How To Solve A Differential Equation With Power Series

Transcription

1 Using Series to Solve Differential Equations Many differential equations can t be solved explicitly in terms of finite combinations of simple familiar functions. This is true even for a simple-looking equation like 1 y2xy y 0 But it is important to be able to solve equations such as Equation 1 because they arise from physical problems and, in particular, in connection with the Schrödinger equation in quantum mechanics. In such a case we use the method of power series; that is, we look for a solution of the form The method is to substitute this expression into the differential equation and determine the values of the coefficients c 0, c 1, c 2,.... Before using power series to solve Equation 1, we illustrate the method on the simpler equation yy 0 in Example 1. EXAMPLE 1 Use power series to solve the equation yy 0. SOLUTION y f x We assume there is a solution of the form c n x n c 0 c 1 x c 2 x 2 c 3 x 3 2 y c 0 c 1 x c 2 x 2 c 3 x 3 c n x n We can differentiate power series term by term, so y c 1 2c 2 x 3c 3 x 2 nc n x 3 y 2c 2 2 3c 3 x n n 1 c n x In order to compare the expressions for y and y more easily, we rewrite y as follows: By writing out the first few terms of (4), you can see that it is the same as (3). To obtain (4) we replaced n by n 2 and began the summation at 0 instead of 2. 4 y n 2 n 1 c x n Substituting the expressions in Equations 2 and 4 into the differential equation, we obtain or n 2 n 1 c x n c n x n 0 5 n 2 n 1 c c n x n 0 If two power series are equal, then the corresponding coefficients must be equal. Therefore, the coefficients of x n in Equation 5 must be 0: n 2 n 1 c c n 0 1

2 2 USING SERIES TO SOLVE DIFFERENTIAL EQUATIONS 6 c n c n 1 n 2 n 0, 1, 2, 3,... Equation 6 is called a recursion relation. If c 0 and c 1 are known, this equation allows us to determine the remaining coefficients recursively by putting n 0, 1, 2, 3,... in succession. By now we see the pattern: Put n 0: c 2 c Put n 1: c 3 c Put n 2: c 4 c c c 0 4! Put n 3: c 5 c c c 1 5! Put n 4: c 6 c c 0 4! 5 6 c 0 6! Put n 5: c 7 c c 1 5! 6 7 c 1 7! Putting these values back into Equation 2, we write the solution as y c 0 c 1 x c 2 x 2 c 3 x 3 c 4 x 4 c 5 x 5 c 0 1 x 2 2! x 4 4! x 6 x 2n 6! 1 n 2n! c 1 x x 3 3! x 5 5! x 7 x 2 7! 1 n c 0 For the even coefficients, c 2n 1 n For the odd coefficients, c 2 1 n 1 n x 2n 2n! c 1 1 n Notice that there are two arbitrary constants, and c 1. NOTE 1 We recognize the series obtained in Example 1 as being the Maclaurin series for cos x and sin x. (See Equations and ) Therefore, we could write the solution as y x c 0 cos x c 1 sin x But we are not usually able to express power series solutions of differential equations in terms of known functions. c 0 x 2 c 0 2n! c 1

3 USING SERIES TO SOLVE DIFFERENTIAL EQUATIONS 3 EXAMPLE 2 Solve y2xyy 0. SOLUTION We assume there is a solution of the form y c n x n Then and y nc n x y n n 1 c n x n 2 n 1 c x n as in Example 1. Substituting in the differential equation, we get n 2 n 1 c x n 2x nc n x c n x n 0 n 2 n 1 c x n 2nc n x n c n x n 0 2nc n x n 2nc n x n n 2 n 1 c 2n 1 c n x n 0 This equation is true if the coefficient of is 0: x n n 2 n 1 c 2n 1 c n 0 7 c 2n 1 n 1 n 2 c n n 0, 1, 2, 3,... We solve this recursion relation by putting n 0, 1, 2, 3,... successively in Equation 7: Put n 0: c c 0 Put n 1: c c 1 Put n 2: c c c 0 3 4! c 0 Put n 3: c c c c 1 5! Put n 4: c c ! 5 6 c c 0 6! Put n 5: c c ! 6 7 c c 1 7! Put n 6: c c c 0 8! Put n 7: c c c 1 9!

4 4 USING SERIES TO SOLVE DIFFERENTIAL EQUATIONS In general, the even coefficients are given by and the odd coefficients are given by n 5 c 2n c 0 2n! c n 3 c 1 The solution is y c 0 c 1 x c 2 x 2 c 3 x 3 c 4 x 4 c ! x 2 3 4! x ! c 1 x 1 3! x ! x ! x ! x 8 x ! x 9 or 8 y c ! x 2 c 1 x 3 7 4n 5 2n! n 3 x 2n x 2 NOTE 2 In Example 2 we had to assume that the differential equation had a series solution. But now we could verify directly that the function given by Equation 8 is indeed a solution. NOTE 3 Unlike the situation of Example 1, the power series that arise in the solution of Example 2 do not define elementary functions. The functions y 1 x 1 1 2! x n 5 2n! x 2n 2 T and y 2 x x n 3 x 2 _2 2 T _8 FIGURE 1 15 fi _ are perfectly good functions but they can t be expressed in terms of familiar functions. We can use these power series expressions for y 1 and y 2 to compute approximate values of the functions and even to graph them. Figure 1 shows the first few partial sums T 0, T 2, T 4,... (Taylor polynomials) for y 1 x, and we see how they converge to y 1. In this way we can graph both y 1 and in Figure 2. y 2 NOTE 4 If we were asked to solve the initial-value problem we would observe that y2xy y 0 c 0 y 0 0 y 0 0 c 1 y 0 1 y 0 1 This would simplify the calculations in Example 2, since all of the even coefficients would be 0. The solution to the initial-value problem is FIGURE 2 _15 y x x n 3 x 2

5 USING SERIES TO SOLVE DIFFERENTIAL EQUATIONS 5 Exercises A Click here for answers Use power series to solve the differential equation. 1. yy 0 2. y xy 3. y x 2 y 4. x 3 y2y 0 5. yxy y 0 6. y y x 2 1 yxy y 0 y xy 9. yxy y 0, y 0 1, y 0 0 S Click here for solutions. 10. yx 2 y 0, y 0 1, y yx 2 y xy 0, y 0 0, y The solution of the initial-value problem x 2 y xy x 2 y 0 y 0 1 y 0 0 is called a Bessel function of order 0. (a) Solve the initial-value problem to find a power series expansion for the Bessel function. ; (b) Graph several Taylor polynomials until you reach one that looks like a good approximation to the Bessel function on the interval 5, 5.

6 6 USING SERIES TO SOLVE DIFFERENTIAL EQUATIONS Answers S Click here for solutions. 1. c 0 3. n! c0e x x n c 0 x 3n 3 n n! c0e x c 0 1 n 2 n n! x 2n c 1 2 n n! x 2 7. c 0 c 1x c x c x x 2n 2 n n! e x n n 1 2 3n 1! 1 2n 3! 2 2 n! n 2! x 3 x 2n

7 USING SERIES TO SOLVE DIFFERENTIAL EQUATIONS 7 Solutions: Using Series to Solve Differential Equations P 1. Let y(x) = c nx n.theny 0 P (x) = nc nx n 1 and the given equation, y 0 y =0, becomes n=1 P nc nx n 1 P c nx n P =0.Replacingn by n +1in the first sum gives (n +1)c n+1x n P c nx, n=1 P so [(n +1)c n+1 c n]x. Equating coefficients gives (n +1)c n+1 c, so the recursion relation is c n+1 = cn n +1,, 1, 2,...Thenc1 = c0, c2 = 1 2 c1 = c0 2, c3 = 1 3 c2 = c0 c0 = 2 3!, c4 = 1 c0 c3 = 4 4!,and in general, c n = c0. Thus, the solution is n! y(x) = c nx n = 3. Assuming y(x) = x 2 y = c n x n,wehavey 0 (x) = c n x n+2 = n =2 n =2 c 0 n! xn = c 0 nc n x n 1 = x n n! = c0ex (n +1)c n+1 x n and c n 2 x n. Hence, the equation y 0 = x 2 y becomes P (n +1)c n+1x n P c n 2x n P =0or c 1 +2c 2x + [(n +1)c n+1 c n 2] x. Equating coefficients gives c 1 = c 2 =0and c n+1 = cn 2 n +1 for n =2, 3,...Butc 1 =0,soc 4 =0and c 7 =0and in general n =2 c 3n+1 =0. Similarly c 2 =0so c 3n+2 =0. Finally c 3 = c0 3, c6 = c3 6 = c 9 = c 6 9 = c = c !,...,andc3n = c 0. Thus, the solution is 3 n n! y (x) = c nx n = c 3nx 3n = c 0 3 n n! x3n = c 0 c0 6 3 = c !, x 3n 3 n n! = c0 x 3 /3 n = c 0e x3 /3 n! 5. Let y (x) = P cnxn y 0 (x) = P n=1 ncnxn 1 and y 00 (x) = P (n +2)(n +1)cn+2xn.The differential equation becomes P (n +2)(n +1)cn+2xn + x P n=1 ncnxn 1 + P cnxor P [(n +2)(n +1)c n+2 + nc n + c n ]x n since P n=1 nc nx n = P nc nx n. Equating coefficients gives (n +2)(n +1)c n+2 +(n +1)c, thus the recursion relation is c n+2 =, 1, 2,... Then the even coefficients are given by c 2 = c0 2, c4 = c2 4 = (n +1)cn (n +2)(n +1) = cn n +2, c0 2 4, c6 = c4 6 = c , and in general, c 2n =( 1) n c n = ( 1)n c 0. The odd coefficients are c 2 n 3 = c1 n! 3, c 5 = c3 5 = c1 3 5, c1 c 7 = c5 7 = 3 5 7, and in general, c 2n+1 =( 1) n c (2n +1) = ( 2)n n! c 1. The solution is (2n +1)! X ( 1) n ( 2) n n! y (x) =c 0 2 n n! x2n + c 1 (2n +1)! x2n+1.

8 8 USING SERIES TO SOLVE DIFFERENTIAL EQUATIONS 7. Let y (x) = P cnxn.theny 00 = P n (n 1) cnxn 2, xy 0 = P ncnxn and x 2 +1 y 00 = P n (n 1) c nx n + P (n +2)(n +1)c n+2x n. The differential equation becomes P [(n +2)(n +1)cn+2 +[n (n 1) + n 1] cn] (n 1) xn cn =0. The recursion relation is c n+2 =, n +2, 1, 2,...Givenc 0 and c 1, c 2 = c0 2, c 4 = c2 4 = c !, c 6 = 3c4 3c 0 6 =( 1) !,..., c 2n =( 1) n (2n 3) c 0 2 n n! =( 1) n 1 (2n 3)! c 0 2 n 2 n 2 n!(n 2)! =( 1)n 1 (2n 3)! c 0 2 2n 2 n!(n 2)! for n =2, 3,... c 3 = 0 c1 =0 c 2n+1 =0for, 2,... Thus the solution is 3 x 2 y (x) =c 0 + c 1 x + c c ( 1) n 1 (2n 3)! 0 x 2n. 2 2n 2 n!(n 2)! n=2 9. Let y(x) = y 00 (x) = P c n+2 = P c n x n.then xy 0 (x) = x nc n x n 1 = nc n x n = (n +2)(n +1)c n+2 x n, and the equation y 00 xy 0 y =0becomes [(n +2)(n +1)c n+2 nc n c n ]x. Thus, the recursion relation is nc n x n, nc n + c n cn(n +1) = (n +2)(n +1) (n +2)(n +1) = cn for, 1, 2,... One of the given conditions is n +2 y(0) = 1. Buty(0) = c 6 = c4 6 = y 0 (0) = n=1 c n(0) n = c = c 0,soc 0 =1. Hence, c 2 = c 0 2 = 1 2, c4 = c 2 4 = 1 2 4,,..., c2n = 1 2 n n!. The other given condition is y0 (0) = 0. But nc n(0) n 1 = c = c 1,soc 1 =0. By the recursion relation, c 3 = c 1 3 c 2n+1 =0for, 1, 2,... Thus, the solution to the initial-value problem is =0, c5 =0,..., y(x) = c nx n = c 2nx 2n = x 2n 2 n n! = X x 2 /2 n n! = e x2 /2 P 11. Assuming that y(x) = c n x n P,wehavexy = x c n x n = x 2 y 0 = x 2 nc nx n 1 = nc nx n+1, y 00 P (x)= n(n 1)c nx n 2 = n =2 n= 1 P =2c 2 + (n +3)(n +2)c n+3x n+1, c n x n+1, (n +3)(n +2)c n+3x n+1 [replace n with n +3]

9 USING SERIES TO SOLVE DIFFERENTIAL EQUATIONS 9 and the equation y 00 + x 2 y 0 P + xy =0becomes 2c 2 + [(n +3)(n +2)c n+3 + nc n + c n] x n+1 =0. So c 2 =0and the recursion relation is c n+3 = ncn cn (n +1)cn =,, 1, 2,... (n +3)(n +2) (n +3)(n +2) But c 0 = y(0) = 0 = c 2 and by the recursion relation, c 3n = c 3n+2 =0for, 1, 2,.... Also, c 1 = y 0 (0) = 1,so c 4 = 2c1 4 3 = 2 5c , c7 = =( 1) =( 1)2,..., 7! c 3n+1 =( 1) n (3n 1) 2. Thus, the solution is (3n +1)! y(x) = c n x n = x + ( 1) n (3n 1) 2 x 3n+1 (3n +1)!