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## Calculus and Analytic Geometry

by: Zechariah Hilpert

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# Calculus and Analytic Geometry MATH 222

Zechariah Hilpert
UW
GPA 3.8

Staff

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COURSE
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Staff
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2
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KARMA
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## Popular in Mathematics (M)

This 2 page Class Notes was uploaded by Zechariah Hilpert on Thursday September 17, 2015. The Class Notes belongs to MATH 222 at University of Wisconsin - Madison taught by Staff in Fall. Since its upload, it has received 13 views. For similar materials see /class/205291/math-222-university-of-wisconsin-madison in Mathematics (M) at University of Wisconsin - Madison.

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Date Created: 09/17/15
fI he method of Undetermined Coef cients lOur text contains two techniques for nding a particular solution to a nonhomogeneous lineail differential equation with constant coef cients in 1772 The second one is called Variation of Parameters This is a owerful tool and you should de n el r39 learn to use Variation of P ameter if you intend to go on in subjects where such equations come up frequently But it is also morel complicated to apply in many situations The other technique which works to solve some suchl problems is called Undetermined Coefficients and when it does work it is generally much easier td use In our class I will only expect you to solve problems where this technique can be used Youl are welcome to use the more powerful Variation of Parameters technique on these problems but I am guaranteeing that any nonlmmou eneon linear differential equation with constant coef rents that you need to solve for this class will be one that be done using Undetermined Coef cients Before proceeding remember what the goal is for either Undetermined Coef cients or Variationl of Parameters in our somewhat restricted circh quottame quot e have learned how to find solutionsl of an equation like ay by39 cy or in theory higher order equations which have I constant coefficients a b and c on y and its derivatives and j3931 0 for the other side ie the equation is homogeneous We want to handle the case where f L is not zero but as a step we still nd all solutions to the homogeneous equation that results from replacing by 0 The text calls this equation where f has been replaced by J the complementary equation I will use ml to denote the solutions to the homogeneous complementary equation We have learned that if we can nd any one solution to the real equation where f is not zero I will use 21 to denote that particular solution we can combine it with the general solution y to the complementary equation and get all solutions to the non homogeneous equation we really want to solve In summary w nd all solutions to ay by cy 0 and call that yh and any one solution to ay by 6 and call it 21 and then 1 yp will represent all solutions to the equation ay by y f that we want to solve The problem we still have to solve is nding that particular solution 11p Since any solution any function that solves ay by cy f r is OK any way of nding 11 is ne including a luckyl guess I11 a real worldquot7 problem you might well have an idea for a solution based 011 what you know about the problem But since luck is not always with you we need a more quot tematic way td nd yp and that is what both Variation of Parameters and Undetermined Coefficients attempt td provide You could describe Undetermined Coefficients as guided guessing Here is how we carry out the guessing If we lucky and f L is particularly simple and it does not interact in bad ways with the solutions to the homogeneous equation we may be able tol nd a solution 11 as follows We will assume that f being simple means it is some combination of terms like 6 cosk r sinkr and polynomials 112 dr If both cosine and sine terms are present the same k2 must appear in all of them We construct a candidate for 11 using the table on the next page There are extended versions of the table that would work for other functions but if you get that deeply involved you will also have Variation of Parameters in your tool kit NOTE There is a typo in the table as it appears in our textbook It says k is not a rootquot wherd it should say k1 is not a root How we use this is best shown by examples As a simple rst example we will nd the solutionsl of 2 21 7 3y 6 The auxiliary or characteristic equation is r2 27 7 3 0 with roots 7 73 and 7 1 Hence we can write 7167 726139 The rinht side of the e uation in this cas 6 a polynomial of degree zero Since 0 is not a root of the characteristic equation we try for yp a generic polynomial of degree zero ie a constant Let yp F where F represents a constant yet to be determined Then yz and y are both 0 Plugging this into the equation for yp to be a solution we must have 0 0 3F 2 6 Hence the only possibility is F 2 but that is good enough The complete solution is obtained by adding yh Clea 3 026quot 2 as the general solution For a term in f If Then use a term like A F which is a multiple of represent numbers to be de termined sinka or coska M is not a root of the characteristic equation Acoska B sinka M is a root of the characteristic equation A11 coska Ba sinka 6quot n is not a root of the characteristic equation 06quot n is a single root of the characteristic equation Ca 6quot n is a double root of the characteristic equa 0132 6quot tion A polynomial 0432 O is not a root of the characteristic equation a polynomial D1132 Ea F g y of degree at of the same degree as 0432 most 2 g y 0 is a single root of the characteristic equation a polynomial D333Eaj2Faj of degree one more 0 is a double root of the characteristic equa a polynomial D1134 E1133 tion F1132 of degree two more Another example Solve y y 5695 sin21 The characteristic equation is 7 2 7 0 with roots 7 O and 7 1 The complementary equation will thus have solutions yh Clem 026133 01 l 0263 The terms in f are 5695 and sin2z Start with 5695 which is a constant times 6quot where n 1 Since 1 is a single root of the characteristic equation we put the term Ca 6 into yp Now sin2z is a multiple of sincz where k 2 and 2 is not a root of the characteristic equation so we include A cos2zB sin2z in our candidate solution Thus we arrive at yp A cos2zB sin2zCz e and have to determine the coef cients A B and C Then yz 2A sin2z 23 cos21 0690 Ca egg and y 4A cos21 4B sin212Cex 051 ex Putting these into the equation y y 5695 sin2z we get 4A cos21 4B sin2z 2Ce C39a ex2A sin2z 2B cos21 Cex Ca 6 3 5633 sin2z We collect together the terms from both sides with cos21 and get 4A 2B 2 O From the sin2z terms we get 4B 2A 2 1 The 5136 terms cancel out a consequence of the fact that 1 was a root of the characteristic equation and the 6 3 terms give C 5 Solving we get A 110 and B 15 Hence our particular solution is yp cos 21 sin 21 513673 You can check that yp actually does work in y y 5695 sin2z Combining this with the general solution to the complementary equation yh we get cos 21 i sin 21 5513690 Cl 02696 Note that this method does have limitations As given here for example it would not apply to Problems 18 and 28 on section 17 16 in our text since the limited table I included above does not suggest what to use for yp when f includes the tangent function or has a product of two of the functions that are in the table It would also not work for problem 29 if you were asked to nd yp but in that problem you are given yp It will work for all the problems I will expect you to do Bob Wilson July 28 2008

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