# Find the form of a second linear independent solution when the two roots of indicial equation are different by a integer

Consider the differential equation $$x^2y''+3(x-x^2)y'-3y=0$$

$(a)$ Find the recurrence equation and first three nonzero terms of the series solution in powers of $$corresponding to the larger root of the indicial equation. (b) What would be the form of a second linearly independent solution of this differential equation? I found the indicial equation to be r(r-1)+3r-3 = 0, so the two roots are r_1=-3 and r_2=1. And the recurrence relation is a_n = \dfrac{3(n-1+r)a_{n-1}}{(n+r)(n+r-1)+3(n+r)-3} set r=1, then a_n = \dfrac{3na_{n-1}}{(n+3)(n+1)-3}. And I can figure out first linearly independent solution. But for r=-3,then a_n = \dfrac{3(n-4)a_{n-1}}{(n-3)(n-1)-3}, if let a_0 be an arbitrary constant, then a_1 =\dfrac{3(-3)a_0}{(-2)(0)}, which doesn't work. Then how do I figure out the second linearly independent solution? • @Amzoti I edited my post and included the second case root, which doesn't work. – user59036 Jan 26 '14 at 20:50 • @Amzoti No, in fact the lecture didn't even talk about the case where the two roots are different by a integer. But can you tell me about the procedure using partial derivatives of the first solution? – user59036 Jan 26 '14 at 21:03 • @Amzoti That would be great, thank you very much! – user59036 Jan 26 '14 at 21:06 ## 3 Answers Here is the procedure. Denote the two roots by r_1 and r_2, with r_1 \gt r_2. The Method of Frobenius will always generate a solution corresponding to r_1, but may generate a solution for the smaller second root r_2 of the indicial equation. If the method fails for r_2, then an approach is to keep the recursion solution in terms of r and use it to find the coefficients a_n (for n \ge 1), in terms of both r and a_0, where a_0 \ne 0. For ease, a_0 is typically chose to be one. Using this more general form and the coefficients, the two independent solutions can be written as:$$y_1(r, x) = x^r \sum_{n=0}^\infty ~ a_n(r)x^n = \sum_{n=0}^\infty ~ a_n(r)x^{n+r} \\y_2(r,x) = \dfrac{\partial}{\partial r}[(r - r_2)y_1(r, x)]~\Bigr|_{r=r_2}$$You should be able to use this approach and show:$$y(x) = y_1(x) + y_2(x) = \dfrac{c_1(3x(3x^2 + 3x+ 2) + 2)}{x^3} + \dfrac{c_2e^{3x}}{x^3}$$• Could I ask you some question? Where can I read about this part:$$y_2(r,x) = \dfrac{\partial}{\partial r}[(r - r_2)y_1(r, x)]~\Bigr|_{r=r_2}$$In my class the lecture didn't cover about this. In my class, when two roots differ by an integer, the second solution is assumed to be$$ y_{2}(x)=C_1y_{1}(x)\ln{x} + x^{r_{2}}\sum_{n=0}^{\infty} b_nx^{n} $$And then I would just plug this back into the original equation and find the coefficients. I've never seen your method before. Could you shed some light on this? I'd be glad if you could help me understand various ways of solving this. Thanks. – IgNite May 1 '16 at 8:00 • @IgNite This method to compute y_2 is explained in 34-39 of Nielsen's slides on the Frobenius method: creighton.edu/fileadmin/user/CCAS/departments/Math/docs/DE/… – shamisen May 23 '18 at 15:43 Let y=\sum\limits_{n=0}^\infty a_nx^{n+r} , Then y'=\sum\limits_{n=0}^\infty(n+r)a_nx^{n+r-1} y''=\sum\limits_{n=0}^\infty(n+r)(n+r-1)a_nx^{n+r-2} \therefore x^2\sum\limits_{n=0}^\infty(n+r)(n+r-1)a_nx^{n+r-2}+3(x-x^2)\sum\limits_{n=0}^\infty(n+r)a_nx^{n+r-1}-3\sum\limits_{n=0}^\infty a_nx^{n+r}=0 \sum\limits_{n=0}^\infty(n+r)(n+r-1)a_nx^{n+r}+\sum\limits_{n=0}^\infty3(n+r)a_nx^{n+r}-\sum\limits_{n=0}^\infty3(n+r)a_nx^{n+r+1}-\sum\limits_{n=0}^\infty 3a_nx^{n+r}=0 \sum\limits_{n=0}^\infty(n+r+3)(n+r-1)a_nx^{n+r}-\sum\limits_{n=1}^\infty3(n+r-1)a_{n-1}x^{n+r}=0 (r+3)(r-1)a_0x^r+\sum\limits_{n=1}^\infty((n+r+3)(n+r-1)a_n-3(n+r-1)a_{n-1})x^{n+r}=0 The indicial equation is (r+3)(r-1)=0 , the two roots are r_1=-3 and r_2=1 . The recurrence relation is (n+r+3)(n+r-1)a_n-3(n+r-1)a_{n-1}=0 , i.e. (n+r+3)(n+r-1)a_n=3(n+r-1)a_{n-1} Each root from the indical equations may not always only find one group of the linearly independent solutions, sometimes we can find more than one group of the linearly independent solutions at the same time. When we choose r=-3 , the recurrence relation becomes n(n-4)a_n=3(n-4)a_{n-1} , which has a value of n at n=4 so that it is independent to the recurrence relation. Besides a_0 can chooes arbitrary, a_4 can also chooes arbitrary. This makes the effect that we can find two groups of the linearly independent solutions that basing on n=0 to n=3 and n=4 to n=+\infty respectively. Let me rewrite your ODE as follows:$$L[y] = p_0 y'' + p_1 y' + p_2 y = 0,$$where p_i(x) are the coefficients of the equation. If you know one of the two linear indepent solutions of the homogenous part (indeed the whole ode is homogenouse), say y_1, you can obtain y_2 (y = A y_1+B y_2) by using the method of variation of parameters. Let me show you. Set y = A(x)y_1 and substitute back into the original ode:$$p_0 (A''y_1 + 2A'y_1' + A y_1'') + p_1 (A'y_1 + A y_1') + p_2 A y_1 = 0,$$arrange terms in order to get a ODE-1 for A(x):$$ A'' + \left(\frac{2y'_1}{y_1} + \frac{p_1 y_1}{p_0} \right) A' + \frac{1}{p_0} L[y_1] A= 0,$$since L[y_1]=0, A can be obtained by solving this equation in terms of an integrating factor as follows:$$\frac{d}{dx} (u A') = u \cdot 0 = 0,$$where u = e^{\int \left(\frac{2y'_1}{y_1} + \frac{p_1 y_1}{p_0} \right) \, dx}. It yields:$$uA' = a_2 \Rightarrow A = a_1 + a_2 \int \frac{dx}{u},

and hence the solution, $y = a_1 y_1 + a_2 y_1 \int \frac{dx}{u}$, being $a_i$ constants of integration.

I hope this is useful to you.

Cheers!

• I believe this is called reduction of order, not variation of parameters. – Radon Rosborough Apr 4 '16 at 21:26