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?

  • $\begingroup$ @Amzoti I edited my post and included the second case root, which doesn't work. $\endgroup$
    – user59036
    Jan 26, 2014 at 20:50
  • $\begingroup$ @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? $\endgroup$
    – user59036
    Jan 26, 2014 at 21:03
  • $\begingroup$ @Amzoti That would be great, thank you very much! $\endgroup$
    – user59036
    Jan 26, 2014 at 21:06

3 Answers 3


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:

$$y(x) = y_1(x) + y_2(x) = \dfrac{c_1(3x(3x^2 + 3x+ 2) + 2)}{x^3} + \dfrac{c_2e^{3x}}{x^3}$$

  • 1
    $\begingroup$ 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. $\endgroup$
    – IgNite
    May 1, 2016 at 8:00
  • $\begingroup$ @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/… $\endgroup$
    – shamisen
    May 23, 2018 at 15:43

Let $y=\sum\limits_{n=0}^\infty a_nx^{n+r}$ ,

Then $y'=\sum\limits_{n=0}^\infty(n+r)a_nx^{n+r-1}$


$\therefore x^2\sum\limits_{n=0}^\infty(n+r)(n+r-1)a_nx^{n+r-2}+3(x-x^2)\sum\limits_{n=0}^\infty(n+r)a_nx^{n+r-1}-3\sum\limits_{n=0}^\infty a_nx^{n+r}=0$

$\sum\limits_{n=0}^\infty(n+r)(n+r-1)a_nx^{n+r}+\sum\limits_{n=0}^\infty3(n+r)a_nx^{n+r}-\sum\limits_{n=0}^\infty3(n+r)a_nx^{n+r+1}-\sum\limits_{n=0}^\infty 3a_nx^{n+r}=0$



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.


  • 1
    $\begingroup$ I believe this is called reduction of order, not variation of parameters. $\endgroup$ Apr 4, 2016 at 21:26

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