Find the first three nonzero terms in each of TWO solutions (not multiples of each other) about x = 0

Question

anonymous · Answer

$$xy''+y'-y = 0$$


The first solution I already have, which is:

$$y_{1}(x) = 1+ x + \frac{1}{4}x^{2}+\frac{1}{36}x^{3}+...$$

This problem has a double root of r = 0. Supposedly, the form of a second solution when you have a double root is:

$$y_{2}(x) = y_{1}(x)\ln|x| + |x|^{r_{1}}\sum_{n-1}^{\infty}b_{n}(r_{1})x^{n}$$and the solution in the back of the back for this second solution is:

$$y_{2}(x)= y_{1}(x)lnx - 2x - \frac{3}{4}x^{2} - \frac{11}{108}x^{3}+...$$I have NO idea how to go and get this second solution. Many kind people have posted in a previous question I had for this, but I still have no idea how to determine this second solution. Can anyone show me how to derive the 2nd solution I have posted? Assuming a solution has the form of
$$y = \sum_{n=0}^{\infty}a_{n}x^{r+n}$$simply gives you exactly what I already have, a double root of r = 0 and the first solution posted, so there has to be a DIFFERENT way to get the 2nd solution. Any ideas? x_x

compassionate · Answer

@satellite73 , @mathmale , @ganeshie8 , @goformit100 , @Luigi0210 , @Abhisar , @jim_thompson5910 , @Hero

anonymous · Answer

Im not currently in a class. This is self-study right now.

jtvatsim · Answer

yikes, this one is a step above my pay grade... if I find out anything, I'll be sure to let you know though. :)

anonymous · Answer

Thank you for looking it over, though : )

anonymous · Answer

How did you get that first solution ? Using series method or something else ?

anonymous · Answer

I got it through two methods. One was assuming the solution was of the form
$$y = \sum_{n=0}^{\infty}a_{n}x^{r+n}$$and substituting it into the given DE. I was able to determine the double root of r through that method and it gave me the recurrence relation $$a_{n}=\frac{a_{n-1}}{(r+n)^{2}}$$. Using r = 0, I can get those first solutions for y1. I also was able to get that solution using a formula in my text. With a DE in the form of 
$$P(x)y''+Q(x)y'+R(x)y = 0$$, let $$p_{0}= \lim_{x \rightarrow 0}\frac{xQ(x)}{P(x)}$$and let
$$q_{0}=\lim_{x \rightarrow 0}\frac{x^{2}R(x)}{P(x)}$$ find the roots of the equation using $$F(r) = r(r-1)+p_{0}r+q_{0}$$ using this F(r) also gives me my double root of 0. To get the an's, I put this all together in this formula from the book:
$$F(r+n)a_{n} + \sum_{k=0}^{n-1}a_{k}[(r+k)p_{n-k}+q_{n-k}]=0 $$ ao is assumed to be arbitrary. So yeah, plugging everything into that recurrence relation looking thing Im able to get the same numbers for the first solution.

anonymous · Answer

So you want to know how to get that series form of the second solution. Right ? It is quite difficult to type the mathematics here. Do you have Mathematical Methods for Physicists 8th edition by Arfken ? If you have, you'll find the method on page 364.

anonymous · Answer

Yeah basically how to find the bn coefficients in that 2nd series solution form. No matter what I try, Im not able to come up with the -2x -3x^2/4 - 11x^3/108 terms. I dont have that textbook, though. The textbook this comes from is Elementary Differential Equations and Boundary Value Problems, 10th edition by Boyce and DiPrima.

anonymous · Answer

For  a differential equation of the form 
$$y'' + P(x)y' + Q(x)y = 0$$
if the indicial equation has equal roots, then by Fuch's theorem the ODE has only one series solution. Second solution can be obtained by using Wronskian condition of independent solutions.
If the Ist series solution is $$y_{1}(x)$$ then the second solution is 
$$y _{2}(x) = y _{1}(x) \int\frac{ e ^{-\int P(x)dx} }{ [y _{1}(x)]^2} dx$$

Now the given equation is $$y''+\frac{ 1 }{ x} y' -\frac{ 1 }{ x} y = 0$$ This equation has a regular singularity at x = 0 and the indicial equation has two equal roots i.e r =0 and 0.
The first series solution as you've already calculated is 
$$y _{1}(x) = 1+x+\frac{ 1 }{ 4} x^2 + \frac{ 1 }{ 36} x^3 + ...$$
Here $$P(x) = \frac{1}{x} $$ So the second solution is 
$$y _{2}(x) = y _{1}(x) \int\frac{ e ^{-\int x^{-1} dx} }{ [y _{1}(x)]^2} dx$$ 
Let's write 
$$y _{1}(x) = 1+x+\frac{ 1 }{ 4} x^2 + \frac{ 1 }{ 36} x^3 + ... = 1 + z$$ 
So
$$y _{2}(x) = y _{1}(x) \int\frac{ \frac{1}{x}} { (1+z)^2} dx$$ 
Since we want our solution in the neighborhood of x = 0, so z < 1
and we can apply binomial expansion for 
$$(1+z)^{-2} $$ which is $$(1+z)^{-2}  = 1 -2z+3z^2 -4z^3+...$$
$$y _{2}(x) = y _{1}(x) \int\frac{1}{x} (1 -2z+3z^2 -4z^3+... ) dx$$ 
Putting$$ z = x+\frac{ 1 }{ 4} x^2 + \frac{ 1 }{ 36} x^3 $$
$$y _{2}(x) = y _{1}(x) \int\frac{1}{x} (1-2x + \frac{5}{2}x^2 -.... ) dx$$ 
$$y _{2}(x) = y _{1}(x) (\ln x -2x +\frac{5}{4}x^2-.... )$$ 
Distributing y1(x) and expanding it for the terms except first, we get
$$y _{2}(x) = y _{1}(x) \ln x -2x -\frac{3}{4}x^2-.... $$ 

Hope this helps.

anonymous · Answer

@ShailKumar Alright, I managed to do that expansion and come up with the correct terms. Last question is how do I know how far to expand it? Since I had terms in a first solution up to x^3, I assume the binomial expansion I use is also expanded to x^3? And same if I had the first 4 terms, I would expand the binomial to 4 terms? 

My textbook doesn't mention this, it looks like a simple linear ODE when you make it like that. Let me see if I can find another double-root problem and see if that will still hold. I think I have 2 other methods apart from that, but pretty neat thing you came up with there :D

anonymous · Answer

"Last question is how do I know how far to expand it?"
That depends on the required precision of the solution. In the above method, if you want second solution correct up to x^n, you should take first solution up to x^n. 
In the solution above, I've taken first solution up to x^3 and got second solution up to x^2 because I didn't bother to calculate the coefficient of x^3 term. Had I taken 1st solution up to x^2, it would be enough to get 2nd solution correct up to x^2.

How long to do the binomial expansion ? That you have to see up to which term you get the term you want. 
So, for (1+x + x^2)^(-2), if you want expansion up to x^3, you see the lowest nonzero  power of x in  (1+x + x^2) is x^1. So to get x^3, you should expand up to  (x+x^2)^3 term.
Again, for $$(1+x^{3/2} + x^{5/2})^{-2}$$ if you want expansion up to x^3, you see the lowest nonzero  power of x in  $$(1+x^{3/2} + x^{5/2})$$ is x^(3/2). So to get x^3, you need to expand it up to $$(x^{3/2} + x^{5/2})^{2}$$
You don't need go for the next $$(x^{3/2} + x^{5/2})^{3}$$ term as this term contributes nothing to x^3.
I hope this will help.