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# INTRO TO MATHEMATICAL REASON MATH 300

Virginia Commonwealth University

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This 6 page Class Notes was uploaded by Paul Spencer on Wednesday October 28, 2015. The Class Notes belongs to MATH 300 at Virginia Commonwealth University taught by Staff in Fall. Since its upload, it has received 18 views. For similar materials see /class/230623/math-300-virginia-commonwealth-university in Applied Math at Virginia Commonwealth University.

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Date Created: 10/28/15

Lecture Notes in Differential Equations Dr Hassan Sedaghat Importantl These instructor notes are made available for convenience to reduce the amount of note taking in class and also in case it is not possible to attend class due to circumstances beyond control eg illness These notes are not intended as a replacement for class attendance and participation nor as a substitute for using the textbook The class presentation may differ from these notes and further these notes do not contain solutions for homework problems which may be discussed in class Section 44 Undetermined Coef cients ln this section we discuss a procedure for solving nonehomogeneous linear DE with constant coef cients 71 anyquot an71yquot azy a1y my 906 1 a0 a1 an are given real numbers with an y 0 From Section 41 we recall that the general solution of 1 is written as y ye l 917 where the complementary solution ya is the solution of the homogeneous part of 1 ie with the right side set equal to 0 So ya is the superposition of n linearly independent solutions y1 y2 yquot ie ya C1y1 C2y2 qtyn The solution ya is obtained as in Section 43 using the auxiliary equation Note that for each linear DE of order 71 there are n arbitrary constants 01 cn and in the nunihamugeneuus case these canstants all appear in ya The function y7 is a particular solution of 1 ie any one function satisfying ln this section we discuss a method for computing yp UC functions gme The procedure discussed in this section applies only to certain types of functions not to any arbitrary function We call these special function types for gm UCfunetz39zms We use the UC function gm given in a DE to construct the particular solution yp First we consider the basic must cummun UCfunetz39ans UC function gm Sample Transforms to yp Constant eg 72 A Polynomial degree 1 eg 2m 7 1 Ax B Polynomial degree 2 eg m2 7 390 A962 Bar C Polynomial degree 3 eg m3 1 A963 B962 Car D Trigonometric 3coshx andor 3sinhx Acoshx B sinhx Exponential 35 A51 The basic function types in the above Ucetable may be combined to create new UC functions as follows Addition If 996 is a sum of two or more function types in the rst column in the above UCitdble then we add the corresponding entries in the third column For example 9 m 5x 7 2 cos 3x 7 2sinx gives degi polynomial trig triple mg single y7 Ax B Ccos3x Dsin3x Ecosx Fsinx Multiplication If 996 is a product of two or more function types in the rst column in the above UCitdble then we multiply the corresponding entries in the third column For example 990 5125721 gives y7 A962 Bar Ce 2quot 990 3572 sin or gives y7 Asin m B cos 572 Remarks 1 In the case of the exponential function it is not necessary to incorporate its constant since it will be redundant For example for 5125721 above if we picked y7 1412 Bar CDe 2quot then the extra constant D could be absorbed into the rst 3 without loss of information y7 4ng BDm CDe 2 14962 Bloc C e 2quot where it would only be necessary to nd three undetermined coef cients A B and C 2 There are some familiar functions that are not of UC type They include tan 96 sec or ln or 719 for h l 2 3 lf 996 is one of these types of functions then the UC method does not apply we need to use a different method such as variation of parameters in Section 46 The method of Undetermined Coe icients or the UC method If 990 in l is a UC function then we determine the particular solution y7 as above and nd the numbers A B C etc by substitution in the DE and matching similar functions on the two sides of the equation The numbers A B C etc are the undetermined coef cients which once calculated give yp We also nd yc as in Section 43 to complete the general solution Example 1 Find the general solution of the non7homogeneous DE y 7y 7 2y 630 First we nd yc by computing the roots of the auxiliary equation m27m720gtm72m10gtm712 These distinct real roots give two linearly independent solutions 57quot and 52quot Therefore by Sections 41 43 we have ya 015 0252 Next 990 696 is a polynomial of degree 1 so from the UC7table y7 AxB We calculate the values of A and B for this problem by substituting y7 into the given DE and matching coef cients on two sides ypAxBgty7A yg0 so 07A72AxB6x 72Ax7A72B6x0 11 72A6 7A72B0 A 73 B 7142 32 With the coef cients A B now determined we have y7 73m 32 so the general solution of the non7homogeneous DE is 3 yycypcle x0252x73m Example 2 Find the general solution of the non7homogeneous DE y 7 y 7 2y 65quot First we nd yc as in Example 1 above Next since 990 65quot is of exponential type we have y7 As from the UC7table To nd the coef cient A we use the DE ypAe gtyygAex so Aea 7Aequot72Ae 6e gt72A6gtA73 Thus y7 7357 and the general solution is found to be 9 96 l 917 01571 l 02521 i 351 Example 3 Find the general solution of the non7homogeneous DE y 7 y 7 2y 6sin2m First we nd yc as in Example 1 above Next since 990 6sin2x from the UC7table y7 Acos2x B sin2x and we must calculate the coef cients A and B using the DE y 7 72A sin2x 2B cos2x y 74A cos2x 7 4B sin2x 74A cos2x 7 4B sin2x 7 72A sin2x 2B cos 76A 7 2Bcos2x 24 7 6B sin matching coef cients A 296 7 2Acos2x Bsin2x 6sin2x 2x 0cos2m 6sin2x 76A72B0 2A76B6 B73A 2A7673A6gtA 3 1373 20 10 10 A Therefore the general solution is y ya yp Clef 0252 03 cos2x 7 09sin2m A glitch in the UC method When the complementary solution ya and the choice y7 from the UC7table contain a common term we say that there is an overlap between ya and yp then the UC method breaks down unless the choice of y7 is modi ed to remove the overlap For instance if in Examaple 2 above we had 990 65 then choosing y7 Ae gives yll 7 yl 7 651 A57 7 71457 7 2457 657x 0 65 which is not possible This problem occurs because ya 015 0252 and y7 Ae both contain 57quot This overlap must be removed and we do that using the following rule Removing the overlap Multiply the initial Choice Of y7 by the smallest pU LUET m 30 that the mudi ed 07 new y7 no longer shares cummun terms with ya Example 4 Find the general solution of the non7homogeneous DE y 7 y 7 2y 657x As we saw above we must modify yp Multiplying by 96 gives y7 Axe and this no longer overlaps ya me is a different functiun than 57quot despite the common factor 5 Now using the product rule we compute the derivatives and then use the DE to calculate A as follows y A57 7 Axe y 72Ae x A3657 so 72Ae x Axe x 7 Ae x 7 Axe x 7 2Axe x 657x 73Ae 65 gt 73A 6 gt A 72 Thus y7 72x57 and the general solution comes out as y ya y7 015 0252 7 2965 Example 5 Find the general solution of the non7homogeneous 3rd order DE y 7 y 7 2y 696 First we nd yc using the auxiliary equation m37m272m0gtmm1m720gtm0712 These distinct real roots give three linearly independent solutions yl e7 yg 52quot and ye 50quot 1 Therefore as in Sec43 we have ya 016 0262 03 Here as in Example 1 990 696 Our rst choice from the UC7table is y7 Ax B However both ya and this rst choice of y7 overlap in a constant the arbitrary constant 03 can in particular be equal to the undetermined but xed constant B Multiplying by 96 gives the new function y7 A962 Bx It is clear that there is no longer any overlap between ya and y7 so we proceed to calculate the numbers A and B y2AmB yg2A y 0 so 072A722AxB6m 74Ax72A7QB6x0 74A6 72A72B0gtA7 B7 Thus y7 715962 1596 and general solution is found to be y ya y7 015 0252 03 7 15962 1596 Example 6 Find the unique solution of the initial value problem y 2y y 65 y0 1 y 0 1 First we nd ya using the auxiliary equation m22m10gtm120gtm7171 As in Case H in Sec43 repeated root 71 gives ya 015 029657 We are given 996 65 Choosing y7 A57 would overlap ya multiplying this rst choice by 96 gives A9657 However this choice still overlaps yc So we multiply by m2 to get y7 A962 57quot Since there are no further overlaps we proceed to compute A as usual y 2141 571 7 1430257 y 2457 7 449657 1430257 so 2457 7 4AIL 571 Axge x 22Ame x 7 Axge x Amge x 657x 2Ae x6e xgt2A6gtA3 Therefore y7 3x257 and the general solution is found to be y yc yp 015 0290571 3x257 Finally we use the given initial values to calculate 01 and 02 Since y 7015 025 7 023057 6965 7 3x257 we have 1 y0 01500001 1y 0 7015002507007070102 So 01 1 and 02 2 Therefore the unique solution of the lVP is y 57quot 2965 3x257 e 3x2 2x 1 More on removing overlaps In general the particular solution may consist of two or more complete entries from the UC7 table For example we can have y7 y71 y72 where each of y71 and y72 is a complete entry from the UC7table In this case if only one of y71 or y72 overlaps yc then the overlap is removed by only multiplying that part by a power of x

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