of the function, Cœ/ 7B and its derivatives and the good 'ol quadratic equation from algebra.

Transcription

1 Linear Differential Equations with Constant Coefficients Section 7.1 In this Chapter we deal with a specific type of DE. Specifically, it is linear, homogeneous, ordinary, and has constant coefficients. These limitations result in a DE which can e solved y a little insight into the nature of the function, Cœ/ and its derivatives and the good 'ol quadratic equation from algera. We will find that the solution to our DE will always e of the asic form, will e explained. The chart elow provides an outline of the chapter: Cœ/. Variations in this Distinct 7 ( 8 of them) Cœ-/ " -/ â-/ 8 " 8 Repeated roots ( 8of them) 8" Cœ-/ " -B/ -B/ $ â-b 8 / Imaginary roots ( 7œ+, 3Ñ +B +B Cœ-/ " a-9= a,b3=38 a,b-/ a-9= a,b3=38 a,b +B +B œ-/ -9= a,b-/ =38 a,b $ % The last section introduces the hyperolic trigonometric functions, -9=2aB and =382aB. 1

2 The Auxiliary Equation: Distinct Roots Section 7.2 Several methods for solving differential equations with constant coefficients are presented in this text. Chapters 7 and 8 show the classical treatment, Chapter 12 shows matrix techniques, and Chapters 14 and 15 show Laplace Transform techniques. Any linear homogeneous differential equation with constant coefficients, can e written as 8 8". C. C.C + o + â+ +Cœ!ß.B 8.B 8".B " 8" 8 (1) where 0aH is a linear differential operator. 0HCœ!ß a (2) As we saw in the previous chapter, 0aH is generally a polynomial in H. Furthermore, if 7 is any root of the algeraic equation, 0a7 œ! then 0H/ a œ!. This suggests that Cœ/ is the solution to our differential equation. Hence, we need only sustitute / in our DE for C and an algeraic equation 0a7 will e generated. This equation, 0a7 œ!, is called the auxiliary equation associated with (1) or (2). The auxiliary equation for (1) is degree 8. Let its roots e 7ß7ßâ ", 78. If all of these roots are real and distict then the 8 solutions to the DE are: C œ/ " ß C œ/ ß C œ/ ßâßC œ/ is $ 8 $ 8 ". These solutions are linearly independent so the general solution to (1) " $ 8 " $ 8 Cœ-/ -/ -/ â-/. 2

3 www ww w Example Solve C 'C ""C 'C œ!þ www ww w C 'C ""C 'C œ! let Cœ/ ß C œ7/ ß C œ7/ ß C œ7/ w ww www $ $ 7/ '7 / ""7/ '/ œ! ˆ $ 7 '7 ""7'/ œ! ˆ $ 7 '7 ""7'/ œ! solutions: 7œ"ß ß $ B B $B Cœ-/ " -/ -/ $ è www ww w Example Solve C &C C œ! www ww w C &C C œ! let Cœ/ ß C œ7/ ß C œ7/ ß C œ7/ $ 7/ & 7/ 7/ œ! ˆ $ 7 & 7 7/ œ! 7 ˆ 7 & 7 / œ! 77" a a7 / œ! " solutions: 7œ"ß ß w ww www $ " B " B $ B Cœ-/ -/ -/ è 3

4 The Auxiliary Equation: Repeated Roots Section 7.3 This section deals with the situation in which the roots of the auxiliary equation are not distinct. That is, suppose the auxiliary equation is something similar to ˆ 7 %7%/ œ a7 / œ!? If we follow our previous section, we are seeking 2 values 7" and 7 which satisfy the aove equation. Our solution would then e, Cœ-/ " -/ Þ " The difficulty here is ovious. Our roots are not distinct. The solution would then ecome, " " Cœ-/ -/ œ a- - / œ5/ Þ Although this is a solution to the DE, we were expecting 8 (in this case 8œ ) linearly independent functions from which to construct our solution. The resolution to this is to alter the form of our constructed solution. We will assume that if the roots of the auxiliary equation are repeated ( 7 œ7 œ7ß/>-) then we will let the constructed solution for the DE ecome " $ 8" Cœ-/ " -B/ -B/ $ â-b 8 /. Note that our solution is constructed of 8 linearly independent solutions just as efore. 4

5 % $ Example Solve ah $H 'H )H% Cœ! ˆ % $ H $H 'H )H% Cœ! ˆ % $ 7 $7 '7 )7% œ! a7$ a7 œ! 7œ $ß ß ß $B B B B " $ % Cœ-/ -/ -B/ -B/ è $ Example Solve ah $ H Cœ! when Bœ!ß Cœ $ß C w œ!ß C ww œ" ˆ $ H H Cœ! ˆ $ 7 7 œ! 7 a7 œ! 7œ!ß!ß " B $ " a! $ " w B w a! " $ " $ " ww B ww a! Cœ-/ - -B C! a œ-/ - - a! œ $ p- - œ $ C œ - / - C a! œ - / - - a! œ! p- - œ! C œ%- "/ C a! œ%- "/ œ " p%-" œ" -" œ$ - œ ' -$ œ aritary Cœ$/ B '5Bè 5

6 Check the solution: ˆ $ H H ac œ! ˆ $ ˆ B H H $/ '5B œ! we'll need some derivatives: H$/ ˆ B '5B B œ '/ 5 H ˆ B $/ '5B B œ"/ $ H ˆ B $/ '5B B œ %/ now plug them into the DE: ˆ $ B $ B B H H ˆ $/ '5B œh ˆ $/ '5B H ˆ $/ '5B B ˆ B œ %/ "/ œ!r 6

7 D Definition of / for Imaginary D Section 7.4 Since the last several sections have dealt with algeraic equations, we must consider the possiility that our auxiliary equation may have imaginary roots. Therefore, we need to explain the meaning of Cœ / DB when Dis imaginary. Your text explains in sufficient detail the calculus B notation for the infinite series representations for /ß=38aBß and -9= ab. For our purposes, then, Cœ/ D a! œ/ 3 "! œ/ / 3 " œ/! a-9= " 3=38" when! and " are real. With a redefinition of the constant coefficients we will find that imaginary roots to the auxiliary always lead to oth a -9=38/ and a =38/ in our answer. 7

8 The Auxiliary Equation: Imaginary Roots Section 7.5 Consider the auxiliary equation do the DE, 0HCœ! a. We will assume that the coefficients of this algeraic equation are real. We know from algera that if the roots are imaginary, then they will occur in conjugate pairs. Therefore if, (assume,á!) then 7 œ+3, " 7 œ+3,. So if we start with the same form of our solution as efore, we have that Cœ-/ " -/. Sustituting the expressions aove for 7 and 7, yields: " " Cœ-/ œ-/ " " -/ a+3, B a+3, B " -/ +B3,B +B3,B œ-/ " -/ +B +B œ-/ " a-9= a,b3=38 a,b-/ a-9= a,b3=38a,b œ-/ " +B a-9= a,b3=38 a,b-/ +B a-9= a,b3=38 a,b +B +B +B +B œ-/ " -9= a,b3-/ " =38,B-/ -9= a,b3-/ =38 a,b +B +B +B +B œ-/ " -9= a,b-/ -9= a,b3-/ " =38 a,b3-/ =38,B a +B +B œ a-" - / -9= a,b3a-" - / =38 a,b +B +B œ-/ $ -9= a,b-/ % =38 a,b where a- - œ- and 3a- - œ- " $ " % 8

9 $ Example Solve ah $H *H"$ Cœ! ˆ $ H $H *H"$ Cœ! ˆ $ 7 $7 *7"$ œ! 7œ " y inspection long division yields a7" ˆ 7 %7"$ % È"'%" a a"$ % È $' 7œ œ œ $3 B B Cœ-/ " B -/ -9=$B-/ $ =38$Bè 4 Example Solve ah )H "' Cœ! ˆ 4 H )H "' Cœ! ˆ 4 7 )7 "' œ! ˆ 7 % œ! 7 œ % p 7œ 3ß 3ß3ß 3 Cœ c--9= " ab-=38 abdc-b-9= $ ab-b=38 % abd { since 3 œ!3, each term aove has a /!B a component in it.} Cœ a-" -B $ -9= aba- -B % =38 ab Cœ a- -B-9= aba- -B =38 abè " $ % {renamed constants} 9

10 A Note on Hyperolic Functions Section 7.6 In this section the author introduces a new class of trigonometric functions, the hyperolic trigonometric functions. They are encountered so frequently in oth theoretical and applied work that we focus on them in this section. Most students have had some experience with these ut for the sake of those who have not encountered them we riefly explain them here. The trigonometric functions that students are most comfortale with are those called the circular trigonometric functions. They are defined on the unit circle. Another group of function are defined on a hyperola and are consequently called the hyperolic trigonometric functions. and / B. We egin y defining the hyperolic functions in terms of the exponential functions, / B Hyperolic Trigonometric Functions " " B B B B =382aB œ a/ / ; -9=2aB œ a/ / Some properties of these functions are worth noting here as they impact our further study of differential equations. 10

11 The graphs are shown elow: 4 y y=cosh(x) 2 0 y=sinh(x) x -2-4 Some characteristics should e noted. Consider the graphs aove and the respective " " B B B B formulae: =382aB œ a/ / and -9=2aB œ a/ /. Look at the formulae to verify the ehavior of the curves as Bp! and as Bp_ paying attention to the nature of the exponential functions. Next we'll turn our attention to the derivatives of the hyperolic trigonometric functions. Consider the definition of the =382aB. Let's take it's derivative with respect to B. " B Cœ a/ / " C œ a/ / œ-9=2ab B w B B.=382aB.-9=2aB.B.B We see that œ-9=2ab. Similarly, œ=382ab. This last characteristic is important ecause we can now see that oth =382 a+b are each solutions to: -9=2 a+b 11 and

12 ˆ H + Cœ! for +Á!. a3 A question that must e answered is, "Are these functions linearly independent or not?". The Wronskian of the two functions is: solution to W œ º -9=2 a+b =382 a+b + =382 a+b + -9=2 a+b º œ a-9=2 a+ba+ -9=2 a+b a=382 a+b a+ =382 a+b œ+-9=2 a+b+=382 a+b œ+-9=2 a a+b=382 a+b œ+" a œ+ So as long as +Á! the two functions are linearly independent and therefore, the general a$ is: Cœ--9=2 " a+b-=382 a+b. Example Solve ah %Cœ!à when Bœ!ß Cœ!ß C w œ. We can immediately write down the form of our general solution since the DE has the form of ah + Cœ!. Our solution has the form Cœ--9=2 ab-=382 ab. When Bœ!ß Cœ!. This means that we can determine our first constant:!œc! a œ--9=2 " a! -=382 a! œ- " a" - a! œ-" - œ! " " w Our second constant is determined y using the other initial condition, C œ. Our solution is then, Cœ=382 abþ w C œ 2-=382 " ab 2--9=2 ab w œc a! œ 2-=382 " a! 2--9=2 a! œ 2- " a! 2- a" œ 2- - œ 1 12

13 We could just as easily solved this particular DE y the previous method ßCœ/ : ˆ H %Cœ! ˆ 7 % œ! a7 a7 œ! 7œß Cœ-/!œC! a œ-/!œ- - " B -/ B Ð!Ñ Ð!Ñ " -/ " w Ð!Ñ Ð!Ñ œc a! œ- "/ - / œ- a " - "œ- - "!œ-" - " " "œ-" - p-" œ ß - œ " " B B Hence, Cœ / / ut this is =382 ab as we found efore. 13