Frobenius method for a differential equations

[xicor]

Homework Statement

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

Homework Equations

f(x) = a₀ + a₁x + a₂x² + a₃x³ + a₄x⁴

f'(x) = a₁ + 2a₂x + 3a₃x² + 4a₄x³

f'(x) = 2a₂ + 6a₃x + 12a₄x²

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?

 

[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##.

 

[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 a_i 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 won't 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

 

[LCKurtz]

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