# Frobenius method for a differential equations

1. Feb 12, 2013

### xicor

1. The problem statement, all variables and given/known data

The function satisfies the differential equation f''(x) = xf(x) and has boundary conditions

f(0) = 1 and f'(0) = 1

Use Frobenius method to solve for f(x) with a taylor expansion of f(x) up to the quartic term a4x4

2. Relevant equations

f(x) = a0 + a1x + a2x2 + a3x3 + a4x4

f'(x) = a1 + 2a2x + 3a3x2 + 4a4x3

f'(x) = 2a2 + 6a3x + 12a4x2

3. The attempt at a solution

What I have done so far is set the differential equation to zero and applied the series to f''(x) and xf(x) which gave an series of a0x + a1x2 + a2(x3 +2) + a3(x4 + 6x) + a4(x5 + 12x2) = 0. I then set up equations for each power of x which is equal to zero. From the x powers I get a2 = 0, a3 = -a0/6, and a4 = -a4/12. Applying the boundary conditions shows that a0 = 1 and a1 = 1. When I use the values in the equation, the fractions within each power of x cancels each other out and you are left with no series and just a trivial statement of 0 = 0. What am I doing wrong in trying to find a series that describes the function produced by the differential equation?

2. Feb 12, 2013

### LCKurtz

Hard to say where your error is without seeing more steps. I get results like yours but without the minus signs. I get $a_0 = 1,\, a_1 = 1,\, a_2 = 0,\, a_3= 1/6,\, a_4=1/12$ so either you or I have signs wrong. But either way, that won't give you a zero series when you put them in $f(x) = a_0+a_1x +a_2x^2+a_3x^3+a_4x^4$.

3. Feb 12, 2013

### xicor

Well, for example when I do the powers for x, I get the equations

x0: 2a2 = 0 → a2 = 0
x1: a0 + 6a3 =0 → a3 = -a0/6 = -1/6
x0: a1 + 12a4 = 0 → a4 = -a1/12 = -1/12

So when you get the values for the ai constants, you are suppose to apply them to the series for f(x) and not the differential equation? I think my problem is that I applied it to the differential equation and got a series such as:

2a2 + x(a0 + 6a3) +x2(a1+12a4 ) +...

where each fraction in each x power cancels out, whereas they wont if I just apply them to the series for f(x).

For the series f(x), I get 1 + x - (1/6)x3 - (1/12)x4

Last edited: Feb 12, 2013
4. Feb 12, 2013

### LCKurtz

I still think you have the minus signs wrong. Remember your equation is $f''(x) - xf(x)=0$, with a minus sign there.