Find the first 3 nonzero terms in each of two solutions (not multiples of each other) about x =0 (different problem than before)

Question

anonymous · Answer

$$xy''+y = 0$$

I'm having the same trouble I had before, coming up with a 2nd solution. I have the first solution:

$$y_{1}(x) = x-\frac{1}{2}x^{2}+\frac{1}{12}x^{3}-\frac{1}{144}x^{4}+...$$ but the second solution I still am unable to get to.

anonymous · Answer

The roots r1 and r2 are 1 and 0, and for a problem where roots r1-r2 is a positive integer, the form of the second solution is supposed to be:
$$y_{2}(x) = ay_{1}(x)\ln|x| + |x|^{r_{2}}[1+\sum_{n=1}^{\infty}c_{n}(r_{2})x^{n}]$$

The books gives mini formulas to find a and cn (assuming I'm understanding the book correctly), but all give me undefined results. Substituting the form of y2(x) into the original equation comes up with nothing, especially since I do not know what a is. So yeah, kinda lost on what to do.

dan815 · Answer

@wio

dan815 · Answer

@ikram002p remember power series rep of diff eqns?

ikram002p · Answer

its 4:37 AM -_-
eshhh cant open my eyes

dan815 · Answer

@SithsAndGiggles , hey do you remember this stuff?

anonymous · Answer

When I took DE the class only lightly touched on how to find recurrence relations, not so much on these "roots."

anonymous · Answer

Yeah, it's what the book refers to the exponents of x as, "roots" when you have an equation of the form:
$$P(x)y''+Q(x)y'+R(x)y=0$$
The roots come either from this formula:
$$F(r) = r(r-1)+p_{0}r + q_{0}$$Where $$p_{0} = \lim_{x \rightarrow 0}\frac{xQ(x)}{P(x)}$$
$$q_{0}=\lim_{x \rightarrow 0}\frac{x^{2}R(x)}{P(x)}$$
You can also find the roots during simplification of the power series substitutions you do. Often times, you remove terms from a series in order to raise the index. Doing this, you end up removing terms with r in them. Because the terms you remove from the series must equal 0, you're able to solve for r such that you're able to equate it all to zero. 

In this problem you have P(x) = x, Q(x) = 0, R(x) = 1. So p0 = 0 and q0 =0, giving
F(r) = r(r-1). Setting this equal to 0 gives r = 0 and 1. These are the "roots". 

So when doing these problems, you do your normal series substitutions to create your recurrence relation. The r you use is the higher of the two you find. The recurrence relation for this problem turned out to be:
$$a_{n}=\frac{-a_{n-1}}{(r+n)(r+n-1)}$$Using the higher root of one, I'm able to get the first 3 terms of the first solution, as I gave above. I also can come up with the general series. Putting it into the form the book gives, I have:
$$y_{1}(x) = x[1+\sum_{n=1}^{\infty}\frac{(-1)^{n}x^{n}}{n!(n+1)!}]$$This is also what the book says about finding a second solution of this form:

Roots r1 and r2 differing by an integer N
$$y_{2}(x) = ay_{1}(x)\ln|x|+|x|^{r_{1}}[1+\sum_{n=1}^{\infty}c_{n}(r_{2})x^{n}$$
As for finding a and cn, these are the ways it mentions to possibly find them:
$$c_{n}(r_{2})=\frac{d}{dr}[(r-r_{2})a_{n}(r)]$$r = r2 n = 1,2,...where an(r) is determined from this recurrence relation (a0=1):
$$F(r+n)a_{n}+\sum_{k=0}^{n-1}a_{k}[(r+k)p_{n-k}+q_{n-k}=0$$ n>=1 and
$$xp(x) = \sum_{n=0}^{\infty}p_{n}x^{n}$$
$$x^{2}q(x)=\sum_{n=0}^{\infty}q_{n}x^{n}$$
For the a coefficient, this is given:
$$r_{1}-r_{2}=N$$
$$a=\lim_{r \rightarrow r_{2}}(r-r_{2})a_{N}(r)$$
"When r1-r2=N, there are again three ways to find a second solution. First, we can calculate a and cn(r2) directly by substituting (the form of y2) for y into  Eq.1 (which is x^2y''+x[xp(x)]y'+[x^2q(x)]y=0 ). Second, we can calculate cn(r2) and a using (the formulas above). If this is the planned procedure, then in calculating the solution corresponding to r=r1, be sure to obtain a general formula for an(r) rather than just an(r1). The third method is to use the method of reduction of order. 

So that is all the background information and instruction I can gather from the book. Trying to follow what the book says, given I understand it correctly, I'm unable to get the correct answer. The answer I'm supposed to get for the terms of the second solution is:


$$y_{2}(x)=-y_{1}lnx + 1-\frac{3}{4}x^{2}+\frac{7}{36}x^{3}-\frac{35}{1728}x^{4}+...$$

Not sure what else I could give to help solve this problem. I figure either I'm doing something incorrectly or I have a misunderstanding of the instructions. Hopefully this extra information can be enough to help out.

anonymous · Answer

Still havent been able to figure this out :/