Finding a second solution of $xy''+2xy'+6e^xy=0$

Question

I am trying to solve $$xy''+2xy'+6e^xy=0,\space x>0$$ about $x=0$ using Frobenius method. The question is from Boyce & DiPrima.

I have found the exponents at singularity to be $r_1=1$ and $r_2=0$. I have also found the series solution for $r=1$ by substituting $$y=x^r\sum_{n=0}^{\infty}a_nx^n$$ into the ODE and replacing $e^x$ with its Taylor series:

$$x^r\sum_{n=-1}^\infty a_{n+1}(n+r+1)(n+r)x^n+2x^r\sum_{n=0}^\infty a_{n}(n+r)x^n + 6x^r\sum_{n=0}^\infty (a_{n}x^n) \times \sum_{n=0}^\infty \frac{x^n}{n!} = 0$$ $$\dots$$ $$a_1 = -a_0\frac{2r+6}{r(r+1)}$$ $$a_2 = -a_0\frac{-\frac{(2r+8)(2r+6)}{r(r+1)}+6}{(r+2)(r+1)}$$ $$\dots$$

When I set $r=1, a_0=1$ and calculate these coefficients, I get $y_1=x\left(1-4x+\frac{17}{3}x^2-\frac{47}{12}x^3+\dots\right)$ which is correct (according to my book).

Now, to find the second solution (associated with $r_2=0$), I use this equation: $$y_2=c\ln(x)y_1(x) + x^{r_2}\sum{b_nx^n} \tag{1} \label{y2-form}$$

I find $$c = \lim_{r\rightarrow r_2}(r-r_2)a_1(r)=-\frac{2r_2+6}{r_2+1}=-6.$$

This is also correct. But my results for $b_n$ are wrong. To find them, I set $a_0=r-r_2=r$, find $a_n(a_0=r, r)$ (by canceling the factors $r$), and differentiate them with respect to $r$: $$b_n=\left.\frac{d a_n(a_0=r, r)}{d r}\right\vert_{r=0}$$

$$a_0 = r$$ $$b_0 = \left.\frac{d r}{d r}\right\vert_{r=0} = 1$$ $$b_1 = \left.\frac{d}{d r}\left(-\frac{2r+6}{r+1} \right)\right\vert_{r=0} = 4$$ $$b_2 = \dots = -49$$

These are completely wrong. The books says $b_0 = 1, b_1 = 0, b_2 = -33, b_3 = \frac{449}{6}x^3, \dots$

I have seen this method of finding $b_n$ in a proof of Frobenius method in the book An Introduction to Ordinary Differential Equations written by Coddington. Did I get all this "multiplying by $(r-r_2)$ and differentiating" thing wrong? Or maybe I am doing a silly mistake. Boyce & DiPrima suggests putting $\eqref{y2-form}$ in the ODE and calculating the coefficients that way. But I want to understand what is wrong with the other method.

Answer 1

Since $y_1$ and $y_2$ are linearly independent, we can add any multiple of $y_1$ to $y_2$, and get a solution that is still independent with $y_1$. Since $y_1$ is a power series starting with $x^1$, this means that the coefficient $b_1$ in the second solution is arbitrary.

Notice that the difference between the $b_n$ values found in the question and the $b_n$ values in the book are multiples of the coefficients in $y_1$. So we obtain the book's solution by subtracting $4y_1$ from $y_2$. This means that the $y_2$ found here is also correct.

Answer 2

I have one short note related to this problem: If the difference $r_1-r_2$ between the roots of the indicial equation is a positive integer, it is sometimes possible to obtain the general solution using the smaller root alone, without bothering to find explicitly the solution corresponding to the larger root. Indeed, if the difference $r_1-r_2$ is a positive integer, it is a worthwhile practice to work with the smaller root first, in the hope that this smaller root by itself may lead directly to the general solution.

Please for further reading read chapter 6 in Differential equation by "Shepley L.Ross". I think this will help a lot in these types of problems.

Answer 3

Given the linear ODE

$$ a(x)y''+b(x)y'+c(x)y = 0 $$

if $y_1$ is a solution then proposing $y_2=k(x)y_1$ and substituting we get at

$$ (2a(x)y'_1+b(x)y_1) k' + a(x)y_1 k''=0 $$

so calling $u = k'$ we follow with

$$ (2a(x)y'_1+b(x)y_1) u + a(x)y_1 u'=0 $$

here $a(x) = x$ and $b(x) = 2x$ so the question is: is there a form $u = \sum_j\lambda_jx^j$ such that

$$ 2x(y'_1+y_1) u + 2xy_1 u'=0\Rightarrow (y'_1+y_1) u + y_1 u'=0\ \ \text{?} $$