1. Why guessing and checking is alright in solving differentialequations In lecture (and possibly in other courses), you have seendifferential equations solved by looking at the equation, movingparts around, reasoning about it using an analogy witheigenvalue/eigenspaces, and then seeing that the solution that weproposed actually works — i.e. satisfies all the conditions of thedifferential equation problem. This process should have felt a bitdifferent than how you have seen how systems of linear equationsare solved (by doing Gaussian Elimination) where it was clear thatevery step was valid. Indeed, it is different. Although theeigenvalue/eigenspace analogy to differential equations can be madeprecise and rigorous, doing that carefully is beyond the scope ofthis course. In effect, all of that reasoning in between seeing theproblem and checking the solution can be considered a kind ofinspired guessing. This should lead you to a natural question — howcan we be sure that we have found all of the solutions? We’vechecked to see that the solution we found solves the equations, butmaybe there are more solutions that are different. How can we besure? After all, we are using the solution of the differentialequation for its predictive power — for example, we are using thefact of RC time constants to argue that this limits the speed ofdigital computation. Making such inferences is only proper if wehave indeed found the only solution to the differential equation.In the mathematical literature, this is sometimes referred to asthe problem of establishing the “uniqueness” of solutions. Theconcept is also very important for us in engineering contexts. Youhave already seen in EE16A’s touchscreen module that node voltagesneed not be unique, and that is why you need to specify a ground inyour circuit. You also saw this concept in EE16A’s localizationmodule where you learned how to approach inconsistent linearequations by the method of least squares: you started with nosolutions, allowed some error and then got infinitely manypotential solutions with error. To make the solution unique, youhad to specify that you wanted to minimize the size of thehypothesized error. This problem walks you through an elementaryproof of the uniqueness of solutions to a simple scalardifferential equation of the form

d/dt x(t) = ?x(t) (1) with initial condition x(0) =x0  (2)

Being able to do simple proofs is an important skill, not onlyin its own right, but also for the systematic logical thinking thatit exercises. This problem has multiple parts, but the goal issimply to help you see how you could have come up with this proofentirely on your own.

(a) Please verify that the guessed solution xd(t) = x0e ?tsatisfies (1) and (2).

(b) To show that this solution is in fact unique, we need toconsider a hypothetical y(t) that also satisfies (1) and (2)

Our goal is to show that y(t) = x(t) for all t ? 0. (The domaint ? 0 is where we have defined the conditions (1) and (2). Outsideof that domain, we don’t have any constraints. ) How can we showthat two things are equal? In the past, you have probably shownthat two quantities or functions are equal by starting with one ofthem, and then manipulating the expression for it using validsubstitutions and simplifications until you get the expression forthe other one. However, here, we don’t have an expression for y(t)so that style of approach won’t work. In such cases, we basicallyhave a couple of basic ways of showing that two things are thesame. • Take the difference of them, and somehow argue that it is0. • Take the ratio of them, and somehow argue that it is 1. Wewill follow the ratio approach in this problem. First assume thatx0 6= 0. In this case, we are free to define z(t) = y(t) xd (t)since we are dividing by something other than zero. What isz(0)?

(c) Take the derivative d dtz(t) and simplify using (1) and whatyou know about the derivative of xd(t). (HINT: The quotient rulefor differentiation might be helpful since a ratio is involved.)You should see that this derivative is always 0 and hence z(t) doesnot change. What does that imply for y and xd ?

(d) At this point, we have shown uniqueness in most cases. Justone special case is left: x0 = 0. Here, the division approachdoesn’t seem to work because we are not permitted to divide by zeroand xd(t) = 0. However, we want to show that y(t) = 0 here as well.Fundamentally, the argument we want to make is of the “it can’tpossibly be otherwise” variety. Consequently, a proof bycontradiction can be easier to start. In such proofs, we start byassuming the thing that we want to show is not possible. So assumethat y(t) is not identically 0 everywhere for t > 0. What doesthis mean? This means that there is some t0 > 0 for which y(t0)= k 6= 0. (Otherwise, it would be zero everywhere.) We want tocreate a contradiction. It is clear that we will have no easycontradiction if we just move forward for t > t0 because we haveno information given about such solutions y(t) that we cancontradict. What do we know about? We have (2) which says somethingabout y(0). This means, that we need to somehow move backward intime from t0. That way, we can hope to contradict the initialcondition of 0. What do we have to work with? Well, we just didsome work in the previous parts establishing uniqueness ofsolutions assuming nonzero initial conditions. How can we view whathappens at t0 as a kind of nonzero initial condition? Apply thechange of variablest = t0?? to (1) to get a new differentialequation for xe(?) = x(t0??) that specifies how d d? xe(?) mustrelate to xe(?). This should hold for ?? < ? ? t0.

(e) Because the previous part resulted in a differentialequation of a form for which we have already proved uniqueness forthe case of nonzero initial condition, and since ye(0) = y(t0) = k6= 0, we know what ye(?) must be. Write the expressions for ye(?)for ? ? [0,t0] and what that implies for y(t) for t ? [0,t0].