sturm-liouville problem?

Question

Hello, I have a SL problem and I have no idea how to solve it; could someone be kind enough so to tell me how???? please give me a very good explanation about how to solve it because that is not my only one SL problem.
y''+Ly=0 (where L is the greek letter lambda)
y(0)=0
y(1)=0

please help, I know this one is very easy, but I have much more difficult ones so give me an easy explanation please!!!!!!

Answer 1

So y’’=-Ly; Just multiply both parts by y’, then y’y’’=-Lyy’ or (y’^2)’ = -L*(y^2)’, hence y’^2 = -L*y^2 +c^2, c is some constant and c^2 > L*y^2; thence y’ = c*sqrt(1-(L/c^2)*y^2) splittable equation,
i.e. y’/sqrt(1-(L/c^2)*y^2) = (c*t)’; assume sin(u) = (sqrt(L)/c)*y, then y’ = (c/sqrt(L))*cos(u)*du; and the equation is: (c/sqrt(L))*du = c*dt,
hence u = sqrt(L)*t+f, or sin(u) = (sqrt(L)/c)*y = sin(sqrt(L)*t+f) or y = (c/sqrt(L)) * sin(sqrt(L)*t+f); now as y(0)=0, then 0=sin(sqrt(L)*0+f), hence constant f=2pi*k, being k any integer number;
now as y(1)=0 (and did you mean it?!), sqrt(L)+2pi*k = 0, hence your L can be only fixed values L=(2pi*k)^2; thus y(t) = (c/(2pi*k) * sin(2pi*t)

Answer 2

this is easy
I see that sin'' = -sin
so your solution will look like Asin(bt)
Asin(bt)'' = Abcos(bt)' = -Ab^2sin(bt)

so L = -b^2
the rst of thte constants can be found with the boundery conditions

for the complicated ones have a look at http://en.wikipedia.org/wiki/Sturm-Liouville_theory