Series Solutions to Differential Equations | Initial Conditions at x=1

[TranscendArcu]

Homework Statement

The Attempt at a Solution

My trouble is that I don't really understand how to begin this problem. I think I am supposed to center my solution at a value equal to the x given in my initial conditions. That is, choose my centering point as x = 1. Then, I would have a solution: $y = \sum_{n=0}^{∞} a_n (x-1)^n = a_0 (x-1)^0 + a_1(x-1)^1 + ...$. Is this then supposed to imply that $y(1) = 2 = a_0$?

Is this the right beginning to the problem?

 

[Mark44]

Homework Statement

The Attempt at a Solution

My trouble is that I don't really understand how to begin this problem. I think I am supposed to center my solution at a value equal to the x given in my initial conditions. That is, choose my centering point as x = 1. Then, I would have a solution: $y = \sum_{n=0}^{∞} a_n (x-1)^n = a_0 (x-1)^0 + a_1(x-1)^1 + ...$. Is this then supposed to imply that $y(1) = 2 = a_0$?

Is this the right beginning to the problem?

Yes and yes.

 

[TranscendArcu]

so I have: $y' = \sum_{n=1}^∞ a_n(n)(x-1)^{n-1}$. Our initial condition gives y'(1) = 0, which in turn tells me that $a_1 = 0$.

From the original differential equation, I can rewrite the terms to give: $y'' = xy' + y → y''(1) = (1)y'(1) + y(1) = (1)(0) + 2 = 2$. This implies that $a_2 = \frac{2}{2!}$. Taking the derivative of the y'' equation I have:

$y^{(3)} = y' + xy'' +y' → y^{(3)}(1) = y'(1) + (1)y''(1) +y'(1) = 0 + 2 + 0 = 2$. This implies that $a_3 = \frac{2}{3!}$. Taking another derivative:

$y^{(4)} = y'' + y'' + xy''' +y'' → y^{(4)}(1) = y''(1) + y''(1) + (1)y'''(1) +y''(1) = 2+2+2+2 = 8$. This implies that $a_4 = \frac{8}{4!}$.

Is this supposed to be telling me that the first four terms are $y = 2 + \frac{2x^2}{2!} + \frac{2x^3}{3!} + \frac{8x^4}{4!}$?

 

[TranscendArcu]

I think I might need a hint for the second part of the question. How can I see the analytical properties of y?

 

[Mark44]

so I have: $y' = \sum_{n=1}^∞ a_n(n)(x-1)^{n-1}$. Our initial condition gives y'(1) = 0, which in turn tells me that $a_1 = 0$.

From the original differential equation, I can rewrite the terms to give: $y'' = xy' + y → y''(1) = (1)y'(1) + y(1) = (1)(0) + 2 = 2$. This implies that $a_2 = \frac{2}{2!}$. Taking the derivative of the y'' equation I have:

$y^{(3)} = y' + xy'' +y' → y^{(3)}(1) = y'(1) + (1)y''(1) +y'(1) = 0 + 2 + 0 = 2$. This implies that $a_3 = \frac{2}{3!}$. Taking another derivative:

$y^{(4)} = y'' + y'' + xy''' +y'' → y^{(4)}(1) = y''(1) + y''(1) + (1)y'''(1) +y''(1) = 2+2+2+2 = 8$. This implies that $a_4 = \frac{8}{4!}$.

Is this supposed to be telling me that the first four terms are $y = 2 + \frac{2x^2}{2!} + \frac{2x^3}{3!} + \frac{8x^4}{4!}$?

No. Remember that you are finding a series in powers of (x - 1).

 

[Mark44]

I think I might need a hint for the second part of the question. How can I see the analytical properties of y?

You need to find the series representation of y(x), and then use the usual theorems to determine the interval of convergence.

The assumption here is that
$$y(x) = \sum_{n = 0}^{\infty}a_n(x - 1)^n$$
and your DE is y'' - xy' - y = 0

Starting from your assumed solution, calculate y' and y'' and then calculate y'' - xy' - y, as a series. Since this sum is equal to zero, all of its terms must equal zero.

 

[TranscendArcu]

No. Remember that you are finding a series in powers of (x - 1).

Oh, yeah. I forgot. But I only think that changes all of the x coefficients in my series to (x-1) coefficients. My a_n do not change, right?

 

[TranscendArcu]

You need to find the series representation of y(x), and then use the usual theorems to determine the interval of convergence.

The assumption here is that
$$y(x) = \sum_{n = 0}^{\infty}a_n(x - 1)^n$$
and your DE is y'' - xy' - y = 0

Starting from your assumed solution, calculate y' and y'' and then calculate y'' - xy' - y, as a series. Since this sum is equal to zero, all of its terms must equal zero.

So you say I need to find the sigma representation of this series? That seems kind of tricky. Does it even follow an easily-recognized pattern? The 8 in the numerator of the (x-1)^4 term really throws things off for me.

 

[Mark44]

So you say I need to find the sigma representation of this series? That seems kind of tricky.

That's how I would go about it. When I did problems like this I would expand each series (the ones for y, xy', and y'') and write the first few terms of each, plus the general term. The idea is to line up all the terms of the three series so that the exponents match up.

I haven't worked through this, but maybe you could do something like this:

y'' - xy' - y = 0 <==> y'' - (x - 1)y' + y' - y = 0

Just a thought - I don't know if it will be helpful.

Does it even follow an easily-recognized pattern? The 8 in the numerator of the (x-1)^4 term really throws things off for me.

 

[TranscendArcu]

That's how I would go about it. When I did problems like this I would expand each series (the ones for y, xy', and y'') and write the first few terms of each, plus the general term. The idea is to line up all the terms of the three series so that the exponents match up.

I haven't worked through this, but maybe you could do something like this:

y'' - xy' - y = 0 <==> y'' - (x - 1)y' + y' - y = 0

Just a thought - I don't know if it will be helpful.

Right now my series is $y = 2 + \frac{2(x-1)^2}{2!} +\frac{2(x-1)^3}{3!} + \frac{8(x-1)^4}{4!} + ...$. It looks like I can take a 2 out so let's do that $y = 2(1 + \frac{(x-1)^2}{2!} +\frac{(x-1)^3}{3!} + \frac{4(x-1)^4}{4!} + ...)$. But, as I mentioned above, this doesn't seem to follow a pattern that I can easily see. Thus, when you tell me to write out the first few terms, I can do that easily enough; but when it comes to finding the general term you refer to, I hit a wall. And until I have the general terms, I don't see how I can apply the trick you suggest.

Once I have the general terms, I'll be able to shift the series/strip out terms to match up the exponents, and then apply the ratio test (I presume) to determine the interval of convergence.

 

[TranscendArcu]

So after much effort, I determined that there was a recursion relationship of the form $a_{n+2} = \frac{a_n + a_{n+1}}{n+2}$. Unfortunately -- I haven't found this as useful as I thought I would. I can't find anything in this problem that allows me to find the interval of convergence.