JEE MAIN - Mathematics (2024 - 9th April Morning Shift - No. 16)
The solution curve, of the differential equation $$2 y \frac{\mathrm{d} y}{\mathrm{~d} x}+3=5 \frac{\mathrm{d} y}{\mathrm{~d} x}$$, passing through the point $$(0,1)$$ is a conic, whose vertex lies on the line :
$$2 x+3 y=-9$$
$$2 x+3 y=-6$$
$$2 x+3 y=9$$
$$2 x+3 y=6$$
Explanation
$$\begin{aligned}
& 2 y \frac{d y}{d x}+3=5 \frac{d y}{d x} \\
& 2 y d y+3 d x=5 d y \\
& y^2+3 x=5 y+\left.c\right|_{(0,1)} \\
& 1+0=5+c \\
& c=-4 \\
& y^2-5 y=-3 x-4 \\
& y^2-5 y+\frac{25}{4}-\frac{25}{4}=-3 x-4 \\
& \left(y-\frac{5}{2}\right)^2=-3 x+\frac{9}{4} \\
& \left(y-\frac{5}{2}\right)^2=-3\left(x-\frac{3}{4}\right) \\
& \left(\frac{3}{4}, \frac{5}{2}\right) \text { satisfies by } 2 x+3 y=9
\end{aligned}$$
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