Frobenius Series SolutionsDate: 05/01/2008 at 21:32:36 From: Judy Subject: Frobenius series solution Determine whether the differential equation: (x^2)y'' + xy' + ((x^2)-8)y = 0 has two linearly independent Frobenius series solutions. I am confused about the rest of the Frobenius series and how to show that they are linearly independent. I divided through by x^2 to get the general form of the equation: y'' + (y'/x) + [((x^2)-8)y)/x^2] Po=1 and Qo=1 hence the indicial eqn is: r(r-1) + r - 1 (r^2) - 1 = 0 therefore r=1,r=-1 and the Frobenius solutions are y = xsigmaAnx^n. Date: 05/02/2008 at 18:49:15 From: Doctor Jordan Subject: Re: Frobenius series solution Hi Judy, There is a general theorem that tells us that the two functions y_1 and y_2 we get are linearly independent solutions. This theorem tells us what the solutions look like in the three different cases where r_1 - r_2 is not an integer, is 0 or is a positive integer, where r_1,r_2 are the roots of the indicial equation, with r_1 > r_2. p(x) is the coefficient of y' and q(x) is the coefficient of the constant term, thus here p(x) = 1/x q(x) = (x^2-1)/8 p_0 is equal to x*p(x) when we set x=0, thus p_0 = 1. q_0 is equal to x^2*q(x) when we set x=0, thus q_0 = -8. The indicial equation is r(r-1) + p_0*r + q_0 = 0. Thus the indicial equation is r(r-1) + r - 8 =0 i.e. r^2 - 8 = 0 which has the roots sqrt(8) and -sqrt(8). Let r_1 = sqrt(8) and r_2 = -sqrt(8). r_1 - r_2 = sqrt(8) + sqrt(8) = 2*sqrt(8), which is not an integer. Therefore two linearly independent solutions are given by y_1 = sum_{n=0}^infinity a_n x^{n+r_1} y_2 = sum_{n=0}^infinity b_n x^{n+r_n} for x > 0, where a_n = c_n(r_1) and b_n = c_n(r_2). y(x) = sum_{n=0}^infinity c_n(r) x^{n+r}. We put y(x) into the given differential equation (x^2)y'' + xy' + ((x^2)-8)y = 0 and find a formula for c_n(r). Then we take r = sqrt(8) to find the formula for a_n, and r = -sqrt(8) to find the formula for b_n, and this tells us what the series for y_1 and y_2 are. Please write back if you have difficulty finding a formula for c_n(r), or if you have other questions about the Frobenius method or solving differential equations with series. - Doctor Jordan, The Math Forum http://mathforum.org/dr.math/ Date: 05/04/2008 at 00:11:22 From: Judy Subject: Thank you (Frobenius series solution) Thanks for helping me answer this question. I really appreciate it. |
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