Power Series Solutions We use power series to solve second order differential equations

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1 Objectives Power Series Solutions We use power series to solve second order differential equations We use power series expansions to find solutions to second order, linear, variable coefficient equations Introduction We solved second order, linear, homogeneous, constant coefficients equations y + a 1 y + a 0 y = 0, by guessing that the solutions have the form y(x) = e rx, (here we use x for the independent variable instead of t), and finding the appropriate values for the exponent r. Once we have two fundamental solutions, we then have all the solutions to the homogenous equation. This guessing method does not work with variable coefficient equations, y + a 1 (x) y + a 0 (x) y = 0, because the fundamental solutions are too difficult to guess. But, whatever the solution y(x) is, since it is a function, it must have a Taylor series expansion around any point x 0 where y(x 0 ) is defined, y(x) = a n (x x 0 ) n. The idea of the power series method is to put the expression above into the differential equation, and then find the values of the coefficients a n. More specifically, the Power Series method can be summarized as follows: (1) Choose an x 0 and write the solution y as a power series expansion centered at a point x 0, y(x) = a n (x x 0 ) n. (2) Introduce the power series expansion above into the differential equation and find a recurrence relation an equation where the coefficient a n is related to a n 1 (and possibly a n 2 ). (3) Solve the recurrence relation find a n in terms of a 0 (and possibly a 1 ). (4) If possible, add up the resulting power series for the solution y. Page 1 of 5

2 Power Series on First Order Equations The problem below involves a first order, constant coefficient, equation. We use this simple equation to practice the Power Series Method. But recall, this method was created to solve variable coefficient equations. Question 1: Use a power series around the point x 0 = 0 to find all solutions y of the equation y + c y = 0, c R. (1) (2 points) Find the recurrence relation relating the coefficient a n with a n 1. (2) (2 points) Solve the recurrence relation, that is, find a n in terms of a 0. (3) (1 points) Write the solution y as a power series one multiplied by a 0. Then add the power series expression. Note: Using the integrating factor method we know that the solution is y(x) = a 0 e c x, with a 0 R. We want to recover this solution using the Power Series Method. We write a solution of the form y = a n x n y = na n x (n 1). We can start the sum in y at n = 0 or n = 1. We choose n = 1, since it is more convenient later on. Introduce the expressions above into the differential equation, na n x n 1 + c a n x n = 0. Relabel the first sum above so that x n 1 and x n in the first and second sum have the same label, (n + 1)a (n+1) x n + c a n x n = 0 We can now write down both sums into one single sum, " # (n + 1)a(n+1) + c a n x n = 0. therefore, (n + 1)a (n+1) + c a n = 0, n! 0. The last equation is called a recurrence relation among the coefficients a n. The solution is n = 0, a 1 = c a 0 a 1 = c a 0, n = 1, 2a 2 = c a 1 a 2 = c2 2! a 0, n = 2, 3a 3 = c a 2 a 3 = c3 3! a 0, n = 3, 4a 4 = c a 3 a 4 = c4 4! a 0. Page 2 of 5

3 One can check that the coefficient a n can be written as a n = ( 1) n cn n! a 0, which implies that the solution of the differential equation is given by! y(x) = a 0 ( 1) n cn n! xn y(x) = a 0 ( c x) n n! y(x) = a 0 e c x. Page 3 of 5

4 Power Series on Second Order Equations Once again, the problem below involves a second order, constant coefficient, equation. We use this simple equation to practice the Power Series Method. But recall, this method was created to solve variable coefficient equations. Question 2: Use a power series around the point x 0 = 0 to find all solutions y of the equation y + y = 0. (1) (2 points) Find the recurrence relation relating the coefficient a n with a n 2. (2) (2 points) Solve the recurrence relation, which in this case means to find a n in terms of a 0 for n even, and a n in terms of a 1 for n odd. (3) (1 points) Write the solution y as a combination of two power series, one multiplied by a 1, the other multiplied by a 0. Then add both power series expressions. Note: Guessing the fundamental solutions we know that the solution is y(x) = a 0 a 0, a 1 R. We want to recover these solutions using the Power Series Method. cos(x) + a 1 sin(x), with We know that the solution can be found computing the roots of the characteristic polynomial r = 0, which gives us the solutions y(x) = a 0 cos(x) + a 1 sin(x). We now recover this solution using the power series, y = a n x n y = na n x (n 1), y = n(n 1)a n x (n 2). Introduce the expressions above into the differential equation, which involves only the function and its second derivative, n(n 1)a n x n 2 + a n x n = 0. n=2 Relabel the first sum above, so that both sums have the same factor x n. One way is, (n + 2)(n + 1)a (n+2) x n + Now we can write both sums using one single sum as follows, n=2 a n x n = 0. " # (n + 2)(n + 1)a(n+2) + a n x n = 0 (n + 2)(n + 1)a (n+2) + a n = 0. n! 0. Page 4 of 5

5 The last equation is the recurrence relation. The solution of this relation can again be found by writing down the first few cases, and we start with even values of n, that is, n = 0, (2)(1)a 2 = a 0 a 2 = 1 2! a 0, n = 2, (4)(3)a 4 = a 2 a 4 = 1 4! a 0, n = 4, (6)(5)a 6 = a 4 a 6 = 1 6! a 0. One can check that the even coefficients a 2k can be written as a 2k = ( 1)k (2k)! a 0. The coefficients a n for the odd values of n can be found in the same way, that is, n = 1, (3)(2)a 3 = a 1 a 3 = 1 3! a 1, n = 3, (5)(4)a 5 = a 3 a 5 = 1 5! a 1, n = 5, (7)(6)a 7 = a 5 a 7 = 1 7! a 1. One can check that the odd coefficients a 2k+1 can be written as a 2k+1 = ( 1)k (2k + 1)! a 1. Split the sum in the expression for y into even and odd sums. We have the expression for the even and odd coefficients. Therefore, the solution of the differential equation is given by y(x) = a 0! k=0 ( 1) k (2k)! x2k + a 1 k=0 ( 1) k (2k + 1)! x2k+1. One can check that these are precisely the power series representations of the cosine and sine functions, respectively, y(x) = a 0 cos(x) + a 1 sin(x). Page 5 of 5