Reducing $(2x-1)\dfrac{d^2y}{dx^2}-2\dfrac{dy}{dx}=0$ with $y=2$ and $\dfrac{dy}{dx}=3$ at $x=0$ gives the particular solution:

1. Substitute $P=\dfrac{dy}{dx}$ to get the first-order equation $(2x-1)\dfrac{dP}{dx}-2P=0$. 2. Separating and integrating with an integrating factor gives $P=A(2x-1)$, so $y=Ax^2-Ax+C$. 3. This has the general form $y=Ax^2+Bx+C$ with $B=-A$. 4. At $x=0$, $y=2$ fixes $C=2$; and $\dfrac{dy}{dx}=3$ at $x=0$ gives $B=3$, hence $A=-3$. 5. So $y=-3x^2+3x+2=-3(x^2-x)+2$. The sign of $A$ is the whole difference from $3(x^2-x)+2$. _Source: NECTA ACSEE 2023 Advanced Mathematics 142/2, Question 7: Differential Equations_