Consider the following initial-value problem (IVP) defining a function y. Here a > 0 is a constant and f (t) is a given function. (* * ay' (t ) + y(t) = f(t), y(0) = 0. Notice that when a is extremely small, the ODE here is not much different from the identity y(t) = f(t). In what follows we will think of the function f(t) as an "input" and the solution y(t) as an "output" and investigate why this setup is described as a "low-pass filter". (b) Assuming f(t) = 1 for all t in (**), find the unique solution y(t). Hand in both the exact formula, with supporting work, and a reasonable plot of the graph of y(t). (For plotting, use a = 0.1 and consider 0