# Power Series Solution of Differential Equation

1. Mar 25, 2014

1. The problem statement, all variables and given/known data
(x^2)y' = y

2. Relevant equations

3. The attempt at a solution
Plugging in series everywhere I get the equation $\sum na_{n}x^{n+1}$ = $\sum a_{n}x^{n}$. I try to set the coefficients for the corresponding powers equal, but when I do I don't get the correct answer. I also do not know how to handle the sum on the left starting at the x^2 term while the one on the right starts at the constant term.

I hope I made the statement of the problem clearer.

Last edited: Mar 25, 2014
2. Mar 25, 2014

### Ray Vickson

I cannot figure out whether you are making a statement or asking a question; the question mark at the end of your sentence is throwing me off. Anyway, if you show your work we will be better able to comment.

3. Mar 25, 2014

I have edited the write up to hopefully make it more clear

4. Mar 25, 2014

### pasmith

$$x^2 \frac{d}{dx} \sum_{n=0}^\infty a_nx^n = x^2 \sum_{n=0}^\infty na_nx^{n-1} = \sum_{n=0}^\infty na_nx^{n+1} = \sum_{n=1}^\infty na_nx^{n+1}$$
where the last follows because $n = 0$ when $n = 0$, and substituting $k = n +1$ is the next step.

However solving the ODE directly gives
$$y(x) = A\exp\left( -\frac 1{x}\right)$$
where we have the problem at zero that
$$\lim_{x \to 0^{+}} y(x) = \lim_{x \to 0^{+}} y^{(n)}(x) = 0$$
for all $n \geq 1$, so there is no convergent power series about zero unless $A = 0$.

For $x < 0$ it's even worse:
$$\lim_{x \to 0^{-}} y(x) = \begin{cases} \infty & A > 0 \\ 0 & A = 0 \\ -\infty & A < 0 \end{cases}$$

5. Mar 25, 2014

Sorry I'm still confused. I got the series you got for the y' term and then I changed it to $\sum^{\infty}_{n=2}(n-1)a_{n-1}x^{n}$ so I could combine it with the other sum, which I rewrote as a$_{0}$+a$_{1}$x+$\sum^{\infty}_{n=2}a_{n}x^{n}$. Normally I would set all coefficients equal to zero and then find a recursion relation. This did not work this time. If I've understood you're post there is no convergent power series so this method won't work. What should I do differently?

6. Mar 25, 2014

### HallsofIvy

Staff Emeritus
What you have, initially, $\sum_{n= 1}^\infty na_n x^{n+ 1}= \sum_{n= 0}^\infty a_nx^n$.
(Notice that the left sum starts at n= 1. That is because if when n= 0, the term is "$0x^0= 0$.) You say you tried "to set the coefficients for the corresponding powers equal, but when I do I don't get the correct answer" but you don't say what you tried what answer you got or what the correct answer you should have gotten.

You could try writing out the first few terms of each sum: $a_1x^2+ 2a_2x^3+ 3a_3x^4+ \cdot\cdot\cdot$ on the left and $a_0+ a_1x+ a_2x^2+ a_3x^3+ a_4x^4+ \cdot\cdot\cdot$ and since those are equal we must have $a_0= 0$, $a_1= 0$, $a_2= a_1$, $a_3= 2a_2$, $a_4= 3a_3$, etc.

Generally, we can compare corresponding coefficients for $\sum_{n= 1}^\infty na_n x^{n+ 1}= \sum_{n= 0}^\infty a_nx^n$ as you do by letting j= n+ 1 in the left sum and j= n in the sum on the right:
$\sum_{n= 1}^\infty na_n x^{n+ 1}= \sum_{j= 2}^\infty (j-1)a_{j-1}x^j= \sum_{j= 0}^\infty a_j x^j$. Since the left sum does not begin until j= 2, for j= 0 we have $a_0= 0$ and for i= 1,$a_1= 0$. For j> 1, $(j-1)a_{j-1}= a_j$. In particular, $a_2= 1a_1= 0$, $a_3= 2a_2= 2(0)= 0$, etc.

7. Mar 25, 2014

### Ray Vickson

The solution y(x) is non-analytic at x = 0; that means that y(x) is not equal to its Taylor series about x = 0 (i.e., Maclauren series), even though that Maclauren series is convergent on the whole real line! The solution cannot be represented as a series of the form $\sum_{n=0}^{\infty} c_n x^n$. Your power series method yields the correct Maclauren series for y(x), but that is not y(x) itself.

8. Mar 25, 2014

Ok this made a lot of sense. Thank you both very much.

9. Mar 25, 2014

### Ray Vickson

Actually, my previous post is not quite accurate. There are many solutions, depending on conditions at a particular x0 ≠ 0. Your series solution finds one of them (y(x) = 0 for all x); the others cannot be expanded as power series around x = 0.

10. Mar 25, 2014

### HallsofIvy

Staff Emeritus
I assume that you are required to use series. A much simpler solution is to separate variables to get
$$\frac{dy}{y}= \frac{dx}{x^2}$$
so integrating $ln|y|= -\frac{1}{x}+ C$.

And then $y= C' e^{-1/x}$.

Perhaps you can use that as a check.