Find the solution of this differential equation whose graph it is through the point $(1,3e)$.
The general solution to a first-order DE is given by $y(x) = y_h(x) + y_p(x) = cg(x) + y_p(x)$, where $c$ is arbitrary and $g$ is a solution to the associated homogeneous equation. Since you know two solutions $y_1$ and $y_2$, the difference must be proportional to $g$, which means that $g(x) = x^2 + 1$ as you predicted. Moreover, this implies that the particular solution is $y_p(x) = e^x$. Thus, we have $y(x) = c(x^2 + 1) + e^x$ and you can now solve for the constant $c$ using the initial condition $y(1) = 3e$.