If for the solution curve $y=f(x)$ of the differential equation $\frac{\mathrm{d} y}{\mathrm{~d} x}+(\tan x)…

If for the solution curve $y=f(x)$ of the differential equation $\frac{\mathrm{d} y}{\mathrm{~d} x}+(\tan x) y=\frac{2+\sec x}{(1+2 \sec x)^2}$, $x \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right), f\left(\frac{\pi}{3}\right)=\frac{\sqrt{3}}{10}$, then $f\left(\frac{\pi}{4}\right)$ is equal to :
  1. $\frac{\sqrt{3}+1}{10(4+\sqrt{3})}$
  2. $\frac{5-\sqrt{3}}{2 \sqrt{2}}$
  3. $\frac{9 \sqrt{3}+3}{10(4+\sqrt{3})}$
  4. $\frac{4-\sqrt{2}}{14}$

Solution

$\begin{aligned} & \text { If } \mathrm{e}^{\int \tan \mathrm{xdx}}=\mathrm{e}^{\ln (\sec \mathrm{x})}=\sec \mathrm{x} \\ & \therefore \mathrm{y} \cdot \sec \mathrm{x}=\int\left\{\frac{2+\sec \mathrm{x}}{(1+2 \sec \mathrm{x})^2}\right\} \sec \mathrm{xdx} \\ & =\int \frac{2 \cos \mathrm{x}+1}{(\cos \mathrm{x}+2)^2} \mathrm{dx} \text { Let } \cos \mathrm{x}=\frac{1-\mathrm{t}^2}{1+\mathrm{t}^2} \\ & =\int \frac{2\left(\frac{1-\mathrm{t}^2}{1+\mathrm{t}^2}\right)+1}{\left(\frac{1-\mathrm{t}^2}{1+\mathrm{t}^2}+2\right)^2} 2 \mathrm{dt}\end{aligned}$
$\begin{aligned}
& =\int \frac{2-2 t^2+1+t^2}{\left(1-t^2+2+2 t^2\right)^2} \times 2 \mathrm{dt} \\ & =2 \int \frac{3-t^2}{\left(t^2+3\right)^2} \mathrm{dt}
\end{aligned}$
Let $\mathrm{t}+\frac{3}{\mathrm{t}}=\mathrm{u}$
$\begin{aligned}
& \left(1-\frac{3}{\mathrm{t}^2}\right) \mathrm{dt}=\mathrm{du} \\ & =-2 \int \frac{\mathrm{du}}{\mathrm{u}^2}
\end{aligned}$
$\begin{aligned}
& y \cdot(\sec x)=\frac{2}{u}+c \\ & y \cdot \sec x=\frac{2}{t+\frac{3}{t}}+c...(I)
\end{aligned}$
$\text { At } \mathrm{x}=\frac{\pi}{3}, \mathrm{t}=\tan \frac{\mathrm{x}}{2}=\frac{1}{\sqrt{3}}$
$2 \cdot \frac{\sqrt{3}}{10}=\frac{2}{\frac{1}{\sqrt{3}}+3 \sqrt{3}}+\mathrm{c}$
$\begin{aligned} & \text { 2. } \frac{\sqrt{3}}{10}=\frac{2 \sqrt{3}}{10}+\mathrm{c} \Rightarrow \mathrm{C}=0 \\ & \text { At } \mathrm{x}=\frac{\pi}{4}, \mathrm{t}=\tan \frac{\mathrm{x}}{2}=\sqrt{2}-1 \\ & \therefore \mathrm{y} \cdot \sqrt{2}=\frac{2}{\sqrt{2}-1+\frac{3}{\sqrt{2}-1}} \\ & \mathrm{y} \cdot \sqrt{2}=\frac{2(\sqrt{2}-1)}{6-2 \sqrt{2}} \\ & \mathrm{y}=\frac{\sqrt{2}(\sqrt{2}-1)}{2(3-\sqrt{2})}=\frac{1}{\sqrt{2}} \times \frac{2 \sqrt{2}-1}{7} \\ & =\frac{4-\sqrt{2}}{14}\end{aligned}$

Asked in: JEE Main 2025 (29 Jan Shift 2)

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