USing Series solutions to solve DE (please check my work, thanks)

[darryw]

Homework Statement

solve the DE using power series about x_0 = 0

y'' + xy' + 2y = 0

so, i want a solution in form of y = [sum from n=0 to n=infinity]c_n x^n

(i hope this makes sense, i don't know how to get summation symbol)

y = [sum from n=0 to n=infinity]c_n x^n

y' = [sum from n=0 to n=infinity]nc_n x^(n-1)

y'' = [sum from n=0 to n=infinity](n-1)n c_n x^(n-2)

then plug these values into DE..
i just need to know if i am correct so far..then i will proceed..thanks

Homework Equations

The Attempt at a Solution

 

[gabbagabbahey]

Homework Statement

solve the DE using power series about x_0 = 0

y'' + xy' + 2y = 0

so, i want a solution in form of y = [sum from n=0 to n=infinity]c_n x^n

(i hope this makes sense, i don't know how to get summation symbol)

Click on the $\LaTeX$ image below to see the code that generated it:

$$y(x)=\sum_{n=0}^{\infty} c_n x^n$$

And also take a look at my sig

y = [sum from n=0 to n=infinity]c_n x^n

y' = [sum from n=0 to n=infinity]nc_n x^(n-1)

y'' = [sum from n=0 to n=infinity](n-1)n c_n x^(n-2)

then plug these values into DE..
i just need to know if i am correct so far..then i will proceed..thanks

Looks fine so far...

 

[darryw]

I don't understand latex at all.
I enter the exact same code and it gives me two different versions of it..

heres my example.. i want it to read x to the power of (n-1). (please look at the code)

$$x^n-1$$ (1)

so then i change it so that there's a hat after minus sign and a hat after 1,and it gives me desired symbols.

$$x^n^-^1$$ (2)

but then when i copy and paste (2) it reverts back to (1) even though it has the code for (2)
$$x^n^-^1$$ (3)

please click on the code to see what I am talking about.

I totally don't understand the logic of this.. please help.. totally new to latex. thanksEDIT: OK so i noticed that as i submitted reply it modified the code. So is previewing latex unreliable or something? ??

So just to clarify, when i was previewing the message, (3) looked just like (1) even though it had the code for (2). That sounds kinda weird but that is what was happening.. why?

 

[gabbagabbahey]

Just put your entire exponent (or subscript) inside curly brackets...i.e. x^{n-1} comes up as $x^{n-1}$. Also, the images won't always appear properly on your screen when you preview/post them even if you've typed the code in correctly...just hit the refresh button once to see if it comes up properly.

 

[darryw]

OK I've been doing some experimenting and noticed that when i preview a latex image, for some reason it shows the previous latex image rather than the one I am trying to preview. Is this normal?
for example, when i initially tried to preview the image below, it previewed x to the power of (n-1). But when i clicked submit, it shows the correct image.
?

$$\sum_{n=0}^{\infty} c_n x^n^-^1$$

 

[darryw]

ok thanks gabbag, ill try refresh

 

[darryw]

$$y(x)=\sum_{n=0}^{\infty} c_n x^n$$

$$y'(x)=\sum_{n=1}^{\infty} nc_n x^{n-1}$$

$$y''(x)=\sum_{n=2}^{\infty} (n-1)nc_n x^{n-2}$$

but before i insert into DE, I want all the indices (is that right word?) to match (so they all go from n=0 to infinity) so i modifiy y' and y'' to get the following:

$$y'(x)=\sum_{n=1}^{\infty} nc_n x^{n-1}$$
=$$=\sum_{n=0}^{\infty} (n+1)c_{n+1} x^{n}$$

$$y''(x)=\sum_{n=2}^{\infty} (n-1)nc_n x^{n-2}$$
$$=\sum_{n=0}^{\infty} (n+1)(n+2)c_{n+2} x^{n}$$

then, inserting into y'' + xy' + 2y = 0, i get..

$$\sum_{n=0}^{\infty} (n+1)(n+2)c_{n+2} x^{n} + (x) \sum_{n=0}^{\infty} (n+1)c_{n+1} x^{n} + (2) \sum_{n=0}^{\infty} c_n x^n = 0$$

combining sums and factoring out the x^n...

$$\sum_{n=0}^{\infty} x^n [(n+1)(n+2)c_{n+2} + (x) (n+1)c_{n+1} + 2c_n] = 0$$

is this correct so far?? If so, i don't know what to do with the lone x still in the brackets. Thanks for any help

 

[darryw]

wow latex makes it look so much nicer.. and easier to read!

 

[gabbagabbahey]

Careful, if $n\to n+1$, then $c_n\to c_{n+1}$

$$y'(x)=\sum_{n=1}^{\infty} nc_n x^{n-1}=\sum_{n=0}^{\infty} (n+1)c_{n+1} x^{n}$$

And similar results apply to $y''(x)$

 

[darryw]

I edited my post #7 to reflect the changes to C_n
IS it OK now? thanks

 

[gabbagabbahey]

combining sums and factoring out the x^n...

$$\sum_{n=0}^{\infty} x^n [(n+1)(n+2)c_{n+2} + (x) (n+1)c_{n+1} + 2c_n] = 0$$

is this correct so far?? If so, i don't know what to do with the lone x still in the brackets. Thanks for any help

Looks good to me... now use the fact that $a_0+a_1x+a_2x^2+\ldots$ can only be zero for all $x$, if all the $a_n$ are zero.

 

[darryw]

thanks but could you elaborate a little bit on that? I think its the taylor expansion that is confusing me.. As I understand it, i thought everything in the brackets must = 0, so i would solve that equation and find what C_n is, and that is the recursive relation right?
if so, then how does that a_0 + a_1x etc fit into this?? thanks

 

[darryw]

Is this valid thing i can do to the middle term
$$(x) \sum_{n=0}^{\infty} (n+1)c_{n+1} x^{n} = \sum_{n=0}^{\infty} (n+1)c_{n+1} x^{n+1} = \sum_{n=0}^{\infty} (n)c_{n+1} x^{n}$$

?

that way i can get rid of the lone x.

 

[gabbagabbahey]

Is this valid thing i can do to the middle term
$$(x) \sum_{n=0}^{\infty} (n+1)c_{n+1} x^{n} = \sum_{n=0}^{\infty} (n+1)c_{n+1} x^{n+1} = \sum_{n=0}^{\infty} (n)c_{n+1} x^{n}$$

?

that way i can get rid of the lone x.

Your first step is correct, but

$$\sum_{n=0}^{\infty} (n+1)c_{n+1} x^{n+1}=\sum_{n=1}^{\infty} nc_{n} x^{n}\neq \sum_{n=0}^{\infty} (n)c_{n+1} x^{n}$$

 

[darryw]

you saw x at start of equality though, right?

 

[darryw]

please disregard last post...
I am confused about how the subscript of the constant relates to coefficient of constant and the powers of x.
thanks

 

[gabbagabbahey]

please disregard last post...
I am confused about how the subscript of the constant relates to coefficient of constant and the powers of x.
thanks

just plug in numbers for "n" (aka expand each sum)

$$\sum_{n=0}^{\infty} (n+1)c_{n+1} x^{n+1}=c_1x+2c_2x^2+3c_3x^3+\ldots$$

and

$$\sum_{n=1}^{\infty} nc_{n} x^{n}=c_1x+2c_2x^2+3c_3x^3+\ldots$$

so, $$\sum_{n=0}^{\infty} (n+1)c_{n+1} x^{n+1}=\sum_{n=1}^{\infty} nc_{n} x^{n}$$

Or am I misunderstanding your question?

 

[darryw]

OK so after the last correction I am left with:

$$\sum_{n=0}^{\infty} x^n [(n+1)(n+2)c_{n+2} + nc_n + 2c_n] = 0$$

so next i should solve this equation: $$(n+1)(n+2)c_{n+2} + nc_n + 2c_n = 0$$ for $$c_{n+2}$$, correct?
thanks

 

[darryw]

that was kinda my question, but its small thing actually i think i understand that pasrt now.
but am i proceeding correctly so far (up to post #18) ? that is, next i should solve for c_n+2 ?

 

[gabbagabbahey]

OK so after the last correction I am left with:

$$\sum_{n=0}^{\infty} x^n [(n+1)(n+2)c_{n+2} + nc_n + 2c_n] = 0$$

That's not quite correct. The summation in you middle term start at one, not zero. So, you should have:

$$\begin{aligned}0 &= \sum_{n=0}^{\infty} x^n [(n+1)(n+2)c_{n+2} + 2c_n]+\sum_{n=1}^{\infty}nc_nx^n \\ &= 2c_2+2c_0+\sum_{n=1}^{\infty} x^n [(n+1)(n+2)c_{n+2}+nc_n + 2c_n] \end{alinged}$$

 

[darryw]

OK I see what you were saying s few posts ago then..
but given that one of the terms starts at n=1, wouldn't it have been better to just leave the middle term in form $$\sum_{n=0}^{\infty} (n+1)c_{n+1} x^{n+1}$$

i mean, if i change it to n=1, then that means i have to change the other 2 terms to match, right?

 

[gabbagabbahey]

OK I see what you were saying s few posts ago then..
but given that one of the terms starts at n=1, wouldn't it have been better to just leave the middle term in form $$\sum_{n=0}^{\infty} (n+1)c_{n+1} x^{n+1}$$

But then your sum would have terms with both $x^n$ and $x^{n+1}$ in it and you wouldn't simply be able to compare coefficients.

i mean, if i change it to n=1, then that means i have to change the other 2 terms to match, right?

Not necessarily, you can just take out the n=0 term:

$$\sum_{n=0}^{\infty}a_nx^n=a_0+\sum_{n=1}^{\infty}a_nx^n$$

Which is exactly what I did in my last post (that's where the $2c_0+2c_2$ came from).

 

[darryw]

do u think there is there a mistake on this page of lecture? he never changes it to n=1.
(no big deal if u don't want to look at this) thanks

 

[gabbagabbahey]

Yeah, there's definitely an error in that lecture.

 

[darryw]

OK so $$[(n+1)(n+2)c_{n+2}+nc_n + 2c_n] = 0$$
and solve for c_n+2, correct?
if that is right, why solve for c_n+2? (why not c_n or c_n+1 for that matter?)

 

[darryw]

thanks for looking at the notes

 

[gabbagabbahey]

OK so $$[(n+1)(n+2)c_{n+2}+nc_n + 2c_n] = 0$$
and solve for c_n+2, correct?

Right. This is your recursion relationship. It tells you that $c_2=-c_0$ and $c_3=-\frac{1}{2}c_1$, $c_4=-\frac{1}{3}c_2=\frac{1}{3}c_0$ and so on...

Also, after looking again, there is no harm in saying that [tex]\sum_{n=1}^{\infty}nc_nx^n=\sum_{n=0}^{\infty}nc_nx^n[/itex], since the n=0 term is zero anyways. Your lecture is actually fine.

 

[darryw]

Im going to come back to this in couple hours.. Thanks a lot for all the excellent help gabbag..

 

[darryw]

Can someone please clarify something for me?
i found out that c_n+2 = -c_n / (n+1)
Is this called the "recursion relation"? also, i got a bunch of values for c_2 and c_3 and c_4 etc, by just plugging in values for n.

but how do these values equate to y = c_0 (1-x^2 + x^4/3 - x^6/5*3 + ...)

i mean what are the steps to put in in this form??
thanks for any help

 

[gabbagabbahey]

Well, you can express all of the odd coefficients in terms of $c_1$ and all the even coefficients in terms of $c_0$...so you can split your sum into two (even and odd terms) and factor out $c_0$ from the evn sum and $c_1$ from the odd one...