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# Note for MATH 3321 at UH 2

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Date Created: 02/06/15

25 Some Numerical Methods As indicated previously there are only a few types of rst order differential equations for which there are methods for nding exact solutions Consequently we have to rely on numerical methods to nd approximate solutions in situations where the differential equation can not be solved In this section we illustrate two elementary numerical methods Our focus here is on the initial value problem 1 1 y M900 yo 1 where f and 8f8y are continuous functions on arectangle R and mo yo 6 R That is the initial value problem satis es the conditions of the existence and uniqueness theorem EULER7S METHOD Although this method is rarely used in practice we present it because it has the essential features of more advanced methods We begin by setting a step size It gt 0 Then we define m values zk 0 kh where k is a natural number The values zk are the values of z where we try to approximate the solution to Next we give some notation for our approximations to We will use the notation gt 3 ma By the de nition of the derivative we know that Mam 7 24 h 7 which using our notation leads to the approximation MM 3 k 7 k MM 3 2 Substituting this approximation into 1 gives the approximate equations 29 1 yk i i mhyk w1th yo given Rearranging terms gives ka yk hfmk yk for k O 1 where yo is given 3 This method is known as Euler s Method with step size It Example 1 Use Euler s method with a step size of 005 to approximate the solution to the initial value problem y ysinz y01 58 Before we begin7 note that we can give the exact solution to this initial value problem You might naturally ask why we are going to bother approximating the solution if we know how to solve it The answer is simple We want to illustrate how well or poorly Euler s Method works The exact solution is 7 cosz sinz 35 Now we ll use Euler s Method with a step size of h 005 to approximate this solution From 3 we have ka yk 005 iyk sin005klt for k 07 17 where yo 1 Some simple calculations give U1 ug 0904998958 us 0864740681 U20 0686056580 Noting that yzo is supposed to approximate 20 y 1 0702403501 to 9 decimal places7 we can see that our approximation of y1 has an error of y1 2420 07024035017 0686056580 0016346921 Actually7 that s not too bad I IMPROVED EULER7S METHOD Here we give an improvement to Euler s method As above7 we define h gt 0 to be the step size of the method and take zk 0 kh where k is a natural number Also7 we continue to use the notation yk z for the approximations to when x m We can use two different approximations for the derivative Namely7 yk1 7 yk and Ms 1 M was s ti Since was st yltsigtgt and Mam mist mist substitution gives yk1 Ms M h fltk7 240510 59 and 34k 1 yk Jr 3 fk17 ylt k1gtl Adding these equations and solving for ka gives h yk1 yk 5 100 yk fltk17 yk1gtgt Unfortunately this scheme is not easy to implement because ka occurs on both the left and right side of the equation for this reason it is called an implicit scheme We avoid the implicit nature of this equation by replacing ka on the right side by its Euler approximate using yk as a guess That is h i i yk1 yk 5 HM 24k fltk17 M Where M M WWW M and yo IS glVeH 4 This method 4 is commonly known as the Improved Euler s Method with step size It Example 2 Apply the lmproved Euler s Method with step size 005 to the initial value problem y ysinz y01 As noted in Example 1 this initial value problem has the exact solution 7 cos x sin z ge m Using the lmproved Euler s method with a step size of h 005 gives the values yl 0952499479 yg 0909747970 yg 0871504753 2420 0702609956 Again ygo is an approximation to yt20 y1 07024035012 In this case the error at z 1 is given by M1 7 ml z 206454773 x 104 This is much better than our earlier result from Euler s method I The accuracy of these methods can be predicted In general Euler s Method lymk 7 ykl leh2 and Improved Euler s Method 7 ykl szhg Where the constants L1 and L2 are dependent upon the actual solution but independent of the step size It and the number k Notice that the error estimates above imply 60 Euler s Method has an error which is a factor of h over every interval z interval there are essentially k 1h steps of size h across each z interval and the Improved Euler s Method has an error which is a factor of h2 over every m interval It is not hard to see that small values of h should give a much smaller error for Improved Euler s Method than for Euler s Method Exercises 25 1 Use both the Euler and Improved Euler Methods with a step size of h 001 to estimate y2 Where y is the solution of the initial value problem Compare your values with those of the exact solution 2 Use both the Euler and Improved Euler Methods with a step size of h 002 to estimate y1 Where y is the solution of the initial value problem y my y02 Compare your values with those of the exact solution 3 Use both the Euler and Improved Euler Methods with a step size of h 005 to approximate the solution of y y4 i y y0 2 Compare your values with those of the exact solution 4 M95 W for z 120 110 320 1920 1 4 Use the lmproved Euler s Method with a step size of h 01 to approximate y02 Where yt is the unique solution of y sinm7y3 y0 1 61

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