# Solve differential equation using power series

1. Apr 29, 2014

### brkomir

1. The problem statement, all variables and given/known data
Solve $y^{''}+zy=0$ where $y(0)=0$ and $y^{'}(0)=1$

2. Relevant equations

$y(z)=z^r\sum _{k=0}^{\infty } C_kz^k$

3. The attempt at a solution

Well firstly:

$r(r-1)+p_0r+q_0=0$ where obviously $p_0=q_0=0$ so $r_1=0$ and $r_2=1$.

In general $y(z)=\sum _{k=0}^{\infty } C_kz^{k+r}$, meaning $y{'}(z)=\sum _{k=0}^{\infty } (k+r)C_kz^{k+r-1}$ and $y{''}(z)=\sum _{k=0}^{\infty } (k+r)(k+r-1)C_kz^{k+r-2}$. Substituting this into original DE:

$y^{''}+zy=0$

$\sum _{k=0}^{\infty } (k+r)(k+r-1)C_kz^{k+r-2}+z\sum _{k=0}^{\infty } C_kz^{k+r}=0$

Now looking at the lowest exponents:

$z^{r-2}$: $(r-1)rC_0=0$
$z^{r-1}$: $(r+1)rC_1=0$
$z^{r}$: $(r+1)(r+2)C_2=0$
$z^{r+1}$: $(r+2)(r+3)C_3=0$ and
$\sum _{k=4}^{\infty }[(k+r-1)(k+r)C_k+C_{k-4}]z^{k+r-2}$ so $(k+r-1)(k+r)C_k+C_{k-4}=0$

For $r_1=0$:
$C_0\neq 0$ also $C_1\neq 0$ while $C_2=C_3=0$.

using $(k+r-1)(k+r)c_k+C_{k-4}=0$ we get $C_k=\frac{-C_{k-4}}{(k+r-1)(k+r)}=\frac{-C_{k-4}}{(k-1)k}$

Now after some time you find out that there will be two sums:

a) $k=4m$:

$C_k=\frac{-C_{k-4}}{(k-1)k}$

$C_{4m}=\frac{(-1)^mC_{0}}{(4m)!}$

b) $k=4m+1$:

$C_k=\frac{-C_{k-4}}{(k-1)k}$

$C_{4m+1}=\frac{(-1)^mC_{1}}{(4m+1)!}$

Hopefully so far everything is ok?

If yes,than one solution should be: $y(z)=\sum _{m=0}^{\infty }\frac{(-1)^mC_{0}}{(4m)!}z^{4m}+\sum _{m=0}^{\infty }\frac{(-1)^mC_{1}}{(4m+1)!}z^{4m+1}$

While for $r_2=1$:

Only $C_0\neq 0$ while $C_1=C_2=C_3=0$.

so $C_k=C_{4m}=\frac{(-1)^mC_{0}}{(4m)!}$

and finally $y(z)=\sum _{m=0}^{\infty }\frac{(-1)^mC_{0}}{(4m)!}z^{4m+1}$

Ok, hopefully so far I haven't done any bigger mistakes. Now my problems start. Which $y(z)$ is now the solution of my DE? Or is it the linear combination of both? I don't understand that part since we never did that.

2. Apr 29, 2014

### LCKurtz

That is the relevant equation if $z=0$ is a regular singular point. But it isn't. The relevant equation is $$y=\sum _{k=0}^{\infty } c_kz^k$$Since y(0) = 0 and$$y = c_0 +\sum _{k=1}^{\infty } c_kz^k$$ you have $c_0=0$. Also since $y'(0)=1$ and$$y'=c_1+\sum _{k=2}^{\infty } c_kkz^{k-1}$$you have $c_1=1$. Now calculate $y''$ and plug the series into your equation. You should be able to determine all the $c_k$ and get your solution.

Last edited: Apr 29, 2014
3. Apr 29, 2014

### SammyS

Staff Emeritus
I've only looked this far.

The zr+1 term also has a contribution from the second sum, (for k = 0 in that sum).

So you have: $(r+2)(r+3)C_3+C_0=0$

4. Apr 29, 2014

### brkomir

That seems to be something I should have known... With that in mind, here is version 2.0:

$y=C_0+\sum _{k=1}^{\infty }C_kz^k$ and using $y(0)=0$ also $C_0=0$.

$y^{'}=C_1+\sum _{k=2}^{\infty }kC_kz^{k-1}$ and using $y^{'}(0)=1$ also $C_1=1$.

$y^{''}=\sum _{k=2}^{\infty }k(k-1)C_kz^{k-2}$

Plugging that into DE:

$\sum _{k=2}^{\infty }k(k-1)C_kz^{k-2}+z[C_0+\sum _{k=1}^{\infty }C_kz^k]=0$

$\sum _{k=2}^{\infty }k(k-1)C_kz^{k-2}+\sum _{k=1}^{\infty }C_kz^{k+1}=0$

Again for different exponents:

$z^0$: $2C_2=0$
$z^1$: $6c_3=0$ and

$\sum _{k=4}^{\infty }k(k-1)C_kz^{k-2}+\sum _{k=4}^{\infty }C_{k-3}z^{k-3+1}=0$

$\sum _{k=4}^{\infty }z^{k-2}[k(k-1)C_k+C_{k-3}]=0$

So right now we know that $C_0=C_2=C_3=0$ and $C_1=1$ and $C_k=\frac{-C_{k-3}}{k(k-1)}$

Obviously only $C_k=C_{2m+1}\neq 0$ and If I am not mistaken: $C_k=\frac{(-1)^mC_1}{(2m+1)!}$, which leads me to

$y(z)=\sum _{m=0}^{\infty }\frac{(-1)^m}{(2m+1)!}z^{2m+1}$

Right?

5. Apr 29, 2014

### LCKurtz

That's good to here. But the problem begins with the "obviously" below

You are mixing m and k together. What you need to do is write out some of the $c_k$ for $k=1,4,7...$. You will find that $2m+1$ doesn't generate those powers, and you will discover the denominator isn't quite a factorial.

6. Apr 29, 2014

### brkomir

Aham.., hmm, let's take a look.

$C_1=1$

$C_4=\frac{-C_1}{4\cdot 4}=-\frac{1}{4\cdot 3}$

$C_7=\frac{-C_4}{7\cdot 6}=\frac{1}{7\cdot 6 \cdot 4 \cdot 3}$ and one more

$C_{10}=\frac{-C_7}{10\cdot 9}=-\frac{1}{10\cdot 9 \cdot 7\cdot 6 \cdot 4 \cdot 3}$

Ok, $3m+1$ should generate those powers... $(3m+1)=1,4,7,10,13,...$.

Now... Denominator is almost factorial - only $(2+3n)$ numbers are missing, where $n=0,1,2,3,...$. So If I could somehow include that sequence in numerator, that would be perfect.

I'm lost here. Anyhow, wolfram alpha shoots out that $3,4,6,7,9,10,..=\frac{1}{4}(6n-(-1)^n+5)$

I am not really sure if those sequences are any help at all here

7. Apr 29, 2014

### LCKurtz

Usually you just write out the product instead of some obscure formula like Wolfram gave. So, for example, your term with $z^{3m+1}$ would be written$$\frac {(-1)^m z^{3m+1}}{(3m+1)(3m)(3m-2)(3m-3)...(4)(3)}$$How many factors there are in the denominator depends on how large $m$ is. Notice that when $m=0$ which would give the $z^1$ term, the coefficient formula doesn't work. So when you write your final answer you would write $y = z + \text{sum from 1 up}$, keeping the $m=0$ case separate.

Last edited: Apr 29, 2014
8. Apr 29, 2014

### brkomir

Aaaa, ok, I get it!

Solution is than $y(z)=z+\sum _{m=1}^{\infty }\frac {(-1)^m z^{3m+1}}{(3m+1)(3m)(3m-2)(3m-3)...(4)(3)}$

Just to clear things out: Taking $y(z)=z^r\sum _{k=0}^{\infty } C_kz^k$ in this case where $z=0$ is not a regular singular point is WRONG or does it just complicate things? I am asking because in my original post I did find out that $r_1=0$ or $r_2=1$. I can see no reason why would taking $r=0$ bring me to any different result than we just got.

THANK YOU!

9. Apr 29, 2014

### LCKurtz

I suspect the answer is that it just complicates things. I didn't try working it all the way through the way you started, but since r=0 should give the answer, I'm guessing that the $r=1$ case probably gives the same answer. The reason I say that is that since $c_0=0$, then $x$ is a factor of the solution and that's why $r=1$ works. But if $c_0\ne 0$, I'm guessing $r=1$ wouldn't work. I haven't checked that all out though, so don't take this answer too seriously.