JEE Main · 2023 · 12 Apr 2023 · Shift 1 · Differential Equations

Writing the equation as $\frac{dy}{dx}-\frac{x}{1+x^2}y=-\frac{1}{1+x^2}{y}^2$ and dividing by $y^2$, the substitution $t=\frac{1}{y}$ gives the linear equation $\frac{dt}{dx}+\frac{x}{1+x^2}t=\frac{1}{1+x^2}$. The integrating factor is $e^{\int \frac{x}{1+x^2}dx}=\sqrt{1+x^2}$, so $\frac{d}{dx}\left(t\sqrt{1+x^2}\right)=\frac{1}{\sqrt{1+x^2}}$ and therefore $\frac{\sqrt{1+x^2}}{y}=\ln \left(x+\sqrt{1+x^2}\right)+C$. The condition $y(0)=1$ gives $C=1$, so $\frac{\sqrt{1+x^2}}{y}=\ln \left(e\left(x+\sqrt{1+x^2}\right)\right)$. At $x=2\sqrt{2}$, $\sqrt{1+x^2}=3$, hence $\frac{3}{\beta }=\ln \left(e(3+2\sqrt{2})\right)$, which gives $e^{3/\beta }=e(3+2\sqrt{2})$.