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:

Question

Aicharya · Accepted Answer

$\begin{aligned} & (2 y-5) \frac{d y}{d x}=-3 \ & (2 y-5) d y=-3 d x \ & 2 \cdot \frac{y^2}{2}-5 y=-3 x+\lambda\end{aligned}$ $\because$ Curve passes through $(0,1)$ $\Rightarrow \lambda=-4$ $\because$ Curve will be $\left(y-\frac{5}{2}ight)^2=-3\left(x-\frac{3}{4}ight)$ $	herefore$ Vertex of parabola will be $\left(\frac{3}{4}, \frac{5}{2}ight)$ $\because 2 x+3 y=9$

The solution curve, of the differential equation $2 y \frac{\mathrm{d} y}{\mathrm{~d} x}+3=5…

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