Introduction 29
Case 2: λ=0
The general solution is
y=c 1 x+c 2
so that
y′=c 1.
Then,
y′(0) =y′(3) = 0 =c 1.
Therefore, the functiony=c 2 survives both boundary conditions, soλ=0
is an eigenvalue. We write
λ 0 =0
with eigenfunction
y 0 =1.
Case 3: λ> 0 ,λ=k^2 ,k> 0
We have the general solution
y=c 1 cos(kx)+c 2 sin(kx),so that
y′=−c 1 ksin(kx)+c 2 kcos(kx).
Then,
y′(0) = 0 =c 2 k⇒c 2 =0;
y′(3) = 0 =−c 1 ksin(3k)⇒c 1 =0unless
sin(3k)=0,that is, 3k=π, 2 π, 3 π,...
or
k=
nπ
3,n=1, 2 , 3 ,....Therefore, we have eigenvalues
λn=n^2 π^2
9
,n=1, 2 , 3 ,...with associated eigenfunctions
yn=cos
nπx
3,n=1, 2 , 3 ,....Example 3Do the same for
x^2 y′′+xy′−λy=0
y(1) =y(e)=0.