Ordinary Differential Equations What is Differential Equation? An differential equation or Ordinary differential equation that involves one or more derivatives of an unknown function. Taylor's series(pointwise method) The Taylor's series expansion of y(s) about x=x0 is given by y(x)=y(x0)+(xx0)y'(x0)+ or What is solution to DE? A solution of a differential equation is a specific function that satisfies the equation. What is initial value problem? A problem with the given condtion or initial value. Why initial value? x−x 0 2 y''(x0)+... 2! y(x0+h)=y(x1)=y1 =y0+ h h2 h3 y0'+ y0''+ y0''+... 1 ! 2 ! 3 ! In general, if hn+1 and higher power of h are omitted then the error will be proportional to (n+1)th power of the step size. That is, khn+1, k is step size. An arbitrary contant in the solution of DE Problem 01: indicate that a DE does not, in general, dy determine a unique solution function. For Using Taylor's series solve =1+xy with dx this reason, a DE is usually accompanied by y(0)=2. Find y(0.1), y(0.2) and y(0.3) auxiliary conditions that specify the unknown functin precisely. Solution: In this chapter, we concentrate on one type of differential equation and one type of auxiliary condition: the initial value problem for a first order differential equation. The standard form adopted is dy =f x , y with initial conditon y x 0 =x 0 dx the Taylor's series(1st app.,) is y1=y0+ h h2 h3 y0'+ y0''+ y0''+... 1 ! 2 ! 3 ! here intial condition arex0=0, y0=2 and h=0.1 given y'=1+xy y''=y+xy y'''=y'+y'+xy'' = 2y'+xy'' The above system will yield the solutions in for the first approximation, substitute (x0, y0)in the above derivatives. one of the two forms: 1. A series for y interms of powers of x, That is, from which the value of y can be y0'=1+x0y0=1+0*2=1 obtained by direct substitution. y0''=y0+x0y0'=2+0*1=2 2. A set of tabulated values of x and y. y0'''=2y0'+x0y0''=2*1+0*1=2 The methods of Taylor and Picard belong to substitute these values in Taylor's series of class (1), whereas Eulter, RungeKutta 1st approximation belongs to class(2). PES College, PESIT Campus, Department of BCA 1 Ordinary Differential Equations y1=y0+ h h2 h3 y0'+ y0''+ y0''+... 1 ! 2 ! 3 ! 0.1 0.12 0.13 =2+ + ∗2 + ∗2 1 ! 2 ! 3 ! 0.1 0.12 1.4486+ 2.53272+ 1 ! 2 ! y3=2.2430+ 0.13 3.4037 3 ! y(0.3)=2.4011 =2.1103 y(0.1)=2.1103 Problem 02: nd The Taylor's algorithm for the next(2 ) approximation is y2=y1+ h h2 h3 y1'+ y1''+ y1''+... 1 ! 2 ! 3 ! here x1=x0+h=0+0.1=0.1 Solution: Taylor's algorithm is, y1=2.1103 y1'=1+x1y1=1+(0.1)*2.1103=1.21103 y1=y0+ y1''=y1+x1y1'=2.1103+0.1*(1.21103)=2.2314 y1'''=2y1'+x1y1''=2*(1.21103)+0.1*2.2314=2.6452 substitute these values in Taylor's series of 2nd approximation 2 0.1 0.1 y2=2.1103+ 1.21103+ 2.2314+ 1 ! 2 ! 0.13 2.6452 3 ! y(0.2)=2.2430 h h2 h3 y0'+ y0''+ y0'''+... 1 ! 2 ! 3 ! here x0=0, y0=0, h=0.1 y'=12xy y''=2[ yxy' ] y'''=2[ y'+y'+xy'' ]=2[ xy''+2y' ] y0'=12x0y0 = 1 y0''=2[ y0x0y0' ] = 0 y0'''=2[ x0y0''+2y0' ] =4 rd The Taylor's algorithm for the next(3 ) approximation is y3=y2+ Using Taylor's method, find y(0.1) correct to dy 2xy=1, y 0 =0 3 decimal places from dx 2 3 h h h y2'+ y2''+ y2''+... 1 ! 2 ! 3 ! substitute above values in taylor's 1st approximation formula y1=1+0.1+ 2 3 0.1 0.1 0+ x4 = 0.0993 2 ! 3 ! here x2=x1+h=0.1+0.1=0.2 Exercise: y1=2.2430 Using Taylor's find y at x=1.1 and 1.2 by dy 2 2 =x y given y(1)=2.3 solving dx y2'=1+x2y2=1.4486 y2''=y2+x2y2'=2.53272 y2'''=2y2'+x2y2''=3.4037 substitute these values in Taylor's series of 3rd approximation PES College, PESIT Campus, Department of BCA Ans: y(1.1)=3.1209 y(1.2)=4.6823 2 Ordinary Differential Equations Picard's Method 2 4 3 = 1 x x − x − x 2 3∗4 3 In this method, to solve dy =f x , y with initial conditon y x 0 =x 0 dx = 1x x ∫ f x , y dx integrate y=y0+ x 0 x 2 x 4 x3 − − 2 12 3 x rd (3) 3 approximation y =y0+ x0 x0 x the 1st approximation y(1)=y0+ ∫ f x 0, y 0 dx x0 x ∫ f x 0, y 1 0 y =1+ x 3rd approximation y(3)=y0+ x dx x0 ∫ f x 0, y02 dx x0 and so on. ∫ y02−x 2 dx (3) x 2nd approximation y(2)=y0+ ∫ f x 0, y02 dx = 1 + ∫ 0 [ 0 1x y(3) = 1x ] x2 x 4 x3 − − −x 2 dx 2 12 3 x2 x3 x4 x5 − − − 2 6 12 60 put x=0.1 in the above equation, Problem 03: y(0.1)=1+(0.1)+ 0.1 2 0.1 3 0.1 4 0.15 − − − 2 6 12 60 dy = y−x 2 , y(0)=1 by Picard's = 1.1048249 dx method upto the third approximation. Hence put x=0.2 in the above equation, find the value of y(0.1) and y(0.2). 0.2 2 0.2 3 0.24 0.25 y(0.2)=1+(0.2)+ − − − 2 6 12 60 Soultion: = 1.218528 Given that x0=0, y0=1 and f(x,y)=yx2. Solve x st (1) 1 approximation y =y0+ ∫ f x 0, y 0 dx x0 ∫ 1−x 2 dx 0 x x3 x− 3 = 1 + = 1x− 0 x3 3 x 2nd approximation y(2)=y0+ ∫ f x 0, y01 dx x0 x (2) y =1+ ∫ y 1−x 2 dx 0 x = 1+ ∫ 0 [ 1x− dy =x y 2 , y(0)=1 by dx Picard's method upto the 2nd approximation. Hence find the value of y(0.1). [ans: 1.1156255] 1. Solve x y(1)=1+ Exercise: ] x3 −x 2 dx 3 PES College, PESIT Campus, Department of BCA dy =x y , y(0)=1 by Picard's dx method. Hence find the value of y(0.1)&y(0.2) 2. Solve dy =2x− y , y(1)=3 by dx Picard's method. Hence find the value of y(1.1). 3. Solve Note: if the level of approximatiion is not given, assume it according to the marks. 3 Ordinary Differential Equations Euler's Method The Euler's Successive approximation to solve y'=f(x,y) with the initial conditions y(x0)=y0 are as follows: dy =1xy with initial condition dx x0=0 and y0=2. Given DE The Euler's formula for 1st approximation is y1=y0+ h f(x0, y0) y1=y0+ h f(x0, y0)=2+ 0.1 (1+xy) y2=y1+ h f(x1, y1), ... =2+0.1*(1+0) = 2.1 In general, that is, x1=0.1 and y1=2.1. yn+1=yn+ h f(xn, yn), n=0,1,2, ... y2=y1+ h f(x1, y1)=2.1+0.1*(1+0.1*2.1) Problem 04: =2.221 dy =1− y , y(0)=0 at x0.1 and x=0.2 that is, when x1=0.2 and y1=2.221 Solve dx using Euler's method. y3=y2+ h f(x2, y2)=2.221+0.1*(1+0.2*2.221) Solution: dy =f x , y is The solution of dx =2.36542 that is, y(0.3)=2.36542 yn+1=yn+ h f(xn, yn) dy =1− y with initial conditon Modified Euler's Method dx is x0=0, y0=0 and h=0.1 For a reasonable accuracy, we need to take a small value for h and hencethe process y1=y0+ h f(x0, y0)=0+0.1(10)=0.1 will become very slow. Because of this y2=y1+ h f(x1, y1)=0.1+0.1(10.1) = 0.19 restriction on h, the method is not suitable for practical use and a modification of it, hence y(0.1)=0.1 known as the modified Euler's method, y(0.2)=0.19 which gives more accurate reslts. Given DE is Problem 05: dy =1xy dx with y(0)=2. Find y(0.1), y(0.2) and y(0.3). Using Euler's method solve Solution: The solution of dy =f x , y is dx yn+1=yn+ h f(xn, yn) PES College, PESIT Campus, Department of BCA In Modified Euler's method, instead of approximating f(x,y) by f(x0,y0), we 1 approximating it by [f x 0, y 0f x 1, y 1] , 2 which is the mean of the slopes of the tangents at the points corresponding to x=x0 and x=x1. Thus we obtain h y1(1)= y 0 [f x0, y 0 f x 1, y 1] 2 In general, h n y1(n+1)= y 0 [f x0, y 0 f x 1, y 1 ] 2 4 Ordinary Differential Equations Problem 06: Using modified Euler's method solve dy =1xy with y(0)=2. Find y(0.1), y(0.2) dx and y(0.3). Now find y(0.2): h y2(1)= y 1 [f x 1, y 1 f x 2, y 2 ] ... (2) 2 here x1=0.1, y1=2.1105 x2=x1+0.1=0.2, y2 find using Euler's method. Solution: That is, The first approximation is y2=y1+ h f(x1, y1) h y1(1)= y 0 [f x 0, y 0 f x 1, y 1 ] ...(1) 2 =2.1105+0.1 * (1+x1y1) here f(x,y)=1+xy =2.2316 f(x0,y0)=1+x0y0=1+0*2=1 =2.1105+0.1 * (1+0.1*2.1105) f(x1, y1)=(1+x1y1)=(1+0.1*2.1105)=1.21105 next f(x1,y1) f(x2, y2)=(1+x2y2)=1+0.2*2.2316=1.4463 x1=x0+h=0+0.1=0.1 substitute these value in (2), we get y1=?. We need to find y1 using Euler's method. y2(1)= 2.1105 The Euler's formula for 1st approximation is =2.2434 y1=y0+ h f(x0, y0)=2+ 0.1 (1+xy) continuiing this process, 0.1 [1.211051.4463 ] 2 =2+0.1*(1+0) = 2.1 y2(2)=2.2435 that is, x1=0.1 and y1=2.1. y2(3)=2.2434 Again, f(x1,y1)=1+x1y1=1+0.1*2.1=1.21 y2(4)=2.2434 Now apply in equation (1), y1(1)= 2 0.1 [11.21] =2.1105 2 h 1 y1(2)= y 0 [f x0, y 0 f x 1, y 1 ] 2 f x1, y 11 =1+x1y1(1)=1+0.1*2.1105 =1.21105 y1(2)= 2 0.1 [11.21105] =2.1105 2 therefore, the value of y2=2.2434 Similarly, find y(0.3). [Ans: y3=2.4018] Exercise: Given dy y 1 = y 1=1. dx x x 2, Evaluate y(1.3) by modified Euler's method. Hint: here x0=1, take h=0.1 [in case of 7marks question) and find when since differentence between two successive x1=1.1 what is y1? [ans: y1=0.9960] approximation is less than desired accuracy, x2=1.2 what is y2? [ans: y2=0.9857] the final value of y1=2.1105 when x1=0.1 x3=1.3 what is y3? [ans: y1=0.9715] PES College, PESIT Campus, Department of BCA 5 Ordinary Differential Equations RungeKutta Method The Taylor's series method of solving DE numerically is handicapped by the problem of find higher order derivatives. Problem 07: Compute y(0.1) and y(0.2) by RK method dy 2 =xy y , y(0)=1. dx Euler's method is less efficient in practical problems since it requires 'h' to be small for Solution: The formulas for the forth order RK method obtaining resonable accuracy. are The RungeKutta methods do not require the calculations of higher order derivatives and k1=h f(x0, y0) they are designed only the function values at k h k2=h f x 0 , y 0 1 some selected points on the subinterval. 2 2 These methods agree with Taylor's series k h solution upto the terms of hr where r is the k3=h f x 0 , y 0 2 2 2 order of RK method. Runge Kutta(RK) II order k4=h f x 0h , y 0k 3 1 (k1+2k2+2k3+k4) 6 k1=h f(x0, y0) ∆y= k2=h f(x0+h, y0+k1) given f(x, y)= xy+y2; x0=0; y0=1 and h=0.1 y1=y0+ 1 k k 2 1 2 k1=(0.1) (0+1)=0.1 k2=(0.1) f 0 Runge Kutta IV order (referred as RK method) k1=h f(x0, y0) k h k2=h f x 0 , y 0 1 2 2 k h k3=h f x 0 , y 0 2 2 2 k4=h f x 0h , y 0k 3 ∆y= 1 (k1+2k2+2k3+k4) 6 then the 1st approximatiion is given by y1=y0+∆y Similarly, 2nd approximatiion is given by y2=y1+∆y, where x1=x0+h & y0 by y1 In general, yn+1=yn+∆y, PES College, PESIT Campus, Department of BCA 0.1 0.1 ,1 2 2 =f(0.05,0.05) = (0.1) (0.05*0.05+(0.05)2)=0.1155 k3=(0.1) f 0 0.1 0.1155 ,1 2 2 =f(0.05, 0.05775)=0.1172 k4=(0.1) f x 0h , y 0k 3 = (0.1) f(0+0.1, 1+0.1172) = 0.1360 ∆y= 1 (0.1+2(0.1155)+2(0.1172)+0.1360) 6 = 0.1169 Therefore, y(0.1)=1.1169 For 2nd approximatiion, we have x1=0.1; y1=1.1169 and h=0.1 k1=0.1359 k2=0.1582 6 Ordinary Differential Equations k3=0.1610 k4=0.1889 ∆y=0.1605 Therefore, y(0.2)=1.2774 Exercies: 1. Use RK method to find y(0.1) given dy 1 = , y(0)=1 that dx x y [Ans: 1.0914] 2. Find y(0.1), y(0.2) and y(0.3) given dy =1xy , y(0)=2 that dx [ans: 2.1086, 2.2416, 2.3997] dy y 1 = y 1=1. Evaluate dx x x 2, y(1.1) [ans: 0.9957] 3. Given PES College, PESIT Campus, Department of BCA 7

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