 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):The two prescribed values yield $A+B=5$ and $A/2+B/4=3/2$. Thus $2A+B=6$, so $A=1$ and $B=4$. ThereforeDirect substitution gives $y(1)=7$ and $y(2)=9/2$, and the homogeneous solution contributes zero to the left-hand differential operator.