Solving a Frobenius Equation: Finding a Regular Point at x = 0

[01jbell]

hey i am stuck on this question for my ode course its using frobunius

4. show that the equation

yⁱⁱ + 1/x yⁱ + (1-1/(4*x^2))y = 0

has a regual point at x=0
using the method of frobenius assuming a solution of the form

y=[tex]\sum[/tex] a_r x^c+r

show that the idical equation is c^2=1/4


thanks for nay help given

 

[hunt_mat]

What have you managed so far?

 

[01jbell]

i have proved that x=0 is a singular point and i rearanged the ode to get

x^2 yⁱⁱ + x y^ⁱ + x^2 y -1/4 y =0

however trying to work out coeff of x^r , x^r+1 etc etc is the probelm because i get

a₀*(r^2-0.25) = 0
a₁*{(r+1.5)*(r+0.5)}=0

and i thought u would have ot get a a₀ in the equation for a₁

 

[vela]

The r's in your equations should be c's, but otherwise they look okay.

If you solve for the coefficients for r≥2, you'll see they depend on the either a₀ or a₁.

 

[HallsofIvy]

Because this 0 is a "regular singular point" for this problem, you cannot use the standard power series. You will have to use "Frobenious' method"- try something of the form
[tex]y= \sum_n a_nx^{n+c}[/tex]

Choose c so that a₀ is NOT 0.