Series solution near an ordinary point

[mugzieee]

Im given y"-xy'-y=0 at x0=1.

The problem asks for recurrene relation, and the first four terms in each of two linearly independant solutions, and the general term in each solution.

Whats thrwoing me off is the x0=1. I tried doing y= SUM an(x-1)^n, but when i differenetiate and plug in, i get stuck with the xy' part.
I also tried doing y=SUM an+1x^n, and it also doesn't work. I tried many other paths also, but none of them seem to work...

 

[Pyrrhus]

You should have
$$\sum_{n=0}^{\infty} (n+2) (n+1) a_{n+2} (x-1)^{n} - \sum_{n=0}^{\infty} (n+1) a_{n+1} (x-1)^{n} - \sum_{n=0}^{\infty} n a_{n} (x-1)^{n} - \sum_{n=0}^{\infty} a_{n} (x-1)^{n} = 0$$
where the term $\sum_{n=0}^{\infty} n a_{n} (x-1)^{n}$ comes from
$$- (x-1)y' = \sum_{n=1}^{\infty} (n+1) a_{n+1} (x-1)^{n+1} = \sum_{n=0}^{\infty} (n) a_{n} (x-1)^{n}$$

 

[mugzieee]

i don't understand how you got the -(x-1)y'. does the term xy' change to (x-1)y' when the x in the y=SUM an(x-1)^n changes also? I am sorry, but is there a way you can show you got the 2nd and 3rd summation in your solution in more detail please?

 

[mugzieee]

oh wait nvm my last post, i got it, thank you.

 

[mugzieee]

we know that
y'= $$\sum_{n=0}^{\infty} n a_{n} (x-1)^{n-1}$$
y"= $$\sum_{n=0}^{\infty} n (n-1) a_{n} (x-1)^{n-2}$$
when we compute xy', doesn't it become just:
$$\sum_{n=0}^{\infty} n a_{n} (x-1)^{n}$$ ?
i don't see how $$\sum_{n=0}^{\infty} n a_{n} (x-1)^{n}$$ and
- $$\sum_{n=0}^{\infty} (n+1) a_{n+1} (x-1)^{n}$$ were obtained

 

[HallsofIvy]

Another way to do that problem, if it is the x₀= 1 that is throwing you off, is to change variable. Let v= x- 1 so x= v+1. Then $$\frac{dy}{dv}= \frac{dy}{dx}$$ and [tex]\frac{d^2y}{dv^2}= \frac{d^2y}{dx^2} so we can use y" and y' to mean derivatives with respect v also. In terms of the v variable, we have y" (v+1)y'- y= y"- vy'- y'- y= 0 and
v₀= 0.

 

[Pyrrhus]

I simply set x = 1 + (x-1), so you see on the series form above

$$y'' - y'(1 + (x-1)) - y = 0$$

$$y'' - y' - (x-1)y' - y = 0$$

 

[mugzieee]

why was it necessary for you to set x= 1+(x-1)

 

[Pyrrhus]

Look at this

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

right?

so

$$xy'= x \sum_{n=0}^{\infty} (n+1) a_{n+1} (x-1)^{n}$$

You see a problem with that?

 

[mugzieee]

the polynomial is not going to look like the other ones?

 

[mbbjln]

You should have
$$\sum_{n=0}^{\infty} (n+2) (n+1) a_{n+2} (x-1)^{n} - \sum_{n=0}^{\infty} (n+1) a_{n+1} (x-1)^{n} - \sum_{n=0}^{\infty} n a_{n} (x-1)^{n} - \sum_{n=0}^{\infty} a_{n} (x-1)^{n} = 0$$
where the term $\sum_{n=0}^{\infty} n a_{n} (x-1)^{n}$ comes from
$$- (x-1)y' = \sum_{n=1}^{\infty} (n+1) a_{n+1} (x-1)^{n+1} = \sum_{n=0}^{\infty} (n) a_{n} (x-1)^{n}$$

i have this same problem and cannot figure out how you find a_0 and a_1 if all of the indexes are 0 and the (x-1)^n terms are the same as you show. if you solve the equation wouldn't you have the recurrence relation a_n+2 in terms of a_n+1 and a_n for n>or=0? Can you help me with next step?

 

[Pyrrhus]

You should have
$$\sum_{n=0}^{\infty} (n+2) (n+1) a_{n+2} (x-1)^{n} - \sum_{n=0}^{\infty} (n+1) a_{n+1} (x-1)^{n} - \sum_{n=1}^{\infty} n a_{n} (x-1)^{n} - \sum_{n=0}^{\infty} a_{n} (x-1)^{n} = 0$$

where the term $\sum_{n=1}^{\infty} n a_{n} (x-1)^{n}$ comes from

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

Actually there is a small mistake . Now you can solve this by letting n=0 in the first, second and fourth sums.

 

[Sancor]

Hi, I'm actually working on this same problem and I'm not having trouble getting the equation mentioned above, but what I can't seem to figure out is how to extract the recurrence relation from it.

I see that the correct relation is

(n+2)a_n+2-a_n+1-a_n=0

But I don't understand how to get there from the four terms already mentioned above. I've tried writing the equation

a_n+2(n+1)(n+2)-a_n+1(n+1)-a_n-1(n)-a_n=0

which gives the correct answer for n=0 but not for subsequent terms. I've tried toying with ignoring the summation that begins at n=1 but i still can't seem to eliminate the (n+1) factor that isn't in the solution.

 

[Sancor]

Ah I've found my mistake, I failed to correctly shift indices.