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# ADV CALC FOR APPLIC I M 427K

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This 33 page Class Notes was uploaded by Reyes Glover on Sunday September 6, 2015. The Class Notes belongs to M 427K at University of Texas at Austin taught by Staff in Fall. Since its upload, it has received 36 views. For similar materials see /class/181494/m-427k-university-of-texas-at-austin in Mathematics (M) at University of Texas at Austin.

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Date Created: 09/06/15

The Method of Undetermined Coefficients Consider the nth order nonhomogeneous equation with constant coef cients quot71 an1y any bX where a0 a1 an are constant and bX is a nonconstant function of X Let ypX be a particular solution of the nonhomogeneous equation containing no arbitrary constants Let ycX c1f1X sz2X CnfnX where c1 c2 cn are arbitrary constants be the general solution of the corresponding homogeneous equation n71 aoym a1y a0yquota1y an1y any0 Then the general solution of the equation can be expressed as YX ypX t ycX In this section we will discuss how to find the particular solution ypX Remark The kind of functions bX for which the method of undetermined coefficients applies are actually quite restricted De nition A function fX is a UC function if it is either 1 Xquot where n is an integer n 2 0 2 e where a is a constant different from 0 3 sinbX c where b and c are constant and b i 0 4 cosbX c where b and c are constant and b i 0 5 any function that is a finite product of two or more functions of those four types Examples X3 e394x sin5X 7 Xze6x X cos3X eSXSln3X 7 7 cosX sin2X X4e395Xsin2X Remark The method of undetermined coefficients applies when the nonhomogeneous term bX in the nonhomogeneous equation is a linear combination of UC functions Remark Given a UC function fX each successive derivative of fX is either itself a constant multiple of a UC function or a linear combination of UC functions De nition Given a UC function fX We call UC set of fx to the set of all UC functions consisting of fX itself and all linearly independent functions of which the successive derivatives of fX are either constant multiples or linear combinations Example Find the UC set of fX 1 Given fX X5 its derivatives are f X 5x4 f X 20x3 r x 60X2 t 120x t 120 MK 0 ngt5 The linearly independent functions of which the successive derivatives of fX are either constant multiples or linear combinations are X4 X3 X X 1 Then UC set of X5 X5 X4 X3 X2 X l 2 Given fX sin 2X its derivatives are fX 2cos 2X f X 4sin 2X f X 8cos 2X f4 16cos 2X they are either multiples of sin 2X or cos 2X Then UC set of sin 2X sin 2X cos 2X 3 Given fX e its derivatives are fX aeax f X azeax 5 X a equotx They are all multiples of e Then UC set of eax em Remark Given fX and gX UC functions the product function hX fXgX is another UC function The UC set of hX is obtained multiplying each member of the UC set of fX times each member in the UC set of gX Example Let fX X3 and gX cos 2X then hX fXgX X3005 2X UC set X3 Xi X2 X 1 UC set cos 2X cos 2X sin X Then UC set X3005 2X X3cos 2X X3sin 2X X2005 2X X2sin 2X X cos 2X X sin 2X cos 2X sin 2X Remark The table below contains the most common cases of UC sets UC function UC set 1 xquot x xquotquot xvi x l 2 e e 3 sinbx c or sinbx L cosbx 5 cosbx c 4 e x xquotquotequot x 2 xe quot 2 5 xquot sinl7x r or xquot sinbx c 3 050 c xquot cosbx c 7639quot sinbx c xquotquot cosbx r x sin bx c x 050 c sinbx c cosbx 5 6 2 sinbx c or e sinbx r e cosbx 6 e c05bx c 7 g sinbx c or x39equot sinbx c x e C05bx c X cosbx c xquotquote sinbx cxquotquote cosbx 5 x2 sinl7x c are cosbx c e sinbx c 239 CGSUM c Method of Undetermined Coefficient Given the nonhomogeneous LODE With constant coefficients a0yquot a1yquot71 an71y any bX assume that the nonhomogeneous term bX can be eXpressed as a linear combination of UC functions bX A1L11X A2L12X AmumX where 111 uz um are UC functions and A1 A2 Am are known constants Steps to follow 1 Find the complementary solution ycX of the corresponding homogeneous equation determine the fundamental set of solutions F of the corresponding homogeneous equation 2 For each UC function 111 uz um determine its UC set Let s call Sj the UC set corresponding to the UC function uj 3 Compare all the UC sets Ifone UC set say Sr is equal to or included into another UC set say Sk then we disregard Sr from further consideration 4 Consider the UC sets which still remain after step 3 We compare each UC set with the fundamental set of solutions F of the corresponding homogeneous equation Let s say Si includes one or more members of F then we multiply each member of Si by the lowest positive integer power of X so that the resulting reVised UC set Si will contain no member of F We replace Si by the reVised set 85 5 We have at this point a certain of the original UC sets which were neither disregarded in step 3 nor reVised in step 4 b certain reVised UC sets resulting from step 4 Form a linear combination using unknown constant coefficients with the elements in these UC sets and claim it is a candidate to be a particular solution 6 Since particular solutions do not contain arbitrary coefficients to determine the unknown coefficients we substitute the linear combination in the nonhomogeneous LODE and creating a system of linear equations Example Solve the nonhomogeneous LODE l y 7 2y 7 3y 2ex 710 sin X Corresponding homogeneous equation y 7 2y 7 3y 0 Characteristic equation r2 7 2r 7 3 r 3r l 0 Zeros are r1 3 and r2 l Fundamental set of solutions F e3x e39x Complementary solution ycX c1e3x cze39x Nonhomogeneous term is bX 2ex 7 10 sin X it is the linear combination of ex and sin X The UC set ofex is 1 ex The UC set ofsin X is S sin X cos X 81 and S are not equal nor one is included in the other so both set remain Neither S1 nor S2 includes any element of F the fundamental set of solutions of the corresponding homogeneous equation S0 S1 and S2 remain intact We form a linear combination with the element in S1 and S2 using unknown coefficients ypAexBsinXCcos X To determine the unknown coefficient substitute the linear combination in the equation We must compute the first and second derivative yp AeXBcOSX7CsinX yp Aex7Bsin X7Ccos X Replace into the equation Aex7BsinX7Ccos X 2AexBcosX7CsinX73AexBsinXCcos X 2ex 7 10 sin X or 4AeX 4B 2C sin X 4C 7 2B cos X 2ex 7 10 sin X comparing lefthand side to right 7hand side we get the system of linear equations 4B 2C 10 4C 7 2B 0 Solving the system we get A l2 B 2 and C 1 Therefore ypX l2 ex 2 sin X 7 cos X The general solution of the nonhomogeneous equation is yX ycX ypX c1e3x c2e39x l2 ex 2 sin X 7 cos X 2 y 7y 4e39x 3ezx Corresponding homogeneous equation y 7 0 Characteristic equation r3 7 r rrzl rr lr 7 l 0 Zeros are r1 0 r2 l and r3 l Fundamental set of solutions F 1 ex ex Complementary solution ycX c1 c2ex c3e39x Nonhomogeneous term is bX 4e39x 3e2x it is the linear combination of e2x and ex The UC set ofe2x is S1ezx The UC set ofe39x is S2 e39x S1 and S2 are not equal nor one is included in the other so both set remain Since S2 e39x contains elements in F 1 ex ex then we multiply the elements in S2 by X and we obtain S2 Xe39x that does not contain any element of F So replace S2 by S2 Form a linear combination of the elements in S1 and S2 using unknown coefficients ypX AXe39x Be2x To determine the unknown coefficient substitute the linear combination in the equation We must compute the first second and third derivative yp x 7 AXe39x e39x 2Be2x yp x 7 AXe39x 2e39x 4Be2x yp x 7 AXe39x 3e39x 8Be2x Replacing into the equation AXe39x 3e39x 8Be2x AXe39x e39x 2Be2x 7 2Ae39x 6B e 7 4e 3e2x Then 2A4 A2and 6B3B12 Therefore ypX 2Xe39x 12 e2x The general solution of the nonhomogeneous equation is yX ycX ypX c1 czex c3e39x 2Xe39x 12 ezx 3y ysinX Corresponding homogeneous equation y y 0 Characteristic equation r2 l 0 Zeros are r1 i and r2 i Fundamental set of solutions F cos X sin X Complementary solution ycX c1 cos X c sin X Nonhomogeneous term is bX sin X The UC set ofsin X is 1 sin X cos X There is only one UC set Slsin X cos X contains elements of F sin X cos X then we multiply the elements in 81 by X and we obtain 81 X sin X X cos X that does not contain any element of F So replace 81 by 81 Form a linear combination of the elements in 81 using unknown coefficients ypX AX sin X BX cos X To determine the unknown coefficient substitute the linear combination in the equation We must compute the first and second derivative Yp Asin X X cos X Bcos X 7 X sin X Yp AX sin X 2 cos X BX cos X 7 2sin X Replace into the equation AXsinX2cos XBXcOSX2sinXAX sinXBXcOSX2AcOSX2BsinX sin X Then 2A 0 and 2B 1 so A 0 and B l2 Therefore ypX 12 X cos X The general solution of the nonhomogeneous equation is yX ycX ypX c1 cos X c2 sin X 12 X cos X Find the candidate to be a particular solution of the equation do not find the coefficients 4 y4 7 7y 18y 7 20y 8y 7 x3e2x Corresponding homogeneous equation y4 7 7y 18y 7 20y 8y 0 Characteristic equation r4 7 7r 18 r2 7 20r 8 0 Zeros are r1 l and r2 2 is repeated 3 times Fundamental set of solutions F ex ezx Xezx Xze2x Complementary solution ycX c1 ex c c3X c4X2ezx Nonhomogeneous term is bX X ezx The UC set of X3e2x is 1 X3e2x Xzezx X ezx ezx There is only one UC set 1 X3e2x Xzezx X ezx ezx contains elements ofF ex ezx Xezx Xzezxz then we multiply the elements in 81 by X and we obtain 81 X6e2x Xsezx X4e X X3e2x that does not contain any element of F So replace 81 by 81 Form a linear combination of the elements in 81 using unknown coef cients The candidate to be a particular solution is ypX AX6e2x BXSBZX CX4e2x DX3e2x 5 y4 8y 16y X3sin 2X chos 2X Corresponding homogeneous equation y4 8y 16y 0 Characteristic equation r4 8r2 16 r2 42 0 Zeros are r1 2i double zero and r2 2i double zero Fundamental set of solutions F cos 2X X cos 2X sin 2X X sin 2X Complementary solution ycX c1 cos 2X c X cos 2X c3 sin 2X c4 X sin 2X Nonhomogeneous term is bX X3sin 2X chos X The UC set of X3sin 2X is 1 X3sin 2X X3008 2X XZSln 2X chos 2X X sin 2X X cos 2X sin 2X cos 2X The UC set of X2 cos 2X is S Xzsin 2X chos 2X X sin 2X X cos 2X sin 2X cos 2X Since 82 is included in 81 we disregard 2 1 X3sin 2X X3008 2X XZSln 2X chos 2X X sin 2X X cos 2X sin 2X cos 2X contains elements ofF cos 2X X cos 2X sin 2X X sin 2X then we multiply the elements in 81 by X2 and we obtain 81 X sin 2X X5cos 2X X sin 2X X cos 2X X sin 2X X3008 2X XZSln 2X chos 2X that does not contain any element of F So replace 81 by 81 Form a linear combination of the elements in 81 using unknown coef cients The candidate to be a particular solution is ypX AXSSln 2X BXSCOS 2X CX4sin 2X DX4COS 2X EX3sin 2X FX3COS 2X GXZSln 2X Hchos 2X SySEeMS 9 Linger Eimjcx o s We cm Nclke a SySJCQM H Um act015003 I x A gtltHax1 AM X0 a X A2X A n 1 Au1 X710 3 XI AmX quot I g 7 MBA ndong as 0 magm WINE X H7lt WE a WWW ed A 6W Sanbon ED 9W5 Sys rtm S X at E 50 we Look g or A wags 3b de cmxne Qt 9 HAL fouJQG a A A1 rsWM 93am amp SQMQLC waGRCn o are 9m 3950 mgm agar Some n 9AM 9 may be 833 to We 66 ffom 16 degni x o 3 n 9169 I 7pv zzr 3 gig Wt gag meLQS Class H 6A9 is ka 9m 03 wean Looch dabwvahAes a P I as HAM 6 2 eff n this Cam we score ngdd mdxin ms 01611 Hm end MJ 5 655 agmvedvos Emma 58me S W B19 Q33QVNOLIAA S 0 6k VIOLENX A by 51L Wam aaae 0Q lir rI 61th O 41 56W 64W eqmb39on are HM damxralmes quotZ O 63 Pr 3 I O F so Pr r1 H 391 3 F Mm r v 39r L 6 z 4 39F Hkr 76 2 7quot3 Z 2 So hm we have againaw rzrzl 4 For Each Ais nc 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133 0 so 51031413 50 M We m M W eigwaw W8 60 1 2 is Ell 1 Explanation of Methods Used in Solving Systems with Complex Eigenvalues Math 427K October 27 2005 First we need to understand what complex conjugation is The complex conjugate of a number often just called the conjugate is another number and the conjugate of a vector is another vectori Complex conjugation is usually represented with a bar over the number or vectori So the conjugate of 2 9239 is 2 9239 and the conjugate of A is X To nd out what the complex conjugate of a number is all you do is replace 239 with 72quot So 2 9239 2 792quot The same is true for vectorsi If 7 gig then the conjugate of is 2ZZ A number or a vector and its conjugate are called a conjugate pairi Notice that the complex conjugate of a real number is just that number again the conjugate of a sum of numbers or vectors is the sum of their conjugates and doing conjugation twice gives us back the number or vector we started with iiei X A Now if we want to solve the equation t At for the vector function t where A is a given matrix with real number entries then we start by solving for the eigenvalues and eigenvectors of A If A has complex eigenvalues then they will always be a conjugate pair ie the eigenvalues are of the form A1 a and A2 a 7 i6 where B a 0 So A1 T2 and if a is the eigen vector associated with A1 we know that A 7 A11H Now what happens if we take the complex conjugate of this equation We get A 7 A11H Conjugation works the same way for matricies that it does for vectorsi Using this and the fact that the conjugate of a real number is just that real number we can rewrite the equation as A7Ai11 But A2 T1 so what we have is A 7 A2071 Wich means that 17 which is the eigenvector correspond ing to A2 is just So what this means is that whenever the eigenvalues are conjugates of each other the corresponding eigenvectors will also be complex conjugates of each other So what does a general solution look like if we are dealing with a matrix with complex eigenvalues Suppose we have a matrix B whose eigenvalues are a i i6 where a and are real numbers and 6 f 0 And suppose that the a 239 c id 7 eigenvector corresponding to a 26 is w lt where a b c and d are real numbers Then from what we learned in the previous paragraph we know that the eigenvector corresponding to the eigenvalue 1 7 is the conjugate of U which is Z lid t Bt namely So we have two solutions to the differential equation A 7 a ib ai t a 7 a 7 ii a7i t 11t7ltcidgte and 12t7 Ciid e Our differential equation is linear so any linear combination of these solu tions is also a solution Hence letting cl and Cg represent arbitrary constants we can write a general solution as 7gt 7 a ib Dorm a ib a7i t zt7clltcidgte 52 Ciid e But this formula has the imaginary number i all over the place and often we would like to represent our solution using only real numbers ie not involving So we want to nd a convenient choice of cl and Cg that eliminate the number ii First let s look at two speci c linear combinations of I and 13 If I and 7gt 7 I1 7 12 lb 04139 77 047i 7 ltdgteemltdgtee m 7 aib a t t a7ib a t t 7 Cid 6 cos zs1n Ciid 6 cos 72s1n E E gt a L gtcos t7lt 2 gtsin tilt Z gtSinlt5tgtilt 2 gtCos t m Z gtcos t 7 lt 2 gtsin t 7ilt Z gtsin t 7ilt 2 gtCos t 2am Z gtcos t 7 lt S gtsin t a aib cid we can write 604 cos t 7 sin ti A similar calculation to lfwe let and Z then U 7gti h Also the one above will show that I 7 I Ziem Z sin t lt S gt cos t so we can write 7 604 sin t cos ti Letls de ne two new vectors that are both real valued 30 and 70 7 Why are these real valued functions Well 70 em cos t7 sin t which is an expression that does not involve if Similarly t 60 sin t cos t which does not involve if Two other things to notice about 70 and t is that at mt it and at t 7 may 7t 604 cos t 7 sin t iem sin t cos t TED cos t isin t EEO cos t 7 sin t eatcos t isin t iemcos t i sin t 7gtememz i3gtememz 7 ieai t gamma 371 And a similar calculation shows 3t 7 We also know that 3t and t are both linear combinations of the solutions to our differential equation so 70 and t must also be solutions to our differential equation The advantage 3t and t have over our original solutions is that they do not require us to use the imaginary number ii So we want to write our general solution as a linear combination of 70 and 70 instead of and 12 t so we won7t have to worry about 239 showing up in our solution Thus the question now is Can we use 3t and t to write down the general solution77 The answer is Yes and one way to see this is the following The original general solution we found was t 0117er 621305 Now 1711 just pick two new arbitrary constants to replace cl and 02 1711 call them d1 and d2 and de ne them as d1 cl 02 and d2 icl 7 02 If you solve these equations for cl and 52 you7ll nd cli7 and 527i So what happens when we rewrite our general solution in terms of d1 and d2 We ave t amt c2930 lt F16 lt 7 BO 1 gt 2 W dlmt dQWt What that calculation means is that if you can write the general solution in terms of and BO then you can also write it in terms of 3t and 7t you just need to use different but still arbitrary constantsi So now we have a good way to represent our general solution t 537m 54W It s good because it does not involve ii So this explains why7 when solving t Bt when E has complex eigenvalues7 we can use either or 130 to nd the general solution If we use we will nd 70 and t such that 30 and then we can write the general solution as a linear combination of 30 and And if we chose instead7 we would have found and such that it and we would write the general solution in terms of and EX but means t 30 and 7ti So our general solution would have turned out the same M427K7Test 2 Practice Unique 56090 Name USE PROPER NOTATION AND SHOW ALL WORK POINTS WILL BE AWARDED BASED ON YOUR USE OF CALCULUS 1 A mass weighing 161b stretches a spring 3ini The mass is attached to a Viscous damper with a damping constant of 21bsfti If the mass is set in motion from its equilibrium position with a downward velocity of 3ins nd its position u at any time t and determine when the mass rst returns to its equilibrium positioni 2 Find a particular solution of the following differential equation y4 7 y 7 y y t2 4 3 Seek a power series solution of the following differential equation about the given point 10 2 2y I 1W 3y 07 ya 3 42 1A 4 Determine a lower bound for the radius of convergence of series solutions about each given point 10 for the following differential equation 1272173y zy4y0104 10747 100 5 Find all singular points of the following differential equation and determine Whether each one is regular or irregular 1 sin y 3y my 0i 6 Seek a power series solution of the following differential equation about the given point 10 0 21y y my 0A 7 Find the Laplace transform of the following function R 0 tlt2 ft1 tgt2 8 Find the solution of the following initial value problemi y 2y 3y sint6t7 37r7 y0 07 yO 0i M427K7Test 1 Practice Unique 56090 Name USE PROPER NOTATION AND SHOW ALL WORK POINTS WILL BE AWARDED BASED ON YOUR USE OF CALCULUS 1 Consider the initial value problem y y 300st7 y0 2 How can we nd the coordinates of the rst local maximum point of the solution for t gt 0 2 A population is modeled by a function Pt that satis es the logistic differ ential equation g 1 7 i a Consider the solution With the condition P0 15 Is the solution increas ing or decreasing Find lirntn00 Pti b Consider the solution With the condition P0 4 For What value of P is the population growing the fastest 3 Consider a tank of constant volume V and containing at time t an amount Qt of pollutant evenly distributed throughout the lake With a concentration Ct Where Ct Assume that water containing a concentration k of pollutant enters the lake at a rate 7 and that water leaves the lake at the same rate a If at time t 0 the concentration of pollutant is co nd an expression for the concentration Ct at any time What is the limiting concentration as t a 00 b If the addition of pollutants to the lake is terminated k 0 for t gt 0 determine the time interval T that must elapse before the concentration of pollutants is reduced to 50 of its original va uei 4 Determine Without solving the problem an interval in Which the solution of the given initial value preoblem is certain to exist lnty y cot t7 y2 3i 5 Find the value of b for which the given equation is exact and then solve it using that value of 12 W2 bz2ydr I yr2dy 0 6 Assume that p and q are continuous and that the functions yl and y2 are solutions of the differential equation y pty qty 0 on an open interval Ii Prove that if yl and yg are zero at the same point in I then they cannot be a fundamental set of solutions on that interval 7 Find the solution of the given initial value problem y 7 2y 7 3y 07 y0 17 40 0 8 Consider the initial value problem 9y 12y4y 07 y0 a gt 07 yO 71 Find the critical value of a that separates solutions that become negative from those that are always positive

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