Power Series Solution for x^2y'' - y = 0 Expanding about xo = 2: Next Steps?

[Saladsamurai]

Homework Statement

Solve $x^2y'' - y = 0$ using Power Series Solution expanding about x_o = 2.

The Attempt at a Solution

First I expand the coefficient of y" (i.e. x²) about x_o:

$$TS[x^2]|_{x_o=2} = 4+ 4(x - 2) + (x - 2)^2$$

Assuming the solution takes the form:

$$y(x) = \sum_0^{\infty}a_n(x - 2)^n \Rightarrow y''(x) = \sum_2^{\infty}n(n-1)a_n(x - 2)^{n - 2}$$

Plugging the above into the original DE gives:

$$(4+ 4(x - 2) + (x - 2)^2)\sum_2^{\infty}n(n-1)a_n(x - 2)^{n - 2} -<br /> \sum_0^{\infty}a_n(x - 2)^n = <br /> 0$$Distributing we have:

$$4\sum_2^{\infty}n(n-1)a_n(x - 2)^{n - 2} + <br /> 4(x - 2)\sum_2^{\infty}n(n-1)a_n(x - 2)^{n - 2} +<br /> (x - 2)^2)\sum_2^{\infty}n(n-1)a_n(x - 2)^{n - 2} -<br /> \sum_0^{\infty}a_n(x - 2)^n = <br /> 0<br />$$

$$\Rightarrow <br /> 4\sum_2^{\infty}n(n-1)a_n(x - 2)^{n - 2} + <br /> 4\sum_2^{\infty}n(n-1)a_n(x - 2)^{n - 1} +<br /> \sum_2^{\infty}n(n-1)a_n(x - 2)^{n} -<br /> \sum_0^{\infty}a_n(x - 2)^n = <br /> 0<br /> \qquad(1)$$

Now the problem with (1) is twofold:

i) The indices are not the same and ii) the exponents on x are not the same.

I am not sure how to handle this. I was thinking that I could add 2 to the n index in the first term and add 1 to the second and then run the summations from zero. I think that this should work since the factors of 'n' 'n-1' in each term that I augment will cause each term to drop out at n = 0 and n = 1 and hence they will not contribute to the sum.

Edit: Fail. I failed to notice that when I add 2 to the dummy variable 'n' I will no longer have the factor of n to cause the term to drop out at n=0.

Any help on this?

 

[Saladsamurai]

Hate to bump this, but I am kind of in a pinch and would love to be able to move forward on this. If I just work on the first term: $4\sum_2^{\infty}n(n-1)a_n(x - 2)^{n - 2}$ and I move the starting point ahead by 2, I have to move the lower limit of the summation back by two :$4\sum_0^{\infty}(n+2)(n+1)a_{n+2}(x - 2)^{n}$. I feel like this would introduce additional terms though. For example, take the simple case of changing $\sum_2^4 x^n$ to $\sum_0^4 x^{n+2}$. The latter will introduce extra terms unless I reduce the upper limit from 4 to 2. However with an infinite sum, that is meaningless.

Thoughts?

 

[gomunkul51]

Firstly, let us note that this is Euler's ODE:

$$x^{2}y^{''}+0\cdot xy^{'}-y=0$$

and we know how to solve those without using the Series Expansion method.
Explanation: http://www.math.oregonstate.edu/hom...sQuestStudyGuides/ode/second/so_ec/so_ec.htmlSecondly, the way you solve all of this Series Solution expansions is:

First step: you need to change the index is such a way that to leave you with series' that have only powers of n inside.
example:

$$\sum_2^{\infty}n(n-1)a_n(x - 2)^{n - 2} \Leftrightarrow \sum_0^{\infty}(n+2)(n+1)a_{n+2}(x - 2)^{n}$$

Second step: you need to expand the summation in such a way that all of your series' with star at the same index (highest index of the series').
example:

$$\sum_2^{\infty}n(n-1)a_n(x - 2)^{n - 1} \Leftrightarrow \sum_1^{\infty}(n+1)n \cdot a_{n+1}(x - 2)^{n} \Leftrightarrow (1+1)\cdot 1 \cdot a_{1+1}(x - 2)^{1} + \sum_2^{\infty}(n+1)n \cdot a_{n+1}(x - 2)^{n} \Leftrightarrow 2\cdot a_{2}(x - 2) + \sum_2^{\infty}(n+1)n \cdot a_{n+1}(x - 2)^{n}$$

! if you don't know how to proceed and get the final result from here, go to:
http://tutorial.math.lamar.edu/Classes/DE/PowerSeries.aspx

Note: it is also possible you should use the Frobenius Method, if you know what it is (since the solution is around x=2 I don't think you should).

* I hope you understood from my examples how to change the indexes correctly.
** I didn't check you answer so far for computational mistakes.
*** good luck :)

 

[HallsofIvy]

Homework Statement

Solve $x^2y'' - y = 0$ using Power Series Solution expanding about x_o = 2.

The Attempt at a Solution

First I expand the coefficient of y" (i.e. x²) about x_o:

$$TS[x^2]|_{x_o=2} = 4+ 4(x - 2) + (x - 2)^2$$

Assuming the solution takes the form:

$$y(x) = \sum_0^{\infty}a_n(x - 2)^n \Rightarrow y''(x) = \sum_2^{\infty}n(n-1)a_n(x - 2)^{n - 2}$$

Plugging the above into the original DE gives:

$$(4+ 4(x - 2) + (x - 2)^2)\sum_2^{\infty}n(n-1)a_n(x - 2)^{n - 2} -<br /> \sum_0^{\infty}a_n(x - 2)^n = <br /> 0$$


Distributing we have:

$$4\sum_2^{\infty}n(n-1)a_n(x - 2)^{n - 2} + <br /> 4(x - 2)\sum_2^{\infty}n(n-1)a_n(x - 2)^{n - 2} +<br /> (x - 2)^2)\sum_2^{\infty}n(n-1)a_n(x - 2)^{n - 2} -<br /> \sum_0^{\infty}a_n(x - 2)^n = <br /> 0<br />$$

$$\Rightarrow <br /> 4\sum_2^{\infty}n(n-1)a_n(x - 2)^{n - 2} + <br /> 4\sum_2^{\infty}n(n-1)a_n(x - 2)^{n - 1} +<br /> \sum_2^{\infty}n(n-1)a_n(x - 2)^{n} -<br /> \sum_0^{\infty}a_n(x - 2)^n = <br /> 0<br /> \qquad(1)$$

Now the problem with (1) is twofold:

i) The indices are not the same and ii) the exponents on x are not the same.

I am not sure how to handle this. I was thinking that I could add 2 to the n index in the first term and add 1 to the second and then run the summations from zero. I think that this should work since the factors of 'n' 'n-1' in each term that I augment will cause each term to drop out at n = 0 and n = 1 and hence they will not contribute to the sum.

Edit: Fail. I failed to notice that when I add 2 to the dummy variable 'n' I will no longer have the factor of n to cause the term to drop out at n=0.

Any help on this?

Then change the dummy indices:
In
$$4\sum_2^{\infty}n(n-1)a_n(x - 2)^{n - 2}$$
let j= n-2 so that n= j+ 2. When n= 2, j= 0 and n-1= j+ 1. The sum becomes
$$4\sum_0^\infty (j+2)(j+1)a_{j+2}(x- 2)^j$$

In $$4\sum_2^{\infty}n(n-1)a_n(x - 2)^{n - 1}$$
let j= n-1 so that n= j+ 1. When n= 2, j= 1 and n-1= j. The sum becomes
$$4\sum_1^\infty (j+1)j a_{j+1}(x- 2)^j$$

In $$\sum_2^{\infty}n(n-1)a_n(x - 2)^{n}$$
and $$\sum_0^{\infty}a_n(x - 2)^n$$
let j= n so we have basically the same thing-
$$\sum_2^\infty j(j-1)a_j(x- 2)^j$$
and $$\sum_0^\infty a_j(x- 2)^j$$

Putting those into the equations, we have
$$4\sum_0^\infty (j+2)(j+1)a_{j+2}(x- 2)^j+ 4\sum_1^\infty (j+1)j a_{j+1}(x- 2)^j+ \sum_2^\infty j(j-1)a_j(x- 2)^j+ \sum_0^\infty a_j(x- 2)^j= 0$$

Two of those start at j= 0, another at j= 1, and another not until j= 2. Okay, treat j= 0 and j= 1 separately.

When j= 0, we have
$$4(2)(1)a_2+ a_0= 0$$
so that $a_2= -a_0/8$

When j= 1 we have
$$4(3)(2)a_3(x- 2)+ 4(2)(1)a_2(x-2)+ a_1(x- 2)= (24a_3+ 8a_2+ a_1)(x- 2)= 0$$
In order for that to be true for all x, we must have $24a_3+ 8a_2+ a_1= 0$.

Finally, for $j\ge 2$ we have
$$4(j+2)(j+1)a_{j+2}(x- 2)^j+ 4(j+1)j a_{j+1}(x- 2)^j+ j(j-1)a_j(x- 2)^j+ a_j(x- 2)^j$$
[tex]= (4(j+2)(j+1)a_{j+2}+ 4(j+1)j a_{j+1}+ j(j-1)a_j+ a_j)(x- 2)^j= 0[/itex]<br /> In order for that to be true for all x, we must have <br /> $$4(j+2)(j+1)a_{j+2}+ 4(j+1)j a_{j+1}+ j(j-1)a_j+ a_j= 0$$[/tex]

 

[Saladsamurai]

Then change the dummy indices:
In
$$4\sum_2^{\infty}n(n-1)a_n(x - 2)^{n - 2}$$
let j= n-2 so that n= j+ 2. When n= 2, j= 0 and n-1= j+ 1. The sum becomes
$$4\sum_0^\infty (j+2)(j+1)a_{j+2}(x- 2)^j$$

In $$4\sum_2^{\infty}n(n-1)a_n(x - 2)^{n - 1}$$
let j= n-1 so that n= j+ 1. When n= 2, j= 1 and n-1= j. The sum becomes
$$4\sum_1^\infty (j+1)j a_{j+1}(x- 2)^j$$

In $$\sum_2^{\infty}n(n-1)a_n(x - 2)^{n}$$
and $$\sum_0^{\infty}a_n(x - 2)^n$$
let j= n so we have basically the same thing-
$$\sum_2^\infty j(j-1)a_j(x- 2)^j$$
and $$\sum_0^\infty a_j(x- 2)^j$$

Putting those into the equations, we have
$$4\sum_0^\infty (j+2)(j+1)a_{j+2}(x- 2)^j+ 4\sum_1^\infty (j+1)j a_{j+1}(x- 2)^j+ \sum_2^\infty j(j-1)a_j(x- 2)^j+ \sum_0^\infty a_j(x- 2)^j= 0$$

Two of those start at j= 0, another at j= 1, and another not until j= 2. Okay, treat j= 0 and j= 1 separately.

When j= 0, we have
$$4(2)(1)a_2+ a_0= 0$$
so that $a_2= -a_0/8$

When j= 1 we have
$$4(3)(2)a_3(x- 2)+ 4(2)(1)a_2(x-2)+ a_1(x- 2)= (24a_3+ 8a_2+ a_1)(x- 2)= 0$$
In order for that to be true for all x, we must have $24a_3+ 8a_2+ a_1= 0$.

Finally, for $j\ge 2$ we have
$$4(j+2)(j+1)a_{j+2}(x- 2)^j+ 4(j+1)j a_{j+1}(x- 2)^j+ j(j-1)a_j(x- 2)^j+ a_j(x- 2)^j$$
[tex]= (4(j+2)(j+1)a_{j+2}+ 4(j+1)j a_{j+1}+ j(j-1)a_j+ a_j)(x- 2)^j= 0[/itex]<br /> In order for that to be true for all x, we must have <br /> $$4(j+2)(j+1)a_{j+2}+ 4(j+1)j a_{j+1}+ j(j-1)a_j+ a_j= 0$$[/tex]

[tex] <br /> Hi gomunkul51 & Halls! Thanks for taking the time ti reply. <img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f642.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":smile:" title="Smile :smile:" data-smilie="1"data-shortname=":smile:" /><br /> <br /> I am just curious: I believe in class, instead of treating the terms that start at 1 and 2 differently from those that start at 0, we just let <i>all terms</i> start from 0 since each nonzero-starting-term has a factor in it that will cause it to fall out in the j = 0, j = 1 spot. Namely, <br /> <br /> In $4\sum_1^\infty (j+1)j a_{j+1}(x- 2)^j$, at j = 0, the term vanishes. Similarly, in $\sum_2^\infty j(j-1)a_j(x- 2)^j$, at j = 0 and j = 1, the terms vanish.<br /> <br /> Does that sound like something commonly done (at least in engineering)?<br /> <br /> Thanks again guys![/tex]

 

[gomunkul51]

Hi gomunkul51 & Halls! Thanks for taking the time ti reply.

I am just curious: I believe in class, instead of treating the terms that start at 1 and 2 differently from those that start at 0, we just let all terms start from 0 since each nonzero-starting-term has a factor in it that will cause it to fall out in the j = 0, j = 1 spot. Namely,

In $4\sum_1^\infty (j+1)j a_{j+1}(x- 2)^j$, at j = 0, the term vanishes. Similarly, in $\sum_2^\infty j(j-1)a_j(x- 2)^j$, at j = 0 and j = 1, the terms vanish.

Does that sound like something commonly done (at least in engineering)?

Thanks again guys!

1. Did you understand how to solve the question through the help from HallsofIvy and myself and do you have any questions about that?

2. Try to ajust the indexes through the advised way first, then try adjusting it the way you thought in class - do the answers differ?

*Generally, as I wrote before, you need to adjust the indexes so the powers of (x-a) will be the same power ! and only then you adjust the indexes so all the summations start at the same index (look out for non-zero terms).

 

[Saladsamurai]

1. Did you understand how to solve the question through the help from HallsofIvy and myself ... <snip>

I believe so! That is why I am going to go ahead and assume that doing this:

I believe in class, instead of treating the terms that start at 1 and 2 differently from those that start at 0, we just let all terms start from 0 since each nonzero-starting-term has a factor in it that will cause it to fall out in the j = 0, j = 1 spot.

is a perfectly reasonable thing to do.

Thanks!