JEE Main · 2024 · 30 Jan 2024 · Shift 2 · Differential Equations

At a point $(x,y)$ on the curve, the tangent line is $Y-y=Y^{\prime }(x)(X-x)$, where $Y^{\prime }(x)$ denotes $\frac{dy}{dx}$; write $p=Y^{\prime }(x)$.

Its intercepts: setting $Y=0$ gives the $X$-intercept $x-\frac{y}{p}$; setting $X=0$ gives the $Y$-intercept $y-xp$.

The enclosed triangle has area
$\frac{1}{2}\left(x-\frac{y}{p}\right)(y-xp)=\frac{1}{2}\left(2xy-x^{2}p-\frac{y^2}{p}\right)=xy-\frac{x^{2}p}{2}-\frac{y^2}{2p}$.

Setting this equal to $-\frac{y^2}{2p}+1$, the $-\frac{y^2}{2p}$ terms cancel, leaving
$xy-\frac{x^{2}p}{2}=1\Rightarrow x^{2}p=2xy-2\Rightarrow p=\frac{dy}{dx}=\frac{2y}{x}-\frac{2}{x^2}$.

This is linear: $\frac{dy}{dx}-\frac{2}{x}y=-\frac{2}{x^2}$, with integrating factor $x^{-2}$:
$\frac{d}{dx}\left(\frac{y}{x^2}\right)=-\frac{2}{x^4}\Rightarrow \frac{y}{x^2}=\frac{2}{3x^3}+C\Rightarrow y=\frac{2}{3x}+C{x}^2$.

Using $Y(1)=1$: $1=\frac{2}{3}+C\Rightarrow C=\frac{1}{3}$, so $Y(X)=\frac{2}{3X}+\frac{X^2}{3}$.

$Y(2)=\frac{1}{3}+\frac{4}{3}=\frac{5}{3}$, so $12Y(2)=12\times \frac{5}{3}=20$.