Power series - Differential equation

**Mondreus** · Mar 26th 2011, 01:08 AM

Seek the solutions that are power series to the differential equation:
$(1-z)f'(z)=2f(z),\ f(0)=1$

I assume we first need to assume that $|z|<R$ so that $f(z)=\sum_{n=0}^{\infty}c_nz^n$ can be differentiated.

I have that $f'(z)=\sum_{n=1}^{\infty}c_nnz^{n-1}=\sum_{n=0}^{\infty}c_{n+1}(n+1)z^n$

**Prove It** · Mar 26th 2011, 01:45 AM

Do you really need to use power series?

$\displaystyle (1 - z)f'(z) = 2f(z)$

$\displaystyle \frac{f'(z)}{f(z)} = \frac{2}{1 - z}$

$\displaystyle \int{\frac{f'(z)}{f(z)}\,dz} = \int{\frac{2}{1 - z}\,dz}$ .

Go from here...

**Mondreus** · Mar 26th 2011, 01:55 AM

Yes, the question is in the chapter on power series and it explicitly says that the answer should be a power series. The book hasn't covered complex Taylor and Mclaurin series yet, so even if I know that the answer is some branch of -2log(1-z) I can't use that to derive the power series representation of it.

**HallsofIvy** · Mar 26th 2011, 03:47 AM

Originally Posted by Mondreus

Seek the solutions that are power series to the differential equation:
$(1-z)f'(z)=2f(z),\ f(0)=1$

I assume we first need to assume that $|z|<R$ so that $f(z)=\sum_{n=0}^{\infty}c_nz^n$ can be differentiated.

I have that $f'(z)=\sum_{n=1}^{\infty}c_nnz^{n-1}=\sum_{n=0}^{\infty}c_{n+1}(n+1)z^n$ [/QUOTE]
Yes, that derivative, and the change in index, is correct. Now put them into your differential equation:
$(1- z)\sum_{n=0}^\infty c_{n+1}(n+1)z^n= \sum_{n=0}^\infty c_{n+1}(n+1)z^n- \sum_{n=0}^\infty c_{n+1}(n+1)z^{n+1}= \sum_{n=0}^\infty 2c_{n}z^n$

Now you have to change the index in that second sum. let i= n+1. Then
$\sum_{n=0}^\infty c_{n+1}(n+1}z^{n+1}= \sum_{i= 1}^\infty c_i iz^i$ or, changing the dummy index back to n, [tex]\sum_{n= 1}^\infty n c_n z^n[/itex].

Now, we can write the differential equation as
$\sum_{n=0}^\infty c_{n+1}(n+1)z^n- \sum_{n=1}^\infty nc_{n}(n+1)z^{n}= \sum_{n=0}^\infty 2c_{n}z^n$

Since that second sum does not begin until n= 1, we need to do n= 0 separately. If n= 0 then
$c_1= 2c_0$

For n> 0, equating coefficients of the same powers of z,
$(n+1)c_{n+1}- nc_n= 2c_n$ or $c_{n+1}= \frac{3}{n+1}c_n$

That is, [tex]c_1= 2c_0, $c_2= \frac{3}{1+ 1}c_1= \frac{3}{2}c_1= 3c_0$ , $c_3= \frac{3}{2+1}c_2= 3c_0$ , $c_4= \frac{3}{3+1}c_3= \frac{9}{4}c_0$ , etc.

**Mondreus** · Mar 26th 2011, 04:43 AM

You made some small errors, particulary in the last part. It should be: $c_{n+1}=\frac{n+2}{n+1}c_n$ and finally $f(z)=1+2z+3z^2+4z^3+...+(n+1)z^n$

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