frac{1+\sqrt3}{1+2\sqrt3}-\ln\frac12=\ln\frac{2(1+\sqrt3)}{1+2\sqrt3}.$$ Rationalising, $\dfrac{1+\sqrt3}{1+2\sqrt3}=\dfrac{5+\sqrt3}{11}$. Step 4 - Value. $y\left(\frac\pi3\right)=2\ln\dfrac{2\sqrt3+10}{11}$. Answer: (D)