# Homework Help: Series Solution to ODEs

1. Jun 11, 2006

### kape

$$y''-6xy'+(6x^2-2)y=0$$

$$y_{1} = _____________$$

I have to solve the above equation using power series.. but I am stuck. What I have so far is:

$$y=\sum_{m=0}^\infty a_{m}x^{m}$$

$$y'=\sum_{m=1}^\infty ma_{m}x^{m-1}$$

$$y''=\sum_{m=2}^\infty m(m-1)a_{m}x^{m-2} = \sum_{m=0}^\infty (m+2)(m+1)a_{m+2}x^{m}$$

therefore

$$\sum_{m=0}^\infty (m+2)(m+1)a_{m+2}x^{m} - 6\sum_{m=1}^\infty ma_{m}x^{m} + 6\sum_{m=2}^\infty a_{m-2}x^{m} - 2\sum_{m=0}^\infty a_{m}x^{m} = 0$$

(and this is where i'm really not sure about)

and because $$\inline x^{0}$$ occurs in the first and last series..

$$2a_{2} - 2a_{0} = 0$$

$$2(a_{2} - a_{0}) = 0$$

and $$\inline x^{1}$$ occurs in the first, second and last series so..

$$6a_{3} - 8a_{1} = 0$$

but.. how am i supposed to find the a's from here? Am I supposed to do something like the following?

$$y = 2(a_{2} - a_{0}) + (6a_{3} - 8a_{1}) ...$$

I've looked at the textbook and sample problems but I just don't understand..

Can anyone help me??

Last edited: Jun 11, 2006
2. Jun 11, 2006

### HallsofIvy

so, changing index as you correctly do in y" below
$$y'= \sum_{m=0}^\infy (m+1)a_{m+1}x^m$$

No,
$$\sum_{m=0}^\infty (m+2)(m+1)a_{m+2}x^{m} - 6\sum_{m=0}^\infty (m+1_a_{m+1}x^{m} + 6\sum_{m=2}^\infty a_{m-2}x^{m} - 2\sum_{m=0}^\infty a_{m}x^{m} = 0$$
where I've just made the change in y'

Not quite. Since m= 0 occurs in the second sum as well,
$$2a_2- 6a_1- 2a_0= 0$$
or just
$$a_2= 3a_1+ a_0$$
so that $a_2$ is written in terms of $a_0$ and $a_1$.

Again, not quite. Taking m= 1 gives
$$6a_3- 12a_2- 2a_1= 0$$
which you can also write as
$$6a_3- 12(3a_1+a_0)- 2a_1= 0$$
and then
$$a_3= (17/3)a_1+ 2a_0$$
which doesn't look so pretty but is again in terms of $a_1$ and $a_0$.

What you do with the general formula:
$$(m+2)(m+1)a_{m+2}- 6(m+1)a_{m+1}+ 6a_{m-2}- 2a_m= 0$$
depends largely on what you can do!
It's not that difficult to set m= 2, 3, 4, ,5,... and do the calculations:
with m= 2
$$12a_4- 18a_3+ 6a_0- 2a_2= 0$$
so
$$a_4= (3/2)a_3- (1/2)a_0- (1/6)a_2$$
Now, substitute for $a_3$ and $a_2$ from above to get
$a_4$ also in terms of $a_0$ and $a_1$.

Doing that you can get formulas for the first 5 or 6 terms depending upon $a_0$ and $a_1$. Often, that is all you can do but sometimes you can "guess" the general formula and then prove it with induction.

Of course, $a_0$ and $a_1$ are the "undetermined constants" you expect to get from a second order differential equation. They can be determined from the initial conditions: $a_0= y(0)$ and $a_1= y'(0)$.

3. Jun 12, 2006

### kape

Thank you for you detailed answer!

But I am still a little confused about the part where you change the index for y'.. why did you change the index for y'?

Since it is 6xy' wouldn't I have to do add another x to the the series? So that it becomes:

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

In which case, the power of the x's don't match because this one is m+1 so wouldn't I have to change the index back to the original?

$$y' = \sum_{m=1}^{\infty} ma_{m}x^{m}$$

I'm afraid I don't have a grasp of the fundamentals, and I find the textbook quite hard to understand. During the process of trying to figure out what is going on by looking at the textbook and sample problems, I think I make incorrect assumptions.. All I understand is that you change the series so that the power of x is all the same, and then try to figure out the undetermined constants.

I also don't quite understand this part. The answer is supposed to be given in the form:

$$y_{1} = a_{0}( \_\_\_ + \_\_\_x^2 + \_\_\_x^4 ) + a_{1}( \_\_\_x + \_\_\_x^3 ) ...$$

What is $$\inline y_{1}$$ ? Is that the same as y'? Is this the "expansion" (terminology?) for the series:

$$y' = \sum_{m=1}^{\infty} ma_{m}x^{m}$$ ?

I am hopelessly lost and I have so much to do! I hope you reply soon!

Thanks again for all the help.

-------------------------------------------------------------------------------------------------
(Related Topic)

This is related so I thought you could help me out on this too. The textbook says that:

$$y' = 2xy$$

gives

$$1 \cdot a_{1}x^{0} + \sum_{m=2}^{\infty}ma_{m}x^{m-1} = \sum_{m=0}^{\infty}2a_{m}x^{m+1}$$

I understand the right side (the 2xy side) but the left side I don't quite get. First of all, why is it m=2 and not m=1? The rest of that series is for y' but for y' shouldn't it be m=1 and not 2? m=2 is for y'', is it not? And the most puzzling of all is this:

$$1 \cdot a_{1}x^{0}$$

Where did this come from?!?!

(Incidentally, the dot between 1 and a means multiplication, am I correct? I've seen this in other problems and I don't understand why they explicitly leave it in instead of just multiplying it and showing the results. Why is this? Why, for example, in this case did they not simply show it as $$\inline a_{1}x^{0}$$? Is there a purpose to this?)

Last edited: Jun 12, 2006
4. Jun 12, 2006

### HallsofIvy

Oh blast! I misread the equation- I didn't see the "x" in "6xy' ". You are right- that x corrects the exponent so you don't need to change the index.

I'm not sure why they are calling it y1 but that is just the general solution to the equation. Let's go back to your original (correct) equation:
$$\sum_{m=0}^\infty (m+2)(m+1)a_{m+2}x^{m} - 6\sum_{m=1}^\infty ma_{m}x^{m} + 6\sum_{m=2}^\infty a_{m-2}x^{m} - 2\sum_{m=0}^\infty a_{m}x^{m} = 0$$

As you say, when m= 0, the two "surviving" terms give you $2a_2- 2a_0= 2(a_{2} - a_{0}) = 0$ so $a_2= a_0$.

When m= 1, the three terms give you $6a_3- 6a_1- 2a_1= 6a_3- 8a_1= 0$ so $a_3= (4/3)a_1$.

For m>1, we have all four terms:
$$(m+2)(m+1)a_{m+2} - 6ma_{m} + 6a_{m-2} - 2a_{m} = 0$$
which you can solve for $a_{m+2}$:
$$a_{m+2}= \frac{(6m+2)a_m- 6a_{m-2}}{(m+2)(m+1)}$$
What I would now do is make a chart of $a_m$ for as many values of m as I can stand!
$a_2= a_0$, $a_3= (4/3)a_1$, $a_4= \frac{14}{12}a_2- \frac{6}{12}a_0= \frac{2}{3}a_0$, $a_5= \frac{20}{20}a_3- \frac{6}{20}a_1= \frac{4}{3}a_1- \frac{3}{10}a_1= \frac{31}{30}a_1$, etc.
One thing you can clearly see is that, for even n, $a_n$ will be a multiple of $a_0$ and, for odd n, $a_n$ will be a multiple of $a_1$ so the sum breaks easily into even and odd parts:
$$y(x)= a_0(1+ x^2+ \frac{2}{3}x^2+\cdot\cdot\cdot)+ a_1(x+ \frac{4}{3}x^3+ \frac{31}{30}x^5+\cdot\cot\cot$$

In fact, once you see that the sum splits into even and odd parts like that, you can simplify the calculation by first taking $x_0= 1$ and $x_1= 0$ and then taking $x_0= 0$ and $x_1= 1$. (Although, calculating through $a_8= \frac{157}{180}$ and $a_9= \frac{109}{1134}$, I don't see any obvious pattern!)

They have just separated
$$y'= \sum_{m=1}^\infty m a_m x^{m-1}$$
into
$$y'= 1\cdot a_1 x^0+ \sum{m=2}^\infty ma_m x^{m-1}$$
since $a_1 x^0$ is just the m= 1 term in the first sum.
The reason they did that (I think) was that then you can change the index in the sum to get
$$a_1+ \sum_{m=1}^\infty(m+1)a_{m+1}x^m$$
while the right hand side is
[tex]xy= \sum_{m=0}^\infty a_mx^{m+1}[/itex]
and changing the index there gives
[tex]\sum_{m=1}^\infty a_{m-1}x^m[/itex]
so that both sums start at m=1 and have xm. What has happened is the m=0 term has already been separated for you.

They just wanted to make it clear (which it wasn't!) that that term was just $ma_mx^{m-1}$ with m= 1- that's also why they left x0 which is, of course, 1 for all x.