Find the indicial equation?

Question

Find the indicial equation, the recurrence relation, and the roots of the indicial equation for 2xy"+y'+xy=0.

Answer: r(2r-1)=0; a_n= -(a_n-2/((n+r)[2(n+r)-1]; r₁=1/2, r₂=0

I used the substitution y=x^r, y'=rx^r-1, y"=r(r-1)x^r-2.

2x(r(r-1)x^(r-2))+rx^(r-1)+x^(r+1)=0

 

Answer 1

Hello Sun,

These series solution type problems can be a bit annoying, since you have to be very careful not to make a mistake with the indices of summation.  Now, you don't want to assume a solution y=x^r; this form is reserved for the Euler equations themselves.  Here, we are considering a slightly more general equation---one with analytic "coefficient functions".  In other words, we are assuming that an equation of the form

P(x)y'' + Q(x)y' + R(x)y = 0

has, in the neighborhood of a regular singular point, a solution that is almost Euler, if that makes any sense to say.  In fact, we are assuming that the solution has a power series representation (with a₀ ≠ 0) of the form

y = ∑₀ a_nx^r+n

where the summation is over n.  We want to determine for which values r such a solution exists, where such solutions are valid, and also, if possible, a recurrence relation among the coefficients in the power series so that we can take our representation out to as many terms as we desire with minimal cost.

Note that the derivatives are easy to compute:

y' = ∑₀ a_n(r+n)x^r+n-1 and y'' = ∑₀ a_n(r+n)(r+n-1)x^r+n-2.

Your equation is

2xy'' + y' + xy = 0.

Plugging in the above, we get

2∑₀ a_n(r+n)(r+n-1)x^r+n-1 + ∑₀ a_n(r+n)x^r+n-1 + ∑₀ a_nx^r+n+1 = 0

which we can rewrite, taking out the first two terms from each of the first two summations and rewriting the last summation to begin at 2, as

a₀( 2r(r-1) + r )x^r-1 + a₁( 2(r+1)r + (r+1) )x^r + ∑₂ [a_n( 2(r+n)(r+n-1) + (r+n) ) + a_n-2]x^n+r-1 = 0.

Now, the first term, containing x^r-1, is the dominant term, and its coefficient must equal zero in order for the left-hand side to be zero for all x; therefore, we must have

a₀(2r(r-1) + r) = 0     =>     r(2r-1) = 0,

an equation which is called the indicial equation.  The roots of this equation, r₁ = 1/2 and r₂ = 0, are called the exponents of the equation.  You use each of these to write the recurrence relations in terms of n only.  The recurrence relation in full generality comes from setting the coefficient on the general term with x^n+r-1 equal to zero:

a_n(r+n)(2(r+n) - 1) + a_n-2 = 0    =>    a_n = a_n-2/[(r+n)(2(r+n) - 1)].

There is still a bit to do to find the solutions, but your question doesn't seem to ask for the solutions, and I haven't the time to complete the presentation anyways.

Regards,

Hassan H.