Power Series/series solutions near a point

1. Jun 26, 2011

dp182

1. The problem statement, all variables and given/known data
given the equation (2+x2)y''-xy'+4y determine the first four terms of the coresponding independent solutions (a(0),a(1))
given the solution is $\Sigma$a(n)xn

2. Relevant equations
when I calculated the series i got a(n+2)= (n(n-2)+4)/2(n+2)(n+1)

3. The attempt at a solution
so when I used values of 0-7 for n i got a(0),a(1)/4,....ect I'm not sure if my series equations is incorrect any help would be great

2. Jun 27, 2011

lanedance

There should be 2 independent solutions as it is a 2nd order DE. These can be presented in terms of the first coefficients a(0) and a(1)

Generally you should find a general recurrence relation for general n, however you may need specific cases for the first few terms

3. Jun 27, 2011

lanedance

There should be 2 independent solutions as it is a 2nd order DE. These can be presented in terms of the first coefficients a(0) and a(1)

Generally you should find a general recurrence relation for general n, however you may need specific cases for the first few terms

4. Jun 27, 2011

dp182

I find two series with a(0) and a(1) but its wrong and no matter how many times I try to equate my equation for a(n+2) i get the same equation I posted so I'm not sure where I am screwing up

5. Jun 28, 2011

lanedance

I can't be sure either without seeing what you've tried... ;) if you show your attempt I can have a look though

Last edited: Jun 28, 2011
6. Jun 28, 2011

lanedance

also thats not quite an equation, do you mean
(2+x2)y''-xy'+4y=0?

7. Jun 28, 2011

dp182

from the beginning:
since the answer is of the form y=$\Sigma$anxn so the first and second order derivatives are y'=$\Sigma$nanxn-1 and y''=$\Sigma$n(n-1)anxn-2 so plugging those into the original equation we get
(2+x2)$\Sigma$n(n-1)anxn-2-x$\Sigma$nanxn-1+4$\Sigma$anxn and from that equation you can see that the zero terms for the first order and second order for x2 will be zero so you wont need to change the order, and for the 2y'' term setting the sigma to 0 gives the form
2$\Sigma$(n+2)(n+1)an+2xn
so you can then factor out the $\Sigma$ and the xn terms so whats left i used to formulate the equation for an+2 and then subbed in values for n to get terms for a2....a10 in terms of a0 and a1
thats how I did it and it follows my instructors example completely but I am unable to get the correct answer hope this is helpful

8. Jun 28, 2011

lanedance

ok so following along (note you can write a whole equation in tex)

first in these problems its worth including the sum as terms will be a little different at low n as the derivative of a constant is zero
$$y = \sum_{n=0}a_nx^n$$
$$y' = \sum_{n=1}a_nnx^{n-1}$$
$$y'' = \sum_{n=2}a_nn(n-1)x^{n-2}$$

Last edited: Jun 29, 2011
9. Jun 28, 2011

lanedance

the sums and powers can get a little confusing so here's a reasonably foolproof recipe for putting it all together

now calculating each term in the DE
$$4y = 4\sum_{n=0}a_nx^n$$
$$xy' = x\sum_{n=1}a_n n x^{n-1} = \sum_{n=1}a_n n x^{n}$$
$$2y'' = 2\sum_{n=2}a_n n (n-1)x^{n-2}$$
$$x^2y'' = x^2\sum_{n=2}a_nn (n-1)x^{n-2}= \sum_{n=2}a_nn (n-1)x^{n}$$

the next trick I find useful is to re-write each sum so the sum has the power of x expressed in the same form and change the indexes to align.

so we can group three terms together, those with x to the power of n, and these need no change but we change m to n for clarity
$$m = n$$
$$4y = 4\sum_{m=0}a_m x^m$$
$$xy'= \sum_{m=1}a_m m x^{m}$$
$$x^2y'' = \sum_{m=2}a_m m (m-1) x^{m}$$

now for the n-2 power term we do the following shift
$$m = n-2\implies n=m+2$$
$$2y'' = 2\sum_{n=2}a_nn(n-1)x^{n-2}= 2\sum_{m=0}a_{m+2}(m+2)(m+1)x^{m}$$

from here it should be a simple case to substitute into the original and compare terms,

note the sums start at different m values so you may want to write out the terms for m<2 explicitly and have sums form m=2 up

Last edited: Jun 29, 2011
10. Jun 28, 2011

dp182

That is the same as what I did except there are no constants when you took the derivative's of the series how come you dont add any new terms when you took your derivatives

11. Jun 28, 2011

lanedance

sorry i don't understand the question?
$$\frac{d}{dx} (a_0x^0) = \frac{d}{dx} (a_0.1) = 0$$

12. Jun 28, 2011

dp182

when you took the answer to the ode
y=$\sum$anxn and then derived it to get
y'=$\sum$anxn-1 shouldn't you bring down the n as a constant? and if not then wouldn't your solution for am+2=am?

13. Jun 29, 2011

lanedance

woops that one slipped through cutting & pasting tex, sorry, have corrected above

Last edited: Jun 29, 2011
14. Jun 29, 2011

dp182

so then the equation for am+2 would be
am+2=(-m(m-1)+m-4)am/(m+2)(m+1) which would be the same one I got before

15. Jun 29, 2011

dp182

Nvm I am an idiot lol thanks alot you've really helped

16. Jun 29, 2011

lanedance

yeah so i think you got it but it should become
$$(2+x^2)y''-xy'+4y=0$$
$$=2\sum_{m=0}a_{m+2}(m+2)(m+1)x^{m}+ \sum_{m=2}a_m m (m-1) x^{m}+ \sum_{m=1}a_m m x^{m}+ 4\sum_{m=0}a_m x^m$$
$$=2\sum_{m=0}a_{m+2}(m+2)(m+1)x^{m}+ \sum_{m=2}a_m m (m-1) x^{m}+ \sum_{m=1}a_m m x^{m}+ 4\sum_{m=0}a_m x^m$$
$$=2a_{2}+6a_{3}x+ a_1 x+ a_0 +a_1 x+ +\sum_{m=2}(a_{m+2}(m+2)(m+1)+a_m m (m-1)+a_m m +a_m )x^{m} =0$$