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PSU MATH 251 - Undetermined Coefficients Beyond Thunderdome

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Lecture 17. Undetermined Coefficients Beyond Thunderdome1. Recap (again!)2. SumsDifferential EquationsLECTURE 17Undetermined Coefficients Beyond Thunderdome1. Recap (again!)The point of the method of Undetermined Coefficients is to make a guess at the form of aparticular solution Yp(t) of a nonhomogeneous equation based on the form of the nonhomogeneousterm g(t). This method works for exponentials (if g(t) = aeαt, we guess Yp(t) = Aeαt), trigfunctions (if g(t) = a cos(αt) or a sin(αt), we guess Yp(t) = A cos(αt) + B sin(αt)), and polynomials(if g(t) = antn+ an−1tn−1+ . . . + a1t + a0, we guess Yp(t) = Antn+ An−1tn−1+ . . . + A1t + A0).We saw last class that if g(t) is a product of these basic types, our guess for Yp(t) is the productof the guesses for these basic types, with care taken to make sure we don’t have any extraneouscoefficients.Let’s practice writing down the guesses for a couple of these.Example 17.1. Write down the form of the particular solution toy00− 4y0− 12y = g(t)for the following g(t)s:(1) g(t) =9t2− 103tcos(t)Here we’ve got the product of a quadratic and a cosine. The guess for the quadratic isAt2+ Bt + Cand the guess for the cosine isD cos(t) + E sin(t).Multiplying the two guesses givesAt2+ Bt + C(D cos t) +At2+ Bt + C(E sin t)ADt2+ BDt + CDcos t +AEt2+ BEt + CEsin t.Each of our coefficients here is a product of two constants, which is just another constant.So, as before, to simplify everything, we’ll replace each of those with a single constant toyield the following guess.Yp(t) =At2+ Bt + Ccos t +Dt2+ Et + Fsin t.This is indicative of the general rule for a product of a polynomial and a trig function.Write down the guess for the polynomial, multiplied by a cosine, then add to that theguess for the polynomial (with different constants!) multiplied by a sine.(2) g(t) = e−2t(3 − 5t) cos(9t)This nonhomogeneous term has all three things. So, combining the two general rulesfrom before, we getYp(t) = e−2t(At + B) cos(9t) + e−2t(Ct + D) sin(9t). 1Differential Equations Lecture 17: Undetermined Coefficients Beyond Thunderdome2. SumsWe have the following important fact. If Y1satisfiesp(t)y00+ q(t)y0+ r(t)y = g1(t)and Y2satisfiesp(t)y00+ q(t)y0+ r(t)y = g2(t), then Y1+ Y2satisfiesp(t)y00+ q(t)y0+ r(t)y = g1(t) + g2(t).This means that if our nonhomogeneous term g(t) is a sum of terms we know how to deal with,we can write down the guesses for each of those terms and add them together for our guess. Wemay just need to be careful about redundant terms.Example 17.2. Find a particular solution toy00− 4y0− 12y = e7t+ 12.Our nonhomogeneous term g(t) = e7t+ 12 is the sum of an exponential g1(t) = e7t) and a 0thdegree polynomial g2(t) = 12. The guess for g1(t) isAe7twhile the guess for g2(t) isB.Adding these together givesAe7t+ B.This can’t be simplified in any way, so based on our previous fact we’ll go ahead and guessYp(t) = Ae7t+ B. Differentiating and plugging in yields49Ae7t− 28Ae7t− 12Ae7t− 12B = e7t+ 129Ae7t− 12B = e7t+ 12.Setting coefficients equal gives A =19and B = −1, so our particular solution isYp(t) =19e7t− 1. Let’s practice writing down some guesses.Example 17.3. Write down the form of a particular solution toy00− 4y0− 12y = g(t)for each of the following g(t)’s:(1) g(t) = 2 cos(3t) − 9 sin(3t)Our guess for the cosine isA cos(3t) + B sin(3t).Additionally, our guess for the sine isC cos(3t) + D sin(3t).So if we add the two of them together, we obtainA cos(3t) + B sin(3t) + C cos(3t) + D sin(3t) = (A + C) cos(3t) + (B + D) sin(3t).2Differential Equations Lecture 17: Undetermined Coefficients Beyond ThunderdomeBut A + C and B + D are just some constants, so we can replace them as such as we justyet a guess ofYp(t) = A cos(3t) + B sin(3t).(2) g(t) = sin(t) − 2 sin(14t) − 5 cos(14t)Again, we start by writing down the guesses for the individual terms. The guess forsin(t) isA cos(t) + B sin(t).Since they have the same argument, the previous example showed us that we can combinethe guesses for 2 sin(14t) and −5 cos(14t) intoC cos(14t) + D sin(14t).We won’t be able to further combine anything, since the trig functions here have differentarguments, and so we end up with a guess ofYp(t) = A cos(t) + B sin(t) + C cos(14t) + D sin(14t).(3) g(t) = 7 sin(10t) − 5t2+ 4tHere we have the sum of a trig function (7 sin(10t) and a quadratic polynomial (−5t2+4t). The guess for the sine isA cos(10t) + B sin(10t)while the guess for the quadratic isCt2+ Dt + E.Since there’s nothing to be consolidated, our guess isYp(t) = A cos(10t) + B sin(10t) + Ct2+ Dt + E.(4) g(t) = 9et+ 3te−5t− 5e−5tIf we write down the sum of each of the individual guesses, we would obtain a guess ofAet+ (Bt + C)e−5t+ De−5t= Aet+ (Bt + (C + D))e−5t.Of course, C + D is just another constant, so our final guess would beYp(t) = Aet+ (Bt + C)e−5t.We could also have noticed that g(t) could be rewritten asg(t) = 9et+ (3t − 5)e−5t,in which case writing down the sum of our guesses directly works out to what we justobtained through a slightly more indirect approach.(5) g(t) = t2sin(t) + 4 cos(t).Writing down the sum of the guesses gives(At2+ Bt + C) cos(t) + (Dt2+ Et + F ) sin(t) + G cos(t) + H sin(t),which simplifies to(At2+ Bt + (C + G)) cos(t) + (Dt2+ Et + (F + H)) sin(t).Once again, since C + G and F + H are just constants, we end up withYp(t) = (At2+ Bt + C) cos(t) + (Dt2+ Et + F ) sin(t).The last two examples have shown us that if we have two terms in our g(t) whoseguesses differ only by a polynomial factor, then we can just look for the term correspondingto the highest degree polynomial and the guess for that term will include the guess for thesecond one. Without making this observation, of course, we can just do what we did hereand observe that certain coefficients combine together.3Differential Equations Lecture 17: Undetermined Coefficients Beyond Thunderdome(6) g(t) = 3e−3t+ e−3tsin(3t) + cos(3t).We’ll need to start here by writing down the sums of each of the guesses:Ae−3t+ e−3t(B cos(3t) + C sin(3t)) + D cos(3t) + E sin(3t).Notice that we can’t combine anything, since the various terms that look similar differ bynonconstant factors, not just constant coefficients. So our guess is justYp(t) = Ae−3t+ e−3t(B cos(3t) + C sin(3t)) + D cos(3t) + E sin(3t). As we’ve seen, sums are quite straightforward to deal with as long as we’re on the look out forredundant


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