JEE Main 2024 — Differential Equations Question with Solution

$$\begin{array}{rl}& \frac{dy}{dx}=\frac{\sin x+y\cos x}{\sin x\cdot \cos x\left(\frac{1}{\cos x}-\sin x\cdot \frac{\sin x}{\cos x}\right)}\\ & =\frac{\sin x+y\cos x}{\sin x\left(1-{\sin }^{2}x\right)}\\ & \frac{dy}{dx}={\sec }^{2}x+y\cdot 2(\mathrm{cosec}2x)\\ & \frac{dy}{dx}-2\mathrm{cosec}(2x)\cdot y={\sec }^{2}x\\ & \frac{dy}{dx}+p\cdot y=Q\end{array}$$

$$\text{ I.F. }={\mathrm{e}}^{\int p\mathrm{dx}}={\mathrm{e}}^{\int -2\mathrm{cosec}(2\mathrm{x})\mathrm{dx}}$$

Let $$2\mathrm{x}=\mathrm{t}$$

$$2\frac{\mathrm{dx}}{\mathrm{dt}}=1$$

$$\mathrm{dx}=\frac{\mathrm{dt}}{2}$$

$$={\mathrm{e}}^{-\int \mathrm{cosec}(\mathrm{t})\mathrm{dt}}$$

$$={\mathrm{e}}^{-\ln |\tan \frac{t}{2}|}$$

$$={\mathrm{e}}^{-\ln |\tan x|}=\frac{1}{|\tan x|}$$

$$\begin{array}{rl}& y(\text{ IF })=\int Q(IF)dx+c\\ & \Rightarrow y\frac{1}{|\tan x|}=\int {\sec }^{2}x\cdot \frac{1}{|\tan x|}+c\\ & y\cdot \frac{1}{|\tan x|}=\int \frac{dt}{|t|}+c\text{ for }\tan x=t\\ & y\cdot \frac{1}{|\tan x|}=\ln |t|+c\\ & y=|\tan x|(\ln |\tan x|+c)\\ & \text{ Put }x=\frac{\pi }{4},y=2\\ & 2=\ln 1+c\Rightarrow c=2\\ & y=|\tan x|(\ln |\tan x|+2)\\ & y\left(\frac{\pi }{3}\right)=\sqrt{3}(\ln \sqrt{3}+2)\end{array}$$