JEE Main 2025 Jan 29 Shift 2, Mathematics Q20: Methods of Solving First Order, First Degree Differential Equation

If $e^{\int \tan xdx}=e^{\ln (\sec x)}=\sec x$

$\therefore y\cdot \sec x=\int \left\{\frac{2+\sec x}{(1+2\sec x{)}^2}\right\}\sec xdx$

$=\int \frac{2\cos x+1}{(\cos x+2{)}^2}dx$

Let $\cos x=\frac{1-t^2}{1+t^2}$.

$=\int \frac{2\left(\frac{1-t^2}{1+t^2}\right)+1}{{\left(\frac{1-t^2}{1+t^2}+2\right)}^2}\cdot 2dt$

$=\int \frac{2-2t^2+1+t^2}{(1-t^2+2+2t^2{)}^2}\times 2dt$

$=2\int \frac{3-t^2}{(t^2+3{)}^2}dt$

Let $t+\frac{3}{t}=u$.

$\left(1-\frac{3}{t^2}\right)dt=du$

$=-2\int \frac{du}{u^2}$

$y\cdot (\sec x)=\frac{2}{u}+c$

$\boxed{y\cdot \sec x=\frac{2}{t+\frac{3}{t}}+c}$

Initial condition. At $x=\frac{\pi }{3}$, $t=\tan (\pi /6)=\frac{1}{\sqrt{3}}$, so $t+3/t=\frac{1}{\sqrt{3}}+3\sqrt{3}=\frac{10}{\sqrt{3}}$. Also $\sec (\pi /3)=2$, so

$\frac{\sqrt{3}}{10}\cdot 2=\frac{2\sqrt{3}}{10}+C\Rightarrow C=0$.

Evaluate at $x=\pi /4$. $t=\tan (\pi /8)=\sqrt{2}-1$.

$t+\frac{3}{t}=(\sqrt{2}-1)+\frac{3}{\sqrt{2}-1}=(\sqrt{2}-1)+3(\sqrt{2}+1)=4\sqrt{2}+2$.

$y\sec (\pi /4)=\frac{2}{4\sqrt{2}+2}=\frac{1}{2\sqrt{2}+1}=\frac{2\sqrt{2}-1}{7}$.

Dividing by $\sec (\pi /4)=\sqrt{2}$:

$y=\frac{2\sqrt{2}-1}{7\sqrt{2}}=\frac{4-\sqrt{2}}{14}$.