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Solve for the characteristic eqn. So r^2+lambda=0 so r1,2= +/- sqrt(-lambda)
Now since lambda can vary you have to do it for 3 cases because each case you would get a different solution to the differential equation. It's basically just solving 3 2nd order linear d.e. Case 1 would be if lambda is < 0 giving 2 real solutions. So since the solutions are real, you solve the 2nd order linear d.e accordingly. Differentiate to get y' since your BV involves y'. Now try to solve for c1 and c2 and lambda using your BV. Now if the only way to satisfie the BV is for c1=c2=0, then the trivial solution is obtained and there is no eigenvalue for that case. Repeat these steps for case 2 where lambda=0 to giving 1 root and case 3 where lambda>0 giving 2 non real roots. You want to find the specific instances where there is a way to satisfy the BV WITHOUT having c1=c2=0. For example unrelated to this problem, you could end up with c1lambda+ c2sinlambdapi and know for sure c1 equals 0 from one of the BV. Then you have to solve for when sinlambdapi=0. So lambdapi=n pi, pi's cancel leaving lambda=n for n=1,2,3... You get the idea.