Question

DE series @ganeshie8

Answer 1

If I'm given a differential equation and have to solve it via power series, would I then just use Airy's equation, \[y = \sum_{n=0}^{\infty} a_n(x-1)^n\] as my DE is \[y''+\frac{ 2 }{ (4-x^2) }y=0\]

Answer 2

Or wait I guess I can use \[\sum_{n=0}^{\infty} a_n x^n\] damn I'm dumb xD

Answer 3

That's like the assumption I should always be making it looks like haha

Answer 4

Then \[y = \sum_{n=0}^{\infty} a_n x^n\] \[y' = \sum_{n=1}^{\infty} a_{n+1} nx^{n-1}\]
\[y''= \sum_{n=2}^{\infty} a_nn(n-1)x^{n-2}\] ehh ehhh now plug into the equation

Answer 5

Oh \[x_0 = 0 \]

Answer 6

\[\sum_{n=2}^{\infty} a_n n(n-1)x^{n-2}+\frac{ 2 }{ (4-x^2) }\sum_{n=0}^{\infty} a_nx^n\] what is this

Answer 7

Ok shift index now here we go, this is where I'll make a mistake

Answer 8

i = n-2 guaranteed I'm calling it then n = i+2

Answer 9

dont reply im gonna see this hold onnn

Answer 10

noo dont

Answer 11

leave

Answer 12

yep just make the indeces match up

Answer 13

Oh yeah I think it's looking rather fancy, lets see if you agree

Answer 14

\[\sum_{n=0}^{\infty} a_{n+2}(n+2)(n+1)x^n+\frac{ 2 }{ (4-x^2) } \sum_{n=0}^{\infty} a_n x^n\]

Answer 15

err

Answer 16

I need a recurrence relationship

Answer 17

What to do with 2/(4-x^2)

Answer 18

oh I guess, \[\sum_{n=0}^{\infty} [a_{n+2}(n+2)(n+1)+a_n \frac{ 2 }{ (4-x^2) }]x^n=0\]

Answer 19

omggg you dummy

Answer 20

don't need to put it in standard form

Answer 21

!@%$!@$!@$

Answer 22

m totally blank i jst knw how to solve this kinda things-
y'-y=0
lol

Answer 23

I have to erase some stuff now lol, and it's ok

Answer 24

It was making no sense, because there shouldn't be any x's err ok let me see again

Answer 25

hey still working on this ?

Answer 26

Yup, there's a few parts to this problem, I see my mistake so I'm trying to fix that right now

Answer 27

\[(4-x^2)\sum_{n=2}^{\infty} a_n n(n-1)x^{n-2}+2 \sum_{n=0}^{\infty} a_n x^n=0\] this is what I have, I'm trying to find a recurrence relationship, so lets see what I get haha

Answer 28

Ok I think I have to split the first series in two because that -x^2 will mess things up

Answer 29

Oh perfect, I think that does work!

Answer 30

\[4\sum_{n=2}^{\infty}a_nn(n-1)x^{n-2}-\sum_{n=2}^{\infty} a_nn(n-1)x^{n-2+2} + 2 \sum_{n=0}^{\infty} a_n x^n=0\] ok lol

Answer 31

Now I can shift the index for those two series

Answer 32

I hope that looks right

Answer 33

Looks good to me...
you ust need to shift the index to first term

Answer 34

\[4\sum_{n=2}^{\infty}a_nn(n-1)x^{n-2}-\sum_{n=2}^{\infty} a_nn(n-1)x^{n-2+2} + 2 \sum_{n=0}^{\infty} a_n x^n=0\]

is same as

\[4\sum_{n=2}^{\infty}a_nn(n-1)x^{n-2}-\sum_{n=0}^{\infty} a_nn(n-1)x^{n-2+2} + 2 \sum_{n=0}^{\infty} a_n x^n=0\]

Answer 35

Notice that \[ \sum\limits_{n=0}^5 n(n-1) = \sum\limits_{n=2}^5 n(n-1) \]

Answer 36

Yeah I see, that's what I sort of noticed from shifting the index, you should actually get the same result for when you change the index as you would with the original series, just hard to tell sometimes without plugging in random numbers haha

Answer 37

Well I guess it's not very random...

Answer 38

But how did you know right away the first one was correct :O

Answer 39

idk the recurrence relation looks nasty..

Answer 40

lol ok one second let me see what I get

Answer 41

\[4(n+1)(n+2)a_{n+2}-[n(n+1)-2]a_n=0\]

Answer 42

do double check...

Answer 43

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Answer 44

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Answer 45

\[4a_{n+2}(n+2)(n+1)-a_nn(n-1)+2a_n=0\]

Answer 46

ya thats right

Answer 47

Yeah so -2 because when you factor out the a_n ok ok

Answer 48

ya

Answer 49

So this would be the recurrence relation given the point x0

Answer 50

yes just solve it and you're done ! ;)

Answer 51

Hey, so these are the solutions, and we don't have to use like variation of parameter to find the general solution, none of that characteristic equation business eh

Answer 52

It is a very simple linear equation, you could have solved it by that integration factor thingy...

but you wanted to solve it using power series..

Answer 53

Yeah haha ok, it just wanted me to find the recurrence relation now 4 terms...blah blah, then find general term in each solution

Answer 54

Hey quick question, remember the method of undetermined coeff, the guess is basically like using an integrating factor, I think that's why you can't have duplicate stuff, thought I'd let you know if you didn't already...or you can ignore it if it doesn't make sense

Answer 55

for lienar equations, integrating factor is a clever way to make use of reverse product rule...

what has that anythign to do with the guess for a particular solution in nonhomogeneous eqn ?

Answer 56

here is the solution generated by wolfram...
this time, its really a very smart solution!

Answer 57

I mean if you have something like y'+y=2e^-t, which then gives you \[y=2te^{-t}+ce^{-t}\] and this gives you a clue on your guess, just by looking at this solution

Answer 58

Oooh

Answer 59

got you :)

Answer 60

I am looking at the problem I might have found an alternate way to an answer but it's quite hairy in its own way lol

Answer 61

I don't think I've used integrating factor like this on a second order? That's pretty cool

Answer 62

hey me neither, i got confused earlier...

Answer 63

Looks like it has the same amount of work as series

Answer 64

yeah

Answer 65

So to find the first four terms, I have \[a_{n+2} = \frac{ a_n[n(n-1)-2] }{ 4(n+2)(n+1) }\] lol whyyyy

Answer 66

Tell me why this doesn't work cause it seems too simple to do it this way:

Start from the DE:
\[(x^2-4)y''-2y=0\]
substitute \(x^2-4=u^2\)
\[u^2y''-2y=0\]
Now we see that this is an Euler equation, \(y=u^n\) we solve for n,
\[n(n-1)u^n-2u^n=0\]
\[n^2-n-2=(n-2)(n+1)=0\]
so we have the general solution
\[y=c_1u^2+c_2u^{-1}\]
substitute in \[u=\sqrt{x^2-4}\]
\[y=c_1(x^2-4) + \frac{c_2}{ \sqrt{x^2-4}}\]

I guess I should check this now.

Answer 67

That's weird, because there's no y' term i think which sort of messes it up

Answer 68

So I don't think Euler's formula will be applicable for this one?

Answer 69

That doesn't matter, I'll show the steps:
\[u^2y''-2y=0\] is the same as
\[u^2y''+0uy'-2y=0\]
Take:
\[y=u^n\]\[y'=nu^{n-1}\]\[y''=n(n-1)u^{n-2}\]
Plug it in accordingly:
\[u^2(n(n-1)u^{n-2})-2u^n=0\]divide out the \(u^n\) term in common to get a quadratic
\[n^2-n-2=(n-2)(n+1)=0\]
so we have the general solution which are the roots
\[y=c_1u^2+c_2u^{-1}\]

Answer 70

wait so is y = u^n like your substitution where you then calculate y' and y'' in terms of dy/dx and d^2y/dx^2 idk what you're doing

Answer 71

qwerty quit logging out haha making me lag when you come as anonymous user

Answer 72

Well since I made the substitution that x is a function of u now, y is a function of u as wel so I'm taking the derivative with respect to u,

\[x(u)\]
\[y(x(u))\]

Well I guess we should just check that \[ y=c_1(x^2-4) + \frac{c_2}{ \sqrt{x^2-4}}\] is a solution to the differential equation before I talk anymore lol

Answer 73

It's strange cause the first part is a solution but the second part doesn't seem to be, I think there's something to do with the square root, perhaps I need to be doing like \(\pm\) in front of it? Interesting.

Answer 74

At any rate, once you have seen that the first one is a solution, we can try looking for another solution that's a multiple of the original solution by plugging in:

\[y=(x^2-4)f(x)\]
this reduces our differential equation (if i didn't mess up) to:

\[4xf'+2(x^2-4)f''=0\]
Now you might notice after you distribute this looks interesting:

\[4xf'+2x^2f''=8f''\]
That left hand side looks an awful lot like the product rule, so with a little playing around you can see that it's just going to be :
\[(2x^2f')'=4xf'+2x^2f''\] so we plug it in:

\[(2x^2f')'=8f''\] Now divide both sides by 2 and let's integrate!

\[x^2f' = 4f' +C\]
This is looking good.

\[f'(x^2-4)=C\]
\[f=\int \frac{C dx}{x^2-4}\]

Yay the rest from here should be pretty straight forward and it looks like it's gonna match. Whew.

Answer 75

Note: Anything that you see here that seems crazy is something I've seen someone else do before. Specifically when I try looking for another solution by simplifying the differential equation with this multiplication, this is a trick and it's sorta related to asymptotic analysis which is just a fancy way of saying you solve the differential equation after making simplifications like as x gets infinite or really small, we might look at:

\[y'' + \frac{2}{4-x^2}y=0\] so as x is large, that 4 in the denominator is less important, so we can solve the much easier equation:
\[y''-\frac{2}{x^2}y=0\] and then taking the solution to this differential equation, multiplying it by something like we did earlier to try to reduce the complexity of the differential equation.

Similarly we could have looked at small x where now the importan term is:

\[y''+\frac{1}{2}y=0\] and done it this way as well.

So that first one which is for large x ends up getting us an euler equation which is why I thought to use this trick while the second one ends up in a sinusoid that didn't take me anywhere.
---

The other trick I used was reversing the product rule weirdly. My partial differential equations prof did this all the time and after a while you get used to it. A pretty awesome trick I like to use is this one:

\[y'' = \frac{d y'}{dx} = \frac{dy'}{dy}\frac{dy}{dx} = \frac{dy'}{dy} y'\] which can make things separable.

Answer 76

Cool stuff man, thanks for sharing haha, I'll have to go over this again, as I'm sort of multitasking and doing my assignment