Power Series/series solutions near a point

[dp182]

Homework Statement

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

Homework Equations

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

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

 

[lanedance]

show your steps

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

 

[lanedance]

show your steps

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

 

[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

 

[lanedance]

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

 

[lanedance]

Homework Statement

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

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

 

[dp182]

from the beginning:
since the answer is of the form y=$\Sigma$a_nxⁿ so the first and second order derivatives are y^'=$\Sigma$na_nx^n-1 and y^''=$\Sigma$n(n-1)a_nx^n-2 so plugging those into the original equation we get
(2+x²)$\Sigma$n(n-1)a_nx^n-2-x$\Sigma$na_nx^n-1+4$\Sigma$a_nxⁿ 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 won't need to change the order, and for the 2y^'' term setting the sigma to 0 gives the form
2$\Sigma$(n+2)(n+1)a_n+2xⁿ
so you can then factor out the $\Sigma$ and the xⁿ terms so what's left i used to formulate the equation for a_n+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

 

[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
$$y = \sum_{n=0}a_nx^n$$
$$y' = \sum_{n=1}a_nnx^{n-1}$$
$$y'' = \sum_{n=2}a_nn(n-1)x^{n-2}$$

 

[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

 

[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 don't add any new terms when you took your derivatives

 

[lanedance]

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

 

[dp182]

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

 

[lanedance]

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

 

[dp182]

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

 

[dp182]

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

 

[lanedance]

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