The intuition of the convolution integral is not the most easy to grasp. Searching the net is rather unhelpful. The discussion on Wikipedia is extensive, but for many of us too complicated. Gilbert Strang has an illuminating and nice example in his book «Calculus». However, his explanation is not «in depth».
The example is as follows: You put mony in the bank as a stream with constant flow rate $r$ (dollars/year). The yearly interest rate is $p$. The interest rate calculated continously is e^pt. Thus, in a infinite short time interval $dt$ your deposit is $dr$.
Now, after $t$ years, how much money do you have in the bank? If a deposit $r$ was made at time $a$, the deposit will, after the time interval $(t-a)$, have increased to $dr\cdot e^{(t-a)}$. This will apply to any depost $dr$ at any time $a$. To obtain the total sum, we just add all deposits and all interest: Thus, we get a convolution integral: $\int (re^{(t-a)} )dt$. Now, here the function $r$ is a constant. However, the reasoning will be the same if $r$ and $p$ varies, i.e. are functions. Thus, the money in the bank after $t$ years is the convolution of the mony that streams in with the interest rate.
In partial differential equations, the integral formulas used when analysing inhomogenous wave and transport equation are convolution integrals. At time t, you just add up all previous effects of the source funcion, just like the interest rate and deposits in the example above, and you use this sum not as a new initial condition at the time t. As these problems are linear, you just add up the convolution at time t with the initial condition at time 0.