Partial Differential Equations with MATLAB

30 An Introduction to Partial Differential Equations with MATLAB©R

First, note that this is a Cauchy–Euler equation, forx>0. We lety=xr
and determine the values ofrthat give us solutions. So

y=xr⇒x^2 r(r−1)xr−^2 +x·rxr−^1 −λxr=0

⇒r(r−1) +r−λ=0

⇒r^2 −λ=0.

Again, we must consider three cases.

Case 1: λ> 0 ,λ=k^2 ,k> 0
We haver=±k, so our two linearly independent solutions arexkandx−k,
giving us the general solution

y=c 1 xk+c 2 x−k.

Then,

y(1) = 0 =c 1 +c 2

y(e)=0=c 1 ek+c 2 e−k

which imply

c 2 =−c 1

c 1 (ek−e−k)=0.

Sincek>0, the latter implies thatc 1 =0,soc 2 = 0 as well, and we have no
positive eigenvalues.

Case 2: λ=0
Inthiscase,wehavetherepeatedrootr= 0, giving us the linearly inde-
pendent solutionsx^0 andx^0 lnx. So the general solution is

y=c 1 +c 2 lnx.

Then,

y(1) = 0 =c 1

y(e)=0=c 1 +c 2

so, again,c 1 =c 2 =0,andλ= 0 is not an eigenvalue.

Case 3: λ< 0 ,λ=−k^2 ,k> 0
Here we have the rootsr=±ikand corresponding linearly independent
solutions cos(klnx) and sin(klnx). The general solution is

y=c 1 cos(klnx)+c 2 sin(klnx).