Math 244 Examples for 5.2 and 5.3

Examples for 5.2 and 5.3 Exercise 3, Section 5.2 Find the recurrence relation and the first four terms of two linearly independent solutions to: y 00 − xy 0 − y = 0 x0 = 1 Substitute the sum: y = ∞ X P an (x − 1)n in for y, y 0 and y 00 : n(n − 1)an (x − 1)n−2 − x n=2 ∞ X ∞ X nan (x − 1)n−1 − n=1 an (x − 1)n = 0 n=0 To simplify the second term, we need to be able to bring in an (x − 1), not just x. Therefore, the middle sum becomes two sums: x ∞ X nan (x − 1)n−1 = (x − 1) ∞ X nan (x − 1)n−1 + = ∞ X nan (x − 1)n−1 n=1 n=1 n=1 ∞ X nan (x − 1)n + ∞ X nan (x − 1)n−1 n=1 n=1 And now there are 4 sums, and they simplify to: ∞ X n−2 n(n − 1)an (x − 1) − ∞ X n nan (x − 1) − nan (x − 1) n−1 − ∞ X an (x − 1)n = 0 n=0 n=1 n=1 n=2 ∞ X Our goal now is to collect all the sums together as a single series of the form: X  (x − 1)k = 0 k To start this process, we want each sum to begin with the same power. The first, third and fourth sums begin with (x − 1)0 , but the second sum does not. In this case, we just need to make the index start at zero instead of 1 (this does not change the sum, since the first term would be 0): ∞ X n(n − 1)an (x − 1)n−2 − n=2 ∞ X nan (x − 1)n − n=0 ∞ X nan (x − 1)n−1 − n=1 ∞ X an (x − 1)n = 0 n=0 Now we put everything into the same sum. We make the following substitutions (in order): k =n−2 n=k+2 To get: ∞ X k=0 k=n n=k k =n−1 n=k+1 k=n n=k ((k + 2)(k + 1)ak+2 − kak − (k + 1)ak+1 − ak ) (x − 1)k = 0 The recurrence relation is found by setting the coefficients to zero (recall that if a polynomial is zero for all values of x, then every coefficient must be zero): ((k + 2)(k + 1)ak+2 − kak − (k + 1)ak+1 − ak = 0 for k = 0, 1, 2, 3, · · · Solve for ak+2 to get the recurrence relation: ak+2 = 1 (ak + ak+1 ) k+2 for k = 0, 1, 2, 3, · · · Now we want to find the first four terms of two linearly independent solutions. Recall from Chapter 3 that we can do this by taking y1 and y2 as solutions to the ODE, with initial conditions (respectively). That is, y1 solves the DE with initial conditions: y 0 (1) = a1 = 0 y(1) = a0 = 1 And y2 solves the DE with initial conditions: y 0 (1) = a1 = 1 y(1) = a0 = 0 We do this for y1 in the first column, y2 in the second: k=0 a2 = 21 (a0 + a1 ) = 1 2 k=0 a2 = 12 (a0 + a1 ) = 1 2 k=1 a3 = 13 (a1 + a2 ) = 1 6 k=1 a3 = 13 (a1 + a2 ) = 1 2 k=2 a4 = 14 (a2 + a3 ) = 1 6 k=2 a4 = 14 (a2 + a3 ) = 1 4 k = 3 a5 = 15 (a3 + a4 ) = 1 15 k = 3 a5 = 15 (a3 + a4 ) = 3 20 and Therefore, 1 1 1 y1 (x) = 1 + (x − 1)2 + (x − 1)3 + (x − 1)4 + . . . 2 6 6 and 1 1 1 y2 (x) = (x − 1) + (x − 1)2 + (x − 1)3 + (x − 1)4 + . . . 2 2 4 Exercise 2, Section 5.3 Determine the values of the first, second, third and fourth derivatives of the solution at the given value of x0 . (Added: Write the corresponding partial Taylor series): y 00 + sin(x)y 0 + cos(x)y = 0 y(0) = 0 y 0 (0) = 1 In this case, it is easy to solve for y 00 and differentiate: y 00 = − sin(x)y 0 − cos(x)y Evaluate at x = 0: y 00 (0) = −0 · 1 − 1 · 0 = 0. Continue differentiating: y 000 = − cos(x)y 0 − sin(x)y 00 + sin(x)y − cos(x)y 0 = −2 cos(x)y 0 − sin(x)y 00 + sin(x)y Evaluate at x = 0: y 000 (0) = −2 · 1 · 1 − 0 · 0 + 0 · 0 = −2 Continuing: y iv = 2 sin(x)y 0 − 2 cos(x)y 00 − cos(x)y 00 − sin(x)y 000 + cos(x)y + sin(x)y 0 Simplify: = 3 sin(x)y 0 − 3 cos(x)y 00 − sin(x)y 000 + cos(x)y Evaluate at x = 0; y iv (0) = 0 · 0 − 3 · 0 − 0 · −2 + 1 · 0 = 0 Writing the power series solution using the Taylor formula, so far we have: y(x) = 0 + 1 · x + 0 · x2 − 2 3 x + 0 · x4 + . . . 3! If we want more constants, we could compute them. For fun, here is the expansion up to the 8th power: 1 1 59 7 x + ... y(x) = x − x3 + x5 − 3 5 2520 Extra Example, Section 5.3 Determine a lower bound for the radius of convergence of the series solutions about each given point x0 for the given differential equation: (x2 − x − 2)y 00 + x2 y 0 − 4y = 0 x0 = 3 x0 = 6 SOLUTION: Dividing by the polynomial in front of y 00 , we have two functions that should be analytic about x0 . Notice that the polynomial factors: x2 (x − 2)(x + 1) 4 (x − 2)(x + 1) We notice that both functions have singularities at x = 2 and x = −1 (these are points at which the denominator is zero). If the power series is based at x0 = 3, there is a distance of |3 − 2| = 1 to the point x = 2 and a distance of |3 − −1| = 4 to the point x = −1. We take the smaller of these as the radius. Conclusion: The lower bound for the radius of convergence is 2. If the power series is based at x0 = 6, the distance between the two “bad points” is 4 and 7, respectively. Take the lower bound for the radius to be 4. Modify the example slightly: If we change the polynomial in front to: (x2 + 2x + 3)y 00 + x2 y 0 − 4y = 0 x0 = 3 √ Using the quadratic formula, we see that x2 +√2x + 3 = 0 if x√= −1 ± 2i. In the complex plane, we would plot these as the points (−1, 2) and (−1, − 2). To find the radius of convergence, we need to find the smaller distance of these points to our point, x0 = 3 + 0i or (3, 0). √ The distance formula for (−1, 2) gives: q √ √ √ (3 − −1)2 + ( 2 − 0)2 = 18 = 3 2 If you compute the distance to the other point, you’ll get the same thing. √ Conclusion: The lower bound for the radius of convergence is 3 2