= Solution
Seek a <particular solution> $y_p=ax+b$. Substitution into the <linear ordinary differential equation> gives $6ax+2b=6x+2$, so $a=b=1$. Add the <complementary solution> from part (a):
$$
y=x+1+\frac A x+\frac B{x^2}.
$$
The two prescribed values yield $A+B=5$ and $A/2+B/4=3/2$. Thus $2A+B=6$, so $A=1$ and $B=4$. Therefore
$$
\boxed{y(x)=x+1+\frac1x+\frac4{x^2}\quad(x>0)}.
$$
Direct substitution gives $y(1)=7$ and $y(2)=9/2$, and the <homogeneous solution> contributes zero to the left-hand differential operator.
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