JEE Main 2025 Jan 22 Shift 2, Mathematics Q10: Methods of Solving First Order, First Degree Differential Equation

Rewrite treating $x$ as the dependent variable and $y$ as independent. From $(1+y^2)+(x-2e^{{\tan }^{-1}y})\frac{dy}{dx}=0$:

$\frac{dx}{dy}=-\frac{1+y^2}{x-2e^{{\tan }^{-1}y}}$. Rearranging:

$\frac{dx}{dy}+\frac{x}{1+y^2}=\frac{2e^{{\tan }^{-1}y}}{1+y^2}$, a linear ODE in $x$.

Integrating factor: $\mu =e^{\int \frac{dy}{1+y^2}}=e^{{\tan }^{-1}y}$.

Therefore: $\frac{d}{dy}\left(x{e}^{{\tan }^{-1}y}\right)=\frac{2e^{2{\tan }^{-1}y}}{1+y^2}$.

Substitute $t={\tan }^{-1}y$, $dt=\frac{dy}{1+y^2}$. Integrate the right side:

$\int 2e^{2t}dt=e^{2t}+C=e^{2{\tan }^{-1}y}+C$.

So $x{e}^{{\tan }^{-1}y}=e^{2{\tan }^{-1}y}+C\Rightarrow x=e^{{\tan }^{-1}y}+C{e}^{-{\tan }^{-1}y}$.

Apply $f(0)=1$, i.e. $y=0$ gives $x=1$: $1=e^0+C{e}^0=1+C\Rightarrow C=0$.

Hence $x=e^{{\tan }^{-1}y}$.

At $y=\frac{1}{\sqrt{3}}$: ${\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)=\frac{\pi }{6}$, so $x=e^{\pi /6}$.