Lecture 18. Undetermined Coefficients IV: The Revenge1. This Is It For Undetermined Coefficients, I Promise2. Variation of ParametersDifferential EquationsLECTURE 18Undetermined Coefficients: The Revenge1. This Is It For Undetermined Coefficients, I PromiseWe ended last class by noticing that if part of our guess for a particular solution Yp(t) coincidedwith a bit of the complimentary solution, we needed to multiply the offending part by t to ensurethat it doesn’t vanish when we plug it into the equation. Let’s look at some examples.Example 18.1. Write down a guess for the form of a particular solution to the followingdifferential equations.(1) y00− 3y0− 28y = 6t + e−4t− 2First, we find the complimentary solution. It isyc(t) = c1e7t+ c2e−4t.Our nonhomogeneous term is g(t) = 6t + e−4t− 2, which we can rearrange to group thepolynomial terms together as g(t) = 6t − 2 + e−4t. Thus our initial guess would beAt + B + Ce−4t.The first two terms aren’t a problem, but the Ce−4tterm also appears in the complimentarysolution. Since Cte−4tdoesn’t show up in the complimentary solution, our final guess isYp(t) = At + B + Cte−4t.(2) y00− 64y = t2e8t+ cos(t)The complimentary solution isyc(t) = c1e8t+ c2e−8t.Our initial guess for a particular solution is(At + Bt + C)e8t+ D cos(t) + E sin(t).If we distributed the exponential through the polynomial, we’d have a Ce8tterm whichalso showed up in our complimentary solution. What we’ll need to do is to multiply theentire first term by t (to see why, just differentiate and plug in...you’ll see that if we don’t,we’ll end up losing a coefficient which we’ll need later). So our final guess isYp(t) = t(At2+ Bt + C)e8t+ D cos(t) + E sin(t).(3) y00+ 4y0= e−tcos(2t) + t sin(2t)The complimentary solution isyc(t) = c1cos(2t) + c2sin(2t).Our first guess for a particular solution would bee−t(A cos(2t) + B sin(2t)) + (Ct + D) cos(2t) + (Et + F ) sin(2t).First, we notice that despite having similar looking terms, we can’t actually combine any-thing, since the similar terms are multiplied by factors which don’t just differ by constantcoefficients. Next, we notice that both the second and third terms involve components ofthe complimentary solution: D cos(2t) and F sin(2t). Thus we’ll need to multiply those1Differential Equations Lecture 18: Undetermined Coefficients: The Revengetwo terms by t. The first term is ok, though, since if we multiplied it out we wouldhave a product of an exponential and a sine or a cosine, and those aren’t terms in thecomplimentary solution. So we end up withYp(t) = e−t(A cos(2t) + B sin(2t)) + t(Ct + D) cos(2t) + t(Et + F ) sin(2t).(4) y00+ 2y0+ 5y = e−tcos(2t) + t sin(2t)Notice that the nonhomogeneous term in this example is the same as in the previousone; we’ve just changed the differential equation. Here, the complimentary solution isyc(t) = c1e−tcos(2t) + c2e−tsin(2t).Our initial guess for the particular solution is the same as in the last example:e−t(A cos(2t) + B sin(2t) + (Ct + D) cos(2t) + (Et + F ) sin(2t).This time, the first term causes the problem, while the second and third are fine just asthey are. So we’ll multiply the first by t:Yp(t) = te−t(A cos(2t) + B sin(2t)) + (Ct + D) cos(2t) + (Et + F ) sin(2t).(5) y00+ 4y0+ 4y = t2e−2t+ 2e−2tHere the complimentary solution isyc(t) = c1e−2t+ c2te−2t.Notice that we can factor out a e−2tfrom our nonhomogeneous term, which then becomesg(t) = (t2+ 2)e−2t. This is the product of a polynomial and an exponential, so our initialguess is(At2+ Bt + C)e−2t.But this Ce−2tterm is the first term in our complimentary solution. Also, we have Bte−2t,which is the second term in the complimentary solution. So this is no good. Next, we trymultiplying by t:t(At2+ Bt + C)e−2t.This still causes problems: the Cte−2tterm is still the second term in our complimentarysolution. If we multiply by t2, though, we have no problems, and so our final guess isYp(t) = t2(At2+ Bt + C)e−2t. So that’s about it for Undetermined Coefficients. As long as you’re comfortable with the guessesfor the basic types, are on the lookout for coefficients that should be combined, and check thatyou’re not replicating any part of the complimentary solution, you can’t go wrong.2. Variation of ParametersThe other major method for determining a particular solution to the linear nonhomogeneousequationp(t)y00+ q(t)y0+ r(t)y = g(t)is Variation of Parameters. Undetermined Coefficients works well enough, but the algebra can bequite messy and it only works for a handful of equations.Variation of Parameters is a much more general method, but it has its drawbacks. We couldsometimes do Undetermined Coefficients without the complimentary solution, but this is neverpossible for Variation of Parameters. Also, while Undetermined Coefficients reduces the problemof finding a particular solution to an algebraic one, Variation of Parameters involves taking someintegrals, which we may not always be able to do. So we’ll always be able to write down a formula2Differential Equations Lecture 18: Undetermined Coefficients: The Revengefor the solution, but we may not be able to explicitly find it. Still, its generality makes it worthdiscussing.How does it work? For this method, we start by dividing through by p(t): we’ll want to startwith y00having a coefficient of 1. So the equation we’re dealing with is really(18.1) y00+ q(t)y0+ r(t)y = g(t).Next, suppose we know the complimentary solution(18.2) yc(t) = c1y1(t) + c2y2(t).Recall that this is the general solution to the homogeneous equationy00+ q(t)y0+ r(t)y = 0.We’re going to see if we can find two functions u1(t) and u2(t) such that(18.3) Yp(t) = u1(t)y1(t) + u2(t)y2(t).This is similar in spirit to doing reduction of order: we know that combinations of y1and y2withconstant coefficients gives solutions to Equation 18.2, so the next thing to try is a combination ofy1and y2involving nonconstant functions.However, to do this, we need to make an assumption. If we differentiate Equation 18.3, weobtainY0p= u01y1+ u1y01+ u02y2+ u2y02.The assumption we will make is thatu01y1+ u02y2= 0.Why do we make this assumption? Unfortunately, I don’t have a good answer for this. It turnsout that it works: it simplifies the expression we’ll end up getting so that we can solve it, and itdoesn’t interfere with obtaining a solution. With this assumption, the first derivative becomesY0p= u1y01+ u2y02.Then the second
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