Solution
= Solution
The <characteristic equation> of the associated <homogeneous linear differential equation> is $m^2+7m+12=(m+3)(m+4)=0$. A constant <particular integral> is $1/2$, so
$$
y=\frac12+Ae^{-3x}+Be^{-4x}.
$$
The first <boundary condition> gives $A+B=-1/2$. The second gives $Ae^{-1}+Be^{-4/3}=-e^{-1}/2$. Eliminating $A$ yields $B(e^{-4/3}-e^{-1})=0$, hence $B=0$ and $A=-1/2$. Therefore
$$
\boxed{y(x)=\frac{1-e^{-3x}}2,\qquad y(1)=\frac{1-e^{-3}}2}.
$$