Question

I'm having trouble solving the 2nd order ODE below. My trouble is that the indicial equation has double roots r=0. When I'm solving it I get y1=-Co which yeilds Co=0 and y2=0 I think I must be doing something wrong. Please help

xy''+(1-x)y'+y=0

Answer 1

btw I've never solved with frobenius equation with equal roots

Answer 2

Converting this into appropriate Frobenius form, we get
\(x^{2}y'' + (1-x)xy' + xy = 0 \text{ where p(x) = 1-x, q(x) = x}\)

\(p(0) = 1, q(0) = 0\)
Indical equation is \(r^{2} = 0\) which gives equal roots \(r_{1} = r_{2}=0\). So only "one" Frobenius power series solution exists

Substituting \[\Large y = x^{r} \sum_{n=0}^{\infty}a_{n}x^{n}\] to get one Frobenius series. Then use reduction of order to get the other solution.

Working on the Frobenius series at the moment,

Answer 3

Just a side note: Do I have to multiply through by x at the beginning?

Answer 4

Thanks for your help

Answer 5

To be honest, I have no prior experience with Frobenius. I had left it out when it was first taught. I'm trying to read my prof's notes again. Linking it for you
http://home.iitk.ac.in/~sghorai/TEACHING/MTH203/ode14.pdf

About multiplying x, I dont know.

Answer 6

Thanks. My professor is a space cadet. I've been trying to find some good notes and/or worked examples but its kinda hard

Answer 7

I find that my recurrence relation is \[C _{n+1}=C _{n}(n+r+1)/(n ^{2}+2nr+2n+2r+r ^{2}+1)\]
so I plug in r=0, then I solve for my constants C1=-Co, C2=0, then everything is zero.

Answer 8

for reduction of order p(x) = (1-x)/x

\[\Large y_{2}(x) = \int\frac{1}{y_{1}^{2}}e^{-\int p dx}dx\]\[\Large y_{2}(x) = \int e^{2x}.\frac{1}{x}e^{x} = \int \frac{e^{3x}}{x}dx\] .. which is non-integrable .... Did i make a mistake anywhere ?

Answer 9

I don't get your a(n) equation. I'm assuming its the supposed to be the same as my C(n)

Answer 10

Isnt the recurrence \(C_{n+1} = -C_{n}/(n+1) \) ? So how do you get \(C_{2} = 0 \)

Answer 11

It is the same thing

Answer 12

Here's my work

Answer 13

Sorry I made a mistake \[C _{n+1}=C _{n}(n+r-1)/(n ^{2}+2nr+2n+2r+r ^{2}+1)\]

Answer 14

when n=1 and r=0 the numerator goes to zero

Answer 15

Ah yes, you are correct .. so taking \(C_{0} = 1\), we get the solution to be
\(y_{1}(x) = 1 - x\)

The other solution can be obtained using reduction of order using \(p(x) = (1-x)/x \)

Answer 16

Ahh yes I see my y1 was wrong. I don't know about this reduction of order. The method Ive seen to solve y2 is to use \[y _{2}=y _{1}\ln(x)+\sum_{n=0}^{\infty}b _{n}x ^{n+r}\] So the same but add this ln term

Answer 17

Which I will try with my new corrected y1=Co(1-x)