xy''-2y'+3y=0 obtain a power series sol'n

• Apr 29th 2009, 04:13 PM
ynotidas
xy''-2y'+3y=0 obtain a power series sol'n
Obtain a power series solution of the equation

xy'' - 2y' + 3y = 0.

I'm not sure where to go with this.

So i just used y= (n=0)SUM(infinity) cnx^(n+r), n got y' and y''.

subbed it into the equation. and simplified to get:

r(r-3)=0 and C[n+1] = 3Cn/(n+r-2)(n+r+1) = 0
r=0 , r=3

is this Frobenius? with difference of positive integer (3)?
• Apr 30th 2009, 12:26 AM
chisigma
Let's write the differential equation as...

$y=\frac{2}{3}\cdot y^{'} - \frac{1}{3}\cdot x\cdot y^{''}$ (1)

Now we suppose that (1) has a solution y(*) that can be expressed as power sum, so that is...

$y(x)= \sum_{n=0}^{\infty} a_{n}\cdot x^{n}$ (2)

In this case the identity (1) becomes [the intermediate computations are not reported...] the following...

$a_{0} + \sum_{n=1}^{\infty} a_{n}\cdot x^{n} = \sum_{n=1}^{\infty} (n-\frac{n^{2}}{3})\cdot a_{n}\cdot x^{n-1}$ (3)

Now we try to compute the $a_{n}$ from (3) imposing that the coefficients ot the terms $x^{n}$ are identical in both terms. Any solution of (1) is defined unless an arbitrary constant, so that we are allowed to set...

$a_{0}=1$

Equating in (3) the coefficients of the term $x^{0}$ we obtain...

$a_{0}= \frac{2}{3}\cdot a_{1} \rightarrow a_{1}= \frac{3}{2}\cdot a_{0}= \frac{3}{2}$

... all right!... equating the the coefficients of the term $x^{1}$ we obtain...

$a_{1}= \frac{2}{3}\cdot a_{2} \rightarrow a_{2}= \frac{3}{2}\cdot a_{1}= \frac{9}{4}$

... all right!... equating the the coefficients of the term $x^{2}$ we obtain...

$a_{2}= 0 \cdot a_{3}$

... ehmm!... there is some minor difficult in computing $a_{3}$ (Thinking) ...

... the conclusion is that our hypothesis is false and doesn't exist any solution of (1) that can be expressed in power series like (2) (Shake)...

Kind regards

$\chi$ $\sigma$