# Solving a second order ode

Gold Member
Homework Statement:
##y"+y'\frac {1}{z}+y[\frac {z^2-n^2}{z^2}]=0##
Relevant Equations:
power series
let ##y= \sum_{k=-∞}^\infty a_kz^{k+c}##
##y'=\sum_{k=-∞}^\infty (k+c)a_kz^{k+c-1}##
##y"=\sum_{k=-∞}^\infty (k+c)(k+c-1)a_kz^{k+c-2}##
therefore,
##y"+y'\frac {1}{z}+y[\frac {z^2-n^2}{z^2}]=0## =##[\sum_{k=-∞}^\infty [(k+c)^2-n^2)]a_k + a_k-2]z^{k+c} ##
it follows that,
##(k+c)^2-n^2)]a_k + a_k-2=0##
if ##k=0, →c^2-n^2=0, a_0≠0##
##c^2=n^2, →c=±n ##(roots)
now considering case 1, if ##c=n##, and given k=1, then ##a_1=0##...this is the part that i am not getting.
i am ok with the rest of the steps...

Homework Helper
therefore,
$$y"+y'\frac {1}{z}+y\Biggl [\frac {z^2-n^2}{z^2}\Biggr ]=0$$ $$\Leftrightarrow \ (?) \\ \sum_{k=-∞}^\infty \Biggl [ (k+c)^2-n^2) a_k + a_k-2\Biggr ] z^{k+c} = 0 \\ \mathstrut \\ \mathstrut \\$$

Doesn't seem right. I suppose you mean ##a_{k-2}##, not ##a_k - 2##. But: where is ##a_{k-1}## ?

 can't get the scroll bar in the quote ! How come ?

Gold Member
yeah, it is supposed to be ##a_{k-2}##, i do not have ##a_{k-1} ## in my working, unless of course you want me to write all my steps...

Homework Helper
do not have ##\ a_{k-1}\ ## in my working
I noticed that and I wondered what made the ##\ y'\over z\ ## series disappear ?

chwala
Homework Helper
By the way, shouldn't it be ##\ a_{k+2}\ ## instead of ##\ a_{k-2}\ ## ?

Homework Helper
And never mind my blabbing about ##\ y'\over z\ ##, it actually seems to fall out.

Homework Helper
I suppose another : Never mind my blabbing about ##a_{k+2}## either. I should retreat and only speak up after some sensible thinking .
Sorry, Chwala !

By the way, I wonder why you sum starting from ##\ -\infty\ ##. Most analyses start at 0 !

chwala
Homework Helper
Took me a while to recognize your equation . Do you know what I mean ?

chwala
Gold Member
Bvu sorry i was in a zoom meeting. I just used the summing formula from my university notes, ...i have ##a_{k-2}## and not ##a_{k+2}##

Gold Member
And never mind my blabbing about ##\ y'\over z\ ##, it actually seems to fall out.
no worries, we are a community i really appreciate your input...

PhDeezNutz and Delta2
Gold Member
The best way to go about this is to multiply the equation by $z^2$ to eliminate any division first. The way you went about it is convoluted and not posting the intermediate steps isn't helping.

Delta2
Gold Member
I can solve the problem to the end...the only area that i do not understand is where i indicated, case 1, ...##k=1, a_1=0## this is the only area that i need help...how is this valid? (the reference that i am using is my undergraduate university notes, that i am trying to refresh on...

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can we say, from ##[(k+c)^2-n^2)]a_k+a_{k-2}##=0, then if ## c=n## & ##k=1## then,
##(1+2n)a_1+a_{-1}=0##
##a_1+2na_1+a_{-1}=0##
is this correct thinking...

Gold Member
The best way to go about this is to multiply the equation by $z^2$ to eliminate any division first. The way you went about it is convoluted and not posting the intermediate steps isn't helping.

and can we solve this problem using laplace?

Gold Member
Series solution is still a valid way to go about it, but I as taught to not be dividing when using a series solution. It looks odd to me.

BvU
Homework Helper
Took me a while to recognize your equation . Do you know what I mean ?

the reference that i am using is my undergraduate university notes
I overlooked that one. Does it say there that you are usibg the Frobenius method to solve the Bessel equation ? (other link , link, solution with pictures, a whole chapter, ...).

And you should really start your power series from ##k=0##, not from ##k=-\infty##.

All this with just a little googling and without sitting down and seriously working out your core question ! Lazy me ...

My own notes are still somewhere in the attic and also in my brain, but under four and a half decades of dust

chwala and PhDeezNutz
Staff Emeritus
Homework Helper
can we say, from ##[(k+c)^2-n^2)]a_k+a_{k-2}##=0, then if ## c=n## & ##k=1## then,
##(1+2n)a_1+a_{-1}=0##
##a_1+2na_1+a_{-1}=0##
is this correct thinking...
The problem is related to what the others have mentioned already a few times. The sum starts from ##k=0##, so the coefficient of the ##k=1## term is ##(1+2n)a_1##. There's no ##a_{-1}##.

Homework Helper
If ##a_k## is only defined for ##k=0,1,2,\ldots## don't you think that for the equation ##\left[ (k+c)^2-n^2\right] a_k+a_{k-2}=0## the only admissible values of ##k## are ##k=2,3,4,\ldots## ? For ##k=2,\text{ and }c=n## this would lead to ##a_2=\tfrac{a_0}{4(n+1)}## and [omitting steps] in general

$$a_{2k}=\tfrac{(-1)^k a_0}{4^k(k!)^2 \binom{n+k}{k} }, \quad k\in\mathbb{Z}^+$$
and
$$a_{2k+1}=\tfrac{(-1)^k a_{1}}{\prod_{j=1}^{k}\left[ (2j+1) (2n+2j+1)\right]}, \quad k\in\mathbb{Z}^+$$

Note: ##a_{2k+1}## is "not pretty" in terms of binomial coefficients, but probably looks ok in terms of double factorials. Just need initial conditions now?

Edit: I messed up the formula for ##a_{2k}## first time around, fixed it!

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chwala
Gold Member
I overlooked that one. Does it say there that you are usibg the Frobenius method to solve the Bessel equation ? (other link , link, solution with pictures, a whole chapter, ...).

And you should really start your power series from ##k=0##, not from ##k=-\infty##.

All this with just a little googling and without sitting down and seriously working out your core question ! Lazy me ...

My own notes are still somewhere in the attic and also in my brain, but under four and a half decades of dust

i guess the lecturer must have mentioned the method but i probably did not indicate the method on my working...

Gold Member
If ##a_k## is only defined for ##k=0,1,2,\ldots## don't you think that for the equation ##\left[ (k+c)^2-n^2\right] a_k+a_{k-2}=0## the only admissible values of ##k## are ##k=2,3,4,\ldots## ? For ##k=2,\text{ and }c=n## this would lead to ##a_2=\tfrac{a_0}{4(n+1)}## and [omitting steps] in general

$$a_{2k}=\tfrac{(-1)^k a_0}{4^k(k!)^2 \binom{n+k}{k} }, \quad k\in\mathbb{Z}^+$$
and
$$a_{2k+1}=\tfrac{(-1)^k a_{1}}{\prod_{j=1}^{k}\left[ (2j+1) (2n+2j+1)\right]}, \quad k\in\mathbb{Z}^+$$

Note: ##a_{2k+1}## is "not pretty" in terms of binomial coefficients, but probably looks ok in terms of double factorials. Just need initial conditions now?

Edit: I messed up the formula for ##a_{2k}## first time around, fixed it!
i am quite fine with the values ##k=2,3,4##... and how to find the solution. At the end we shall have a series solution of form ##y=y_1 + y_2##, alternating for ##k=2,4,6##...and ##k=3,5,7##...my interest on this question is on the value of ##k=1##, i do not understand how ##a_1=0## but of course i may be missing something from the question as indicated by the responses...

Gold Member
The problem is related to what the others have mentioned already a few times. The sum starts from ##k=0##, so the coefficient of the ##k=1## term is ##(1+2n)a_1##. There's no ##a_{-1}##.
I think i may be wrong with my limits, ##k## should start from ##0## and not ##-∞##, as indicated...

Gold Member
...now considering case 1, if ##c=n##, and given ##k=1##, then ##a_1=0##...this is the part that i am not getting.
i am ok with the rest of the steps...

This is the part that i need understanding ...lets assume as you have put it that ##k## starts from ##0→+∞## then how is it that when ##k=1##, that ##a_1=0##
or the other possibility is that my statement (my notes) is/are incorrect.

Homework Helper
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...now considering case 1, if ##c=n##, and given ##k=1##, then ##a_1=0##...this is the part that i am not getting.
i am ok with the rest of the steps...

This is the part that i need understanding ...lets assume as you have put it that ##k## starts from ##0→+∞## then how is it that when ##k=1##, that ##a_1=0##
or the other possibility is that my statement (my notes) is/are incorrect.
Not sure that introducing c helps. Why not just write ##y=\Sigma_0a_kz^k## and allow that a0 and so could be zero?
Then you can write equations for the z0 term, the z1 term, and a general equation for zk, k>1.
The cases to be considered are then n=0, n=1, n=2...; a particular value of k is not a 'case'.
You should find that ak=0 for k<n, an is unconstrained, and for k>n ak is some (varying) multiple of ak-2

Gold Member

Gold Member
i think going with the above attached example, i may have an idea as to why ##a_1=0##, by using the indicial equations (14) and (15) as my comparison,
in my case, i have the indicial equation as,
##[(k+c)^2-n^2)]a_k + a_{k-2}=0## for some reason they chose to use the first part of the indicial equation...
when ##k=0## it follows that ##c=±n##, we may as well choose to ignore the ##-c## and go for ##c=n##
now for ##k=1##, we shall have;
##[(k+c)^2-n^2)]a_k=0## just like in equation(16) of attachment... and ##c=n## therefore,
##[(1+n)^2-n^2)]a_1=0##
##(1+2n)a_1=0## avoiding special case ##n=-0.5##, results into ##a_1=0##

Gold Member
I overlooked that one. Does it say there that you are usibg the Frobenius method to solve the Bessel equation ? (other link , link, solution with pictures, a whole chapter, ...).

And you should really start your power series from ##k=0##, not from ##k=-\infty##.

All this with just a little googling and without sitting down and seriously working out your core question ! Lazy me ...

My own notes are still somewhere in the attic and also in my brain, but under four and a half decades of dust
thats a nice one, you hit me under the belt lol...i need to upgrade my memory lol

Homework Helper
Note: Sorry it took so long, been working on this post off and on all day. I just answered your earlier question about the solution to the recurrence relation, and I'm gonnna let you try to figure the rest out...

Given ##y"+\frac {1}{z} y' +\left(\tfrac{z^2-n^2}{z^2}\right) y=0##

Upon setting ##y = \sum_{k=0}^\infty a_{k}z^{k+c}##, we have:

$$\begin{gathered} \left(\tfrac{z^2-n^2}{z^2}\right) y =\left(\tfrac{z^2-n^2}{z^2}\right) \sum_{k=0}^\infty a_{k}z^{k+c} \\ = \sum_{k=0}^\infty a_{k}z^{k+c-2}(z^2 - n^2) \\ \end{gathered}$$

$$\frac {1}{z} y' = \tfrac{1}{z}\sum_{k=0}^\infty (k+c)a_{k}z^{k+c-1} = \sum_{k=0}^\infty (k+c)a_{k}z^{k+c-2}$$

$$y"=\sum_{k=0}^\infty (k+c)(k+c-1)a_kz^{k+c-2}$$

By the given DE, the sum of these three terms vanishes, to wit

$$\begin{gathered} y"+\tfrac{1}{z} y'+\left( \tfrac{z^2-n^2}{z^2}\right) y = 0 \\ \Rightarrow \sum_{k=0}^\infty [ (k+c)(k+c-1) + (k+c) - n^2 ]a_{k}z^{k+c-2}+ \sum_{k=0}^\infty a_{k}z^{k+c} =0 \\ \Rightarrow [c^2-n^2]a_{0}z^{c-2} +[(c+1)^2 -n^2]a_{1}z^{c-1} + \sum_{k=2}^\infty [ (k+c)^2 - n^2 ]a_{k}z^{k+c}+ \sum_{k=0}^\infty a_{k}z^{k+c} =0 \\ \Rightarrow [c^2-n^2]a_{0}z^{c-2} +[(c+1)^2 -n^2]a_{1}z^{c-1} + \sum_{k=0}^\infty \left\{ [ (k+c+2)^2-n^2)]a_{k+2} + a_{k}\right\} z^{k+c} = 0 \\ \end{gathered}$$

All the coefficients of powers of ##z## must vanish, and ##a_0\neq 0\Rightarrow c^2-n^2=0 \Rightarrow c = \pm n ##. Plugging these roots into the next coefficient gives, ##[(\pm n+1) ^2 -n^2]a_1=0\Rightarrow (1\pm 2n)a_1=0## but ##n\in\mathbb{Z}## so ##a_1=0##. Next plug the roots into the ##\text{k}^{th}## coefficient:

$$[ (k\pm n+2)^2-n^2)]a_{k+2} + a_{k}=0\Rightarrow (k\pm n+2+n)(k\pm n+2-n)a_{k+2} = -a_{k}\Rightarrow a_{k+2}=\tfrac{-a_k}{(k+2)(k\pm 2n+2)}$$

From ##a_0\neq 0## and ##a_1=0## this recurrence simplifies to

$$a_{2k+2}=\tfrac{-a_{2k}}{(2k+2)(2k\pm 2n+2)}=\tfrac{-a_{2k}}{4(k+1)(k\pm n+1)}\wedge a_{2k+1}=0\forall k\in\mathbb{N}$$

Let's determine the pattern for ##a_{2k+2}## for small values of ##k## and then try to guess the solution to the recurrence:

$$\boxed{k=0}:\quad a_2=\tfrac{-a_{0}}{4\cdot 1(\pm n+1)}$$

$$\boxed{k=1}:\quad a_4=\tfrac{-a_{2}}{4\cdot 2(\pm n+2)}=\tfrac{(-1)^2 a_{0}}{4^2\cdot 2\cdot 1(\pm n+2)(\pm n+1)}$$

$$\boxed{k=2}:\quad a_6=\tfrac{-a_{4}}{4\cdot 3(\pm n+3)}=\tfrac{(-1)^3 a_{0}}{4^3\cdot 3\cdot 2\cdot 1(\pm n+3)(\pm n+2)(\pm n+1)}$$

My guess:

$$a_{2k+2}=\tfrac{(-1)^{k+1} a_{0}}{4^{k+1}(k+1)! \prod_{j=1}^{k+1}(\pm n+j)}$$

Whelp, my post has got you to this point

in the Wikipedia article, I ad-libbed bit along the way to take the easier route, note we've already plugged in the roots, but they differ by an integer so you'll need to read on past that point to see how to derive the other linearly independent solution (after you determine which of the roots solves whichever equation (think it was a DE. Good luck!

chwala
Gold Member
Note: Sorry it took so long, been working on this post off and on all day. I just answered your earlier question about the solution to the recurrence relation, and I'm gonnna let you try to figure the rest out...

Given ##y"+\frac {1}{z} y' +\left(\tfrac{z^2-n^2}{z^2}\right) y=0##

Upon setting ##y = \sum_{k=0}^\infty a_{k}z^{k+c}##, we have:

$$\begin{gathered} \left(\tfrac{z^2-n^2}{z^2}\right) y =\left(\tfrac{z^2-n^2}{z^2}\right) \sum_{k=0}^\infty a_{k}z^{k+c} \\ = \sum_{k=0}^\infty a_{k}z^{k+c-2}(z^2 - n^2) \\ \end{gathered}$$

$$\frac {1}{z} y' = \tfrac{1}{z}\sum_{k=0}^\infty (k+c)a_{k}z^{k+c-1} = \sum_{k=0}^\infty (k+c)a_{k}z^{k+c-2}$$

$$y"=\sum_{k=0}^\infty (k+c)(k+c-1)a_kz^{k+c-2}$$

By the given DE, the sum of these three terms vanishes, to wit

$$\begin{gathered} y"+\tfrac{1}{z} y'+\left( \tfrac{z^2-n^2}{z^2}\right) y = 0 \\ \Rightarrow \sum_{k=0}^\infty [ (k+c)(k+c-1) + (k+c) - n^2 ]a_{k}z^{k+c-2}+ \sum_{k=0}^\infty a_{k}z^{k+c} =0 \\ \Rightarrow [c^2-n^2]a_{0}z^{c-2} +[(c+1)^2 -n^2]a_{1}z^{c-1} + \sum_{k=2}^\infty [ (k+c)^2 - n^2 ]a_{k}z^{k+c}+ \sum_{k=0}^\infty a_{k}z^{k+c} =0 \\ \Rightarrow [c^2-n^2]a_{0}z^{c-2} +[(c+1)^2 -n^2]a_{1}z^{c-1} + \sum_{k=0}^\infty \left\{ [ (k+c+2)^2-n^2)]a_{k+2} + a_{k}\right\} z^{k+c} = 0 \\ \end{gathered}$$

All the coefficients of powers of ##z## must vanish, and ##a_0\neq 0\Rightarrow c^2-n^2=0 \Rightarrow c = \pm n ##. Plugging these roots into the next coefficient gives, ##[(\pm n+1) ^2 -n^2]a_1=0\Rightarrow (1\pm 2n)a_1=0## but ##n\in\mathbb{Z}## so ##a_1=0##. Next plug the roots into the ##\text{k}^{th}## coefficient:

$$[ (k\pm n+2)^2-n^2)]a_{k+2} + a_{k}=0\Rightarrow (k\pm n+2+n)(k\pm n+2-n)a_{k+2} = -a_{k}\Rightarrow a_{k+2}=\tfrac{-a_k}{(k+2)(k\pm 2n+2)}$$

From ##a_0\neq 0## and ##a_1=0## this recurrence simplifies to

$$a_{2k+2}=\tfrac{-a_{2k}}{(2k+2)(2k\pm 2n+2)}=\tfrac{-a_{2k}}{4(k+1)(k\pm n+1)}\wedge a_{2k+1}=0\forall k\in\mathbb{N}$$

Let's determine the pattern for ##a_{2k+2}## for small values of ##k## and then try to guess the solution to the recurrence:

$$\boxed{k=0}:\quad a_2=\tfrac{-a_{0}}{4\cdot 1(\pm n+1)}$$

$$\boxed{k=1}:\quad a_4=\tfrac{-a_{2}}{4\cdot 2(\pm n+2)}=\tfrac{(-1)^2 a_{0}}{4^2\cdot 2\cdot 1(\pm n+2)(\pm n+1)}$$

$$\boxed{k=2}:\quad a_6=\tfrac{-a_{4}}{4\cdot 3(\pm n+3)}=\tfrac{(-1)^3 a_{0}}{4^3\cdot 3\cdot 2\cdot 1(\pm n+3)(\pm n+2)(\pm n+1)}$$

My guess:

$$a_{2k+2}=\tfrac{(-1)^{k+1} a_{0}}{4^{k+1}(k+1)! \prod_{j=1}^{k+1}(\pm n+j)}$$

Whelp, my post has got you to this point

View attachment 262799

in the Wikipedia article, I ad-libbed bit along the way to take the easier route, note we've already plugged in the roots, but they differ by an integer so you'll need to read on past that point to see how to derive the other linearly independent solution (after you determine which of the roots solves whichever equation (think it was a DE. Good luck!

Thanks for your insight, i know how to work on the rest of the steps leading to the solution...i keep on repeating this statement. My interest was only on how ##a_1=0##, ...only this part...my interest on this question is only and only on that part of the question...

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Homework Helper
All the coefficients of powers of ##z## must vanish, and ##a_0\neq 0\Rightarrow c^2-n^2=0 \Rightarrow c = \pm n ##. Plugging these roots into the next coefficient gives, ##[(\pm n+1) ^2 -n^2]a_1=0\Rightarrow (1\pm 2n)a_1=0## but ##n\in\mathbb{Z}## so ##a_1=0##.

Like it says, because ##1\pm 2n\neq 0\forall n\in\mathbb{N}## hence we must have ##a_1=0## because there is no other way that the coefficient of ##z^{c-1}## can be zero.

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yeah yeah and i was able to figure that out in post 25. Thank you sir.

Homework Helper
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This is the part that i need understanding ...lets assume as you have put it that ##k## starts from ##0→+∞## then how is it that when ##k=1##, that ##a_1=0##
or the other possibility is that my statement (my notes) is/are incorrect.
I still do not understand your question here, and it looks to me that no one else has understood it either, which is why you are not getting a satisfactory answer.
What do you mean "when k=1"? As I wrote in post #23, that is not a 'case'.
I also suggested dropping c, so let's do that. You have ##z^2y''+zy'+y(z^2-n^2)=0## and ##y=\Sigma_{k=0}a_kz^k##.

##z^0: a_0n^2=0##
##z^1: a_1(1-n^2)=0##
##z^k, k>1: a_k(k^2-n^2)+a_{k-2}=0##

From this we get to deduce:
##a_n## can be chosen arbitrarily.
##a_k=0## if either k<n or k+n is odd.
##a_{n+2}=-\frac{a_n}{4n+4}##
Etc.

chwala
Gold Member
thanks for your asking haruspex, i asked in post 1 on how ##a_1=0##, that was my question, ...and i now know why as shown in my post 25 and also more insight has been given by benorin in post 28. My question has been answered fully sir.

Gold Member
I still do not understand your question here, and it looks to me that no one else has understood it either, which is why you are not getting a satisfactory answer.
What do you mean "when k=1"? As I wrote in post #23, that is not a 'case'.
I also suggested dropping c, so let's do that. You have ##z^2y''+zy'+y(z^2-n^2)=0## and ##y=\Sigma_{k=0}a_kz^k##.

##z^0: a_0n^2=0##
##z^1: a_1(1-n^2)=0##
##z^k, k>1: a_k(k^2-n^2)+a_{k-2}=0##

From this we get to deduce:
##a_n## can be chosen arbitrarily.
##a_k=0## if either k<n or k+n is odd.
##a_{n+2}=-\frac{a_n}{4n+4}##
Etc.