JEE Main 2024 (Online) 9th April Morning ShiftQuestion 1
The solution curve, of the differential equation $2y\frac{dy}{dx}+3=5\frac{dy}{dx}$, passing through the point $(0,1)$ is a conic, whose vertex lies on the line :
Parabola
Solution
$2y\,dy+3\,dx=5\,dy\Rightarrow y^2+3x=5y+c$. Through $(0,1)$: $1=5+c\Rightarrow c=-4$. $y^2-5y+3x+4=0\Rightarrow\left(y-\frac52\right)^2=-3\left(x-\frac34\right)$. Vertex $\left(\frac34,\frac52\right)$: $2x+3y=\frac32+\frac{15}{2}=9$. Option (A).