Solving second order differential equation in a hurry During last test I had to solve second order ODE: $y'' + 3y = e^x + 1$. I managed to write the general solution of homogenous equation (with RHS replaced by zero):
$$y = c_1 \cos \left(\sqrt{3} x\right) + c_2 \sin \left(\sqrt{3} x\right).$$
That wasn't hard. But how to find (at least one) the general solution? Using sosmath (Method of Variation of Parameters) I found out that:
\begin{align}
 u_1 & = \sqrt{3} \int \sin \left(\sqrt{3} x\right) (e^x + 1) \,\textrm{d}x =  \frac{\sqrt{3}}{12} \cdot \left(3e^x \sin \left(\sqrt 3 x\right) - \sqrt 3 (4+3e^x) \cos \left(\sqrt 3 x\right) \right) + C_1 \\
 u_2 & = \sqrt{3} \int \cos \left(\sqrt{3} x\right) (e^x + 1) \,\textrm{d}x = \frac{\sqrt{3}}{12} \cdot \left(3e^x \cos \left(\sqrt 3 x\right) + \sqrt 3 (4+3e^x) \sin \left(\sqrt 3 x\right) \right) + C_2,
\end{align}
after what I don't know how to proceed. There must be a simpler way, because one can guess the solution to be $$y(x) = \frac{4 + 3e^x}{12}.$$

try the method undetermined coefficients; try a particular solution of the form $$ y = ae^x + b, y' = ae^x, y'' = ae^x. $$ sub this in $$y'' + 3y = e^x + 1 \to ae^x + 3(ae^x + b)=e^x + 1 \to a = \frac 14, b = \frac 13 $$  
therefore, a particular solution is $$y_p = \frac 14 e^x + \frac 13, y_g = Ce^{-3x}+y_p  $$
