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:

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

Rearranging: $$(2y - 5)\frac{dy}{dx} = -3$$, so $$\frac{dy}{dx} = \frac{-3}{2y-5} = \frac{3}{5-2y}$$.

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

Integrating: $$x = \frac{5y - y^2}{3} + C$$.

Using $$(0, 1)$$: $$0 = \frac{5-1}{3} + C \implies C = -\frac{4}{3}$$.

$$x = \frac{5y - y^2 - 4}{3} \implies 3x = -y^2 + 5y - 4 = -(y^2 - 5y + 4)$$.

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

Completing the square: $$(y - \frac{5}{2})^2 = -3x + \frac{25}{4} - 4 = -3x + \frac{9}{4}$$

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

This is a parabola with vertex at $$\left(\frac{3}{4}, \frac{5}{2}\right)$$.

Check: $$2x + 3y = 2 \cdot \frac{3}{4} + 3 \cdot \frac{5}{2} = \frac{3}{2} + \frac{15}{2} = 9$$.

The correct answer is Option 1: $$2x + 3y = 9$$.