This paper shows a simple tabular procedure \added{derived from the method of undetermined coefficients} for finding a particular solution to differential equations of the form \sum_{j=0}^m a_j\frac{d^j y}{dx^j} = P(x)e^{\alpha{}x}. This procedure reduces the derivatives of the product of an arbitrary polynomial and an exponential to rows of constants representing the coefficients of the terms. The rows are each multiplied by aj and summed to produce a mth order differential equation such that its solution is the polynomial part of the particular solution of the above equation. Solving this corresponding differential equation determines the coefficients of the polynomial. The underlying algebra of this conversion and its formulaic implication are then discussed. Using the formula derived, the particular solution is found. This procedure is based on but different than the method of undetermined coefficients because while the method of undetermined coefficients requires substitution of a product of a polynomial, Q, and an exponential into the differential equation immediately, this procedure is derived from the examination of the substitution of the product of any function and an exponential. This allows for a richer understanding of the relationship between the differential equation for y and the differential equation for Q. Ultimately this method is better than the method of undetermined coefficients because it is more straightforward. In any case, both methods solve the same problem but KMO is faster

Krohn, David

Mariño-Johnson, Daniel

Ouyang, John Paul

English

Rose-Hulman Institute of Technology: Rose-Hulman Scholar

Rose-Hulman Undergraduate Mathematics Journal Volume 15 Issue 1 Article 8 The KMO Method for Solving Non-homogenous, mth Order Differential Equations David Krohn University of Notre Dame Daniel Mariño-Johnson University of Maryland, College Park, daniel.marino.johnson@gmail.com John Paul Ouyang University of Maryland, College Park Follow this and additional works at: https://scholar.rose-hulman.edu/rhumj Recommended Citation Krohn, David; Mariño-Johnson, Daniel; and Ouyang, John Paul (2014) "The KMO Method for Solving Non-homogenous, mth Order Differential Equations," Rose-Hulman Undergraduate Mathematics Journal: Vol. 15 : Iss. 1 , Article 8. Available at: https://scholar.rose-hulman.edu/rhumj/vol15/iss1/8 Rose-HulmanUndergraduateMathematicsJournalSponsored byRose-Hulman Institute of TechnologyDepartment of MathematicsTerre Haute, IN 47803Email: mathjournal@rose-hulman.eduhttp://www.rose-hulman.edu/mathjournalThe KMO Method for SolvingNon-homogenous, mth OrderDifferential EquationsDavid Krohn a Daniel Mariño-Johnson bJohn Paul Ouyang cVolume 15, No. 1, Spring 2014aDepartment of Political Science, University of Notre DamebDepartment of Mathematics, University of Maryland, College ParkcDepartment of Biology, University of Maryland, College ParkRose-Hulman Undergraduate Mathematics JournalVolume 15, No. 1, Spring 2014The KMO Method for SolvingNon-homogenous, mth Order DifferentialEquationsDavid Krohn Daniel Mariño-Johnson John Paul OuyangAbstract. This paper shows a simple tabular procedure derived from the method ofundetermined coefficients for finding a particular solution to differential equationsof the form:m∑j=0ajdjydxj= P (x)eαx (1)This procedure reduces the derivatives of the product of an arbitrary polynomialand an exponential to rows of constants representing the coefficients of the terms.The rows are each multiplied by aj and summed to produce a mth order differentialequation such that it’s solution is the polynomial part of the particular solution ofequation 1. Solving this corresponding differential equation determines the coeffi-cients of the polynomial. The underlying algebra of this conversion and its formulaicimplication are then discussed. Using the formula derived, the particular solutionfor equation 1 is found. This procedure is based on but different in application thanthe method of undetermined coefficients because while the method of undeterminedcoefficients requires substitution of a product of a polynomial, Q, and an exponen-tial into the differential equation immediately, this procedure is derived from theexamination of the substitution of the product of any function and an exponential.This allows for a richer understanding of the relationship between the differentialequation for y and the differential equation for Q. Ultimately this method is betterthan the method of undetermined coefficients because it is more straightforward. Inany case, both methods solve the same problem but KMO is faster.Acknowledgements: We would like to thank Dr. Bell, our teacher, for his support andencouragement. At the same time, we would like to thank our parents and the HeightsSchool for their patience and generosity.Page 134 RHIT Undergrad. Math. J., Vol. 15, No. 11 IntroductionDifferential equations are a very old field of research. They began with Newton and hisformulation of natural laws using calculus. In particular, they were used to model physicalphenomena. Many famous mathematicians such as Leibniz, Euler, Laplace, and Lagrangealso made contributions to the field[1]. Current researchers focus more on nonlinear equationsdue to the increase in computational abilities and geometric techniques. Linear differentialequations are well understood especially those with constant coefficients. These equationswere used to simplistically model physical systems. For example, non-homogenous secondorder linear equations may be used to model forced springs where the forcing is the nonho-mogeneous part. So for such a system we have equations of the form md2ydt2+γ dydt+ky = F (t)where m is the mass, γ is the dampening constant, k is the spring constant, and F (t) isthe forcing function[1]. Of non-homogenous equations the ones treated in this paper are theones in the form of equation 1. The established method for solving these equations is themethod of undetermined coefficients. This procedure involves the substitution of particularsolution yp = eαxxs∑nk′=1 ck′xk′or yp = eαxQ(x) where s is the number of times α is aroot of the characteristic polynomial. This leads to a to a system of equations which yieldck′ . We instead consider the corresponding differential equation for Q first by means of atabular procedure, then by means of a formula derived from the underlying algebra at playin the tabular procedure. Coupling this formula with triangular matrix inversion and a moregeneral procedure we derive a particular solution of equation 1.First, let us examine the method of undetermined coefficients by using it to solve thefollowing equation:2d2ydx2+ 3dydx+ y = (5x+ 3)e−2x (2)We substitute y = (ax+b)e−2x into the equation to obtain with cumbersome applicationsof the product rule:(3a)xe−2x + (3b− 5a)e−2x = (5x+ 3)e−2xNow we can divide by e−2x and equate 3a with 5 and 3b− 5a with 3. We can then solvethese equations to obtain a and b.In this paper, we develop a technique to solve equation 1 which we call the KMO method.We begin explaining KMO by showing the tabular method from which it is derived. We dothis by using it on the example equation just given. We then discuss the algebra at play inthe tabular method and derive an equation that does the same thing as the table. Finally,we solve non-homogenous equations where the function on the right is a polynomial whichtogether with the aforementioned equation completely solves the problem with our method.Along the way, we test the example equation with this method and compare the solutionwith that of the method of undetermined coefficients.RHIT Undergrad. Math. J., Vol. 15, No. 1 Page 1352 Example Application of the Tabular ProcedureWe consider equation 2. The general solution to this nonhomogeneous equation is found byadding the particular solution to the solution of the corresponding homogenous equation.The homogenous solution is of the form: (with c1 and c2 constants)yh = c1eλ1x + c2eλ2xWhere c1 and c2 are constants. By the quadratic equation, the solutions of the charac-teristic equation may be found:λ1,2 =−3±√32 − 4(2)(1)2(2)The KMO method is concerned exclusively with finding a particular solution. Thushenceforth let y mean yp. Usually y is found with the method of undetermined coefficientsas detailed in [1]. The method of undetermined coefficients is somewhat time consuming asit often involves arduous algebra. The tabular procedure begins with a solution of the formy = Q(x)eαx where Q is a polynomial. In the case of equation 2 this is y = Q(x)e−2x. Usingan arbitrary polynomial eliminates the trial and error that sometimes occurs in the methodof undetermined coefficients. We take derivatives of y:y = Q(x)e−2xdydx=dQdxe−2x − 2Q(x)e−2xd2ydx2=d2Qdx2e−2x − 4dQdxe−2x + 4Q(x)e−2xNow we place the coefficients into a table:Q(x) dQdxd2Qdx2y 1 0 0dydx−2 1 0d2ydx24 −4 1The tabular format is useful for examining the coefficients because when the derivatives ofy are summed along with their coefficients, the coefficients of terms with like order derivativesof Q are summed together. For example, when the y term is added to the first derivativeterms, the coefficient 1 of Q(x)e−2x in y is added to the coefficient −2 of Q(x)e−2x in dydx.However, before the summation the coefficients of the differential equation for y must beaccounted for. Now we multiply each of the rows by the corresponding coefficients of thederivatives of y:Page 136 RHIT Undergrad. Math. J., Vol. 15, No. 1Q(x) dQdxd2Qdx21 · y 1 0 03 · dydx−2 1 02 · d2ydx24 −4 13 −5 2The sum of the components in each column produces the coefficients for the derivativesof Q in the corresponding differential equation for Q. Since all terms contain the sameexponential term, which is never zero, it may be divided out yielding:2d2Qdx2− 5dQdx+ 3Q(x) = 5x+ 3 (3)In equation 3, the Q(x) term is the polynomial of highest degree and by equivalencemust be of first degree. This means with Q(x) = ax+ b, dQdx= a, and d2Qdx2= 0. Substitutioninto equation 3 yields −5a+ 3ax+ 3b = 5x+ 3. Algebraic manipulation of the last equationyields a = 53and b = 349. Thus y = (53x+ 349)e−2x.3 Removal of the ExponentialWe will now show that the numbers in the table follow a certain pattern and thus we mayderive a formula for the coefficients bi of the corresponding differential equation for Q thatwill be useful for the larger problem at hand of solving equation 1.Lemma 3.1. The coefficients of the differential equation for Q,∑mi=0 bidiQdxi= P (x), resultingfrom substitution of Q(x)eαx for y into equation 1 and division by eαx are:bi =m∑j=0(ji)ajαj−iProof. The goal is to find the value of b (where b is a vector whose i’th component is bi) inthe differential equation for Q. Since taking successive derivatives of a product of functionsderives this equation, a rule for the nth derivatives of this product is useful. The Leibnizrule states that dndxnf(x)g(x) =∑ni=0(ni)difdxidn−igdxn−iwhere(ni)is the binomial coefficient. If fis let to be Q and g is let to be eαx, then by the Leibniz rule, djydxj=∑ji=0(ji)diQdxiαj−ieαx.Substitution into the differential equation of Q and division by eαx (a common nonzerofactor) yields the following equation:m∑j=0ajj∑i=0(ji)diQdxiαj−i = P (x)The aj can be moved inside the inner summation since either way all terms with j aresummed by the first summation. The inner summation may be summed to m instead sinceRHIT Undergrad. Math. J., Vol. 15, No. 1 Page 137terms indexed by i > j vanish due to the binomial coefficient. This is useful since as aresult the orders of summation may now be switched due to the commutative property ofsummation. With this new ordering, diQdximay be moved outside of the inner summation toyield the following equation:m∑i=0diQdxim∑j=0(ji)ajαj−i = P (x)The equation may be recognized as the differential equation for Q where bi is the innersummation.3.1 Application to Equation 2The vector a may be quickly seen to be [1, 3, 2], α = −2, and m = 2. Thus the expressionfor bi is as follows:2∑j=0(ji)aj(−2)j−iThe calculations of the coefficients are as follows:b0 = (1 · 1 · 1) + (1 · 3 · (−2)) + (1 · 2 · 4) = 3b1 = 0 + (1 · 3 · 1) + (2 · 2 · (−2)) = −5b2 = 0 + 0 + (1 · 2 · 1) = 2Thus the result is the same as equation 3.4 Particular Solution of Equation 4Before we state the particular solution we find, we establish some convenient notation. Let nbe the degree of P and let p be the index of the first non-vanishing coefficient of the differentialequation in Q. Let T and R be two sets such that T = {i | i ∈ Z, n −m + p < i ≤ n} andR = {i | i ∈ Z, 0 ≤ i ≤ n−m+p}. Also, let kl be the coefficients such that P (x) =∑nl=0 klxl.Finally, let aji =(i+j)!i!bj and nT = m− p.Theorem 4.1. A particular solution is Q(x)eαx where Q(x) =∑n+pk′=pck′xk′. The leadingcoefficient of Q is the leading coefficient of P divided by apn where n is the degree of P ;the remaining coefficients are determined recursively [3] in the following way. For t suchthat n − t ∈ T , cp+n−t = 1apn−t (kn−t −∑ti=1 ap+in−tcp+n−t+i) and for t such that n − t ∈ R,cp+n−t =1apn−t(kn−t −∑nTi=1 ap+in−tcp+n−t+i).Page 138 RHIT Undergrad. Math. J., Vol. 15, No. 1Figure 1: Region over which the summation takes placeProof. Let Pn(x) specify the degree of P as n. The converted differential equation of mthorder may now be explicitly written as the following:m∑j=0bjdjQdxj= Pn(x) (4)The degree ofQ depends on the value of p such that bp is the first non-vanishing coefficient.This may be attributed to the fact that if p=3 then the lowest order derivative is d3Qdx3meaning that if n = 4, d3Qdx3must be of degree 4 since there are no higher degree polynomialsbeing summed. The maximum degree of Q is n + p since the pth derivative of Q reducesto a polynomial of degree n, which is the same degree as P . This also implies ck′ fork′< p can be chosen to be zero since they do not appear in the equation and thereforeare not determined by it. Consequently, Q(x) =∑n+pi=0 cixi where the constant coefficients,ci, are to be found. SincedjQdxj=∑n+p−ji=0(i+j)!i!ci+jxi, substitution into equation 4 yields∑mj=p bj∑n+p−ji=0(i+j)!i!ci+jxi = Pn(x) or more simply:m∑j=pn+p−j∑i=0ajici+jxi = Pn(x) (5)As previously noted, vectors indexed by the dummy variable of the outer sum may bemoved into the inner sum. The product of (i+j)!i!and bj produces aji as previously defined.The sum on the left hand side of equation 5 may be imagined as summation over theparallelogram of integers pictured (see figure 1).The ith row represents the ith degree term of the resulting polynomial on the left handside of the equation. The unknown variable in question is c. As Q is differentiated, the ck′RHIT Undergrad. Math. J., Vol. 15, No. 1 Page 139Figure 2: One-dimensional subregions over which k′is constantcoefficients are pushed back to terms of lower degrees, this means that if a ck′ in column his on row t, in column h+ 1, that ck′ is on row t− 1. Thus the components of c are on thediagonals pictured (see figure 2).Regardless of m, p, or n, apncn+p is not summed with any other coefficient and thus solvingfor cn+p is simple. It is the solution to apncn+p = kn which is cn+p =knapn. There are threepossibilities for the rest of the summation. These possibilities depend on shape of the regionover which the summations take place. If p = 0 and m ≥ n, the region is triangular. Ifp = m, the region is the rectangular strip of ams with s = 0 . . . n. If otherwise, the region istrapezoidal. If p = m it is easy to see that the ck′ are solved in the same way as cn+p. Fortriangles and trapezoids, even though there is a summation taking place, the solution for thethird from last coefficient depends on the last and second from last coefficients. This is dueto the fact that the diagonals depend on the same coefficient. The triangular region T maybe seen to occur from i = n+ p−m+ 1 to n since for i less than that all terms of degree por more are being summed. The rectangular region R is thus i ≤ n+ p−m if it exists. Thesum of a row in T that is t units from the top is the sum of the pth term to the tth term fromit and the sum of these coefficients is the tth term from the leading coefficient of P . So wehave:kn−t = apn−tcn+p−t +t∑i=1ap+in−tcn+p−t+iAs previously discussed, the ck′ in the summation are already known; therefore withalgebra we may solve for cn+p−t. That is, subtraction by the summation and division by apn−tyields:Page 140 RHIT Undergrad. Math. J., Vol. 15, No. 1cp+n−t =1apn−t(kn−t −t∑i=1ap+in−tcp+n−t+i)The expression for ck′ whos indices are in R are similar except that the upper indexof the sum has no dependence on t rather is fixed by the number of js from p such thatp plus this number is m. This is nT . Thus the ck′ may be determined recursively bycp+n−t =1apn−t(kn−t −∑ti=1 ap+in−tcp+n−t+i) and cp+n−t =1apn−t(kn−t −∑nTi=1 ap+in−tcp+n−t+i).Combining the results from section 3 and this section we have the three equations deter-mining the coefficients of the solution:aji =(i+ j)!i!m∑k=0(ki)akαk−i (6)cp+n−t =1apn−t(kn−t −t∑i=1ap+in−tcp+n−t+i) if p+ n− t ∈ T (7)cp+n−t =1apn−t(kn−t −nT∑i=1ap+in−tcp+n−t+i) if p+ n− t ∈ R (8)An application of KMO consists of computing the ajj using equation 6 and then usingeither the equation 7 or both equations 7 and 8 to find the coefficients ci, 0 ≤ i ≤ n + p.The solution is then P (x)eαx.4.1 Application to Equation 2We are solving2d2ydx2+ 3dydx+ y = (5x+ 3)e−2xFirst we note the corresponding differential equation for Q is2d2Qdx2− 5dQdx+ 3Q(x) = 5x+ 3A quick examination of order and vanishing coefficients yields that m = 2, n = 1, andp = 0. As a result, the leading coefficient of Q is c1 and the region over which the summationtakes place is the triangular one pictured (see figure 3).Since cp+n =knapnand the leading coefficient of P is 5, c1 =5a01. Since aji =(i+j)!i!bj andb0 = 3, a01 = 1(3) = 3. Also we see c1 =53. Since cp+n−t =1apn−t(kn−t −∑ti=1 ap+in−tcp+n−t+i),RHIT Undergrad. Math. J., Vol. 15, No. 1 Page 141Figure 3: Region over which the summation takes place for equation 2c0 =1a00(3 − a10(53)). a00 = 3 and a10 = −5 so therefore c0 = 43 . Combining this information,Q(x) = 53x+ 349. The solution is the same as previously shown.5 ConclusionThe KMO method provides a quick alternative to solving differential equations by using theformulas derived. Substitution of parameters a, k and α into the formulae quickly yield aparticular solution to equation 4 in the form of a vector c. Furthermore, these formulae maybe programed into a computer with fractional arithmetic to obtain exact solutions. Includedis the numerical MATLAB code for KMO with triangular summation regions equivalent totriangular matrix inversion.References[1] W.E. Boyce, R.C. DiPrima, Elementary Differential Equations and Boundary ValueProblems. John Wiley and Sons, New York, 6th Edition, 1997.[2] A.B. Israel, R. Gilbert, Computer Supported Calculus, Springer Wein, New York, 1992.[3] R.L. Devaney, A First Course in Chaotic Dynamical Systems, Perseus Books, Boston,1992.Page 142 RHIT Undergrad. Math. J., Vol. 15, No. 1A MATLAB code for KMO1 function [ c ] = KMO(a, k, alpha)2 m = numel(a) − 1;3 n = numel(k) − 1;45 % calculate b6 b = zeros(1, m + 1);7 for i=0:m8 for j=i:m9 b(i+1) = b(i+1) + nchoosek(j,i)*a(j+1)*(alphaˆ(j−i));10 end11 end1213 % get p, nT, and the order of Q14 p = find(b, 1, 'first')−1;15 nT = m − p;16 c = zeros(1,n + p + 1);1718 % calculate c19 c(p + n + 1) = (1/(b(p + 1)*factorial(p + n))/factorial(n))*(k(n + 1));20 for t=1:nT − 121 sum = 0;22 for i=1:t;23 sum = sum + (c(p + n − t + i + 1)*(b(p + 1 + i)*factorial(p + n ...− t + i))/factorial(n − t));24 end25 c(p + n − t + 1) = (1/(b(p + 1)*factorial(p + n − t))/factorial(n − ...t))*(k(n − t + 1)−sum);26 end27 end

The KMO Method for Solving Non-homogenous, mth Order Differential Equations

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The KMO Method for Solving Non-homogenous, mth Order Differential Equations

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