1. Given differential equation

$2y\frac{dy}{dx}+3=5\frac{dy}{dx}$

Rearrange:

$2y\frac{dy}{dx}-5\frac{dy}{dx}=-3$

$\left(2y-5\right)\frac{dy}{dx}=-3$

So,

$\frac{dy}{dx}=\frac{-3}{2y-5}$

Invert to get $\dfrac{dx}{dy}$:

$\frac{dx}{dy}=\frac{5-2y}{3}$

2. Integrate

$dx=\frac{5-2y}{3}\,dy$

Integrating,

$x=\frac{1}{3}\int(5-2y)\,dy$

$x=\frac{1}{3}(5y-y^2)+C$

Thus,

$3x=5y-y^2+C_1$

or

$y^2-5y+3x+C_2=0$

3. Use the point $(0,1)$

Substitute $x=0$, $y=1$:

$1-5+0+C_2=0$

$C_2=4$

Hence the solution curve is

$y^2-5y+3x+4=0$

This is a parabola, hence a conic.

4. Find the vertex

Complete the square in $y$:

$y^2-5y+3x+4=0$

$\left(y-\frac{5}{2}\right)^2-\frac{25}{4}+3x+4=0$

$\left(y-\frac{5}{2}\right)^2+3x-\frac{9}{4}=0$

$\left(y-\frac{5}{2}\right)^2=-3\left(x-\frac{3}{4}\right)$

So the vertex is

$\left(\frac{3}{4},\frac{5}{2}\right)$

5. Check which line passes through the vertex

Compute:

$2x+3y=2\left(\frac{3}{4}\right)+3\left(\frac{5}{2}\right)$

$=\frac{3}{2}+\frac{15}{2}=\frac{18}{2}=9$

So the vertex lies on

$2x+3y=9$

Hence the correct option is C.

6. Comparison with stored answer

Stored correct answer: C

Our derived answer: C

They agree.