Solution
= Solution
For the <general solution> in part (i),
$$
y(0)=A+B=1,\qquad y'(0)=-\frac34-2B=0.
$$
Thus $B=-3/8$ and $A=11/8$, giving the solution of the <initial value problem>
$$
\boxed{y(x)=\frac34x^2-\frac34x+\frac{11}{8}-\frac38e^{-2x}.}
$$
Direct substitution checks both the differential equation and the two initial conditions.