# Thread: Finding Frobenius solutions of a diff equation

1. ## Finding Frobenius solutions of a diff equation

Hi,
I am having a little trouble finding the two linearly independent Frobenius solutions of the following differential equation:
2xy''+(2x+1)y'+2y=0

I have got as far as finding out the recurrence relations:
note ak is a with subscript k and a(k-1) is a with subscript (k-1)
CASE 1:
ak=-(2a(k-1))/(2k-1)

In the solutions I have it says the above equation and then hence:
ak=(-2/(2k-1))(-2/(2k-3))...(-2/1)a0
where a0 is a with subscript 0

and also
CASE 2:
ak=(ak-1)/k

The solutions again have the above line and hence:
ak=(((-1)^k)/k!)a0

However I really do not know how in the solutions they have got from each ak equation to the hence part.

Any help appreciated.

2. Originally Posted by BlueEagle
I have got as far as finding out the recurrence relations:
note ak is a with subscript k and a(k-1) is a with subscript (k-1)
CASE 1:
ak=-(2a(k-1))/(2k-1)
.
Let us assume that you got the recurrence relations correct.

$\displaystyle \boxed{ a_n = - \frac{2a_{n-1}}{2n-1} \ \ n=1,2,...}$.

Let $\displaystyle a_0$ be anything.

Thus, $\displaystyle a_1 = - \frac{2a_0}{1} = -2a_0$.

Thus, $\displaystyle a_2 = - \frac{2a_1}{3} = \frac{2^2 a_0}{1\cdot 3}$.

Thus, $\displaystyle a_3 = - \frac{2a_2}{5} = - \frac{2^3 a_0}{1\cdot 3\cdot 5}$.

In general,
$\displaystyle a_n = (-1)^n \cdot \frac{2^n a_0}{1\cdot 3\cdot ... \cdot (2n-1)}$.

Try doing the other one.

3. Ha ha yeah of course! Thanks for the help. I get the other one now.

a1=-a0/1=-a0
a2=-a1/2=a0/2
a3=-a2/3=(a1/2)/3=a1/6=-a0/6

hence ak=(a0(-1)^k)/k!