Understanding the Frobenius Method for Solving Second Order ODEs

In summary: The initial conditions for this recurrence relation are ##a_0, a_1##. If you want to find ##a_1##, you need to know ##a_{-1}##. But if ##a_k## is only defined for ##k=0,1,2,\ldots##, you can't use ##k=-1## in the recurrence relation, so you're stuck. This is why we don't care about the ##k=1## term; it's not something we
  • #1
chwala
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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...
 
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  • #2
chwala said:
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}## ?[edit] can't get the scroll bar in the quote ! How come ?
 
  • #3
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...
 
  • #4
chwala said:
do not have ##\ a_{k-1}\ ## in my working
I noticed that and I wondered what made the ##\ y'\over z\ ## series disappear :rolleyes: ?
 
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  • #5
By the way, shouldn't it be ##\ a_{k+2}\ ## instead of ##\ a_{k-2}\ ## ?
 
  • #6
And never mind my blabbing about ##\ y'\over z\ ##, it actually seems to fall out.
 
  • #7
:smile: I suppose another o:) : Never mind my blabbing about ##a_{k+2}## either. I should retreat and only speak up after some sensible thinking :nb).
Sorry, Chwala !

By the way, I wonder why you sum starting from ##\ -\infty\ ##. Most analyses start at 0 !
 
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  • #8
Took me a while to recognize your equation :cool: . Do you know what I mean ?
 
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  • #9
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}##
 
  • #10
BvU said:
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...
 
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  • #11
The best way to go about this is to multiply the equation by [itex] z^2[/itex] to eliminate any division first. The way you went about it is convoluted and not posting the intermediate steps isn't helping.
 
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  • #12
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...
 
  • #13
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...
 
  • #14
Dr Transport said:
The best way to go about this is to multiply the equation by [itex] z^2[/itex] 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?
 
  • #15
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.
 
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  • #16
BvU said:
Took me a while to recognize your equation :cool: . Do you know what I mean ?

chwala said:
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 ... :wink:

My own notes are still somewhere in the attic and also in my brain, but under four and a half decades of dust :biggrin:
 
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  • #17
chwala said:
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}##.
 
  • #18
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!
 
Last edited:
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  • #19
BvU said:
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 ... :wink:

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

i guess the lecturer must have mentioned the method but i probably did not indicate the method on my working...
 
  • #20
benorin said:
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...
 
  • #21
vela said:
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...
 
  • #22
...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.
 
  • #23
chwala said:
...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
 
  • #24
1589374007299.png
 
  • #25
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##
 
  • #26
BvU said:
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 ... :wink:

My own notes are still somewhere in the attic and also in my brain, but under four and a half decades of dust :biggrin:
thats a nice one:biggrin::biggrin::biggrin:, you hit me under the belt lol...i need to upgrade my memory lol
 
  • #27
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

FrobeniusMethodPFCalcHW.png


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!
 
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  • #28
benorin said:
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...
 
Last edited:
  • #29
benorin said:
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.
 
  • #30
yeah yeah and i was able to figure that out in post 25. Thank you sir.
 
  • #31
chwala said:
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.

If I have not answered your question, please rephrase it in terms of that analysis.
 
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  • #32
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.
 
  • #33
haruspex said:
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.

If I have not answered your question, please rephrase it in terms of that analysis.
Thanks haruspex, this is more clearer
 
  • #34
@haruspex the problem stated to use the Frobenius method which is where the ##c## comes from.
 
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Related to Understanding the Frobenius Method for Solving Second Order ODEs

1. What is the Frobenius method for solving second order ODEs?

The Frobenius method is a technique used to solve second order ordinary differential equations (ODEs) with variable coefficients. It involves assuming a solution in the form of a power series and then substituting it into the ODE to find the coefficients.

2. When is the Frobenius method used?

The Frobenius method is typically used when the coefficients in the ODE are not constants and cannot be solved using other methods, such as separation of variables or substitution. It is also useful for finding solutions near singular points.

3. How does the Frobenius method work?

The Frobenius method involves assuming a solution in the form of a power series and then substituting it into the ODE. This results in a recurrence relation for the coefficients of the power series, which can be solved to find the values of the coefficients. The solution is then written as a linear combination of the power series terms.

4. What are the advantages of using the Frobenius method?

The Frobenius method allows for the solution of second order ODEs with variable coefficients, which cannot be solved using other methods. It also provides a systematic approach to finding solutions near singular points, which is not possible with other techniques.

5. Are there any limitations to the Frobenius method?

One limitation of the Frobenius method is that it may not always result in a solution that is valid for all values of the independent variable. In some cases, the series may only converge for a limited range of values. Additionally, the method can become more complicated and time-consuming for higher order ODEs.

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