Solution
= Solution
For $y=1+x^2$, the <derivatives> are $y'=2x$ and $y''=2$. Its endpoint values are $y(0)=1$ and $y(2)=5$, so it satisfies both the <ordinary differential equation> and the <boundary conditions>. Moreover any solution of $y''=2$ has the form $x^2+Ax+B$; the <boundary conditions> force $B=1$ and $A=0$. Thus the unique solution of the <boundary value problem> is
$$
\boxed{y(x)=1+x^2.}
$$