I HATE trying to read those Jpg things!
In the first problem, you have
y"+ \lambda y= 0
with boundary conditions y'(0)= 0, y(0)= 0.
You first try \lambda< 0 and determine that there are no non-trivial functions satisfying that so no negative \lambda is an eigenvalue.
You then try \lambda= 0 so your equation is y"= 0 which has general solution y(x)= C1x+ C2. y'= C1 so the boundary conditions give C1= 0. But C2 can be anything so there exist non-trivial solutions of the form y= C. That is,
0 is in fact an eigenvalue. Since the set of all such solutions is an "subspace" of the space of all functions, you can find a basis for it. The simplest is y= 1 so that y= C= C(1). Actually, any number (except 0) would have worked.
If \lamba> 0, then you get general solution
y= C_1 cos(\sqrt{\lambda}x)+ C_2 sin(\sqrt{\lamba x})[/itex]<br />
y&#039;= -C_1/sqrtr{\lambda}sin(\sqrt{\lambda}x)+ C_2\sqrt{\lamba}cos{\lamba x})<br />
The boundary conditions give<br />
y&#039;(0)= C_2= 0<br />
and then <br />
y&#039;(L)= -C_1\sqrt{\lamba}sin(\sqrt{\lambda}L)= 0<br />
which says that \sqrt{\lamba}L) must be a multiple of \pi and so we must have <br />
\sqrt{\lamba}= \frac{n\pi}{L}<br />
so the eigenvalues are of the form <br />
\frac{n^2\pi^2}{L^2}= \(\frac{n\pi}{L}\)^2<br />
for n an integer (that, by the way, includes 0 which you had already found). That means that the &quot;eigenvectors&quot; or solutions are <br />
C_1 cos(\frac{n\pix}{L})<br />
Again, any multiple of a single solution is also a solution so we might as well take C<sub>1</sub>= 1.<br />
The way you&#039;ve written it, you have \lamba_0= 0, y_0(x)= 1 separate but you could as easily take <br />
\lamba_n= \(\frac{n\pi}{L}\)^2 y_n(x)= cos(\frac{n\pix}{L})<br />
with n= 0, 1, 2,... (since cosine is an even function, we don&#039;t get anything new taking negative values for n). <br />
<br />
I think your basic question in both problems is why, after having gotten a non-trivial solution, involving an undetermined constant C<sub>1</sub>, do they then drop the constant. The answer is that the set of all vectors satisfying the equation Ay= \lamba y is a subspace of the set of all vectors and so there exists a basis for it. In particular, in these problems the &quot;eigenspaces&quot; are 1 dimensiona. Since we want a solution y<sub>n</sub> such that any solution can be written y= Cy<sub>n</sub> for some number C, and we already have y= C<sub>1</sub>f(x) we might as well take C<sub>1</sub>= 1 and use that. In fact, we could have as easily let C<sub>1</sub> be any non-zero number.
