I'm not sure I quite understand your question but the y=mx+b seen in the video was a specific solution for the case when lambda=0. When lambda=0, the DE to be solved is y''=0, which gives a solution y=mx+b. So it's just for this particular DE that I'm using y=mx+b. Hope that helps!
A few mistakes towards the end. Yn should be C*sin(n*pi), because the L's should cancel out. Also, the L should be squared in the eigenvalues.
forgot the L^2 at the bottom of the general eigenvalue
When y(-l)=0,y(l)=0 then how can we find the eigen values and eigen functions
can we always use mx+b for eigenvalue problems to find if BC apply or not?
I'm not sure I quite understand your question but the y=mx+b seen in the video was a specific solution for the case when lambda=0. When lambda=0, the DE to be solved is y''=0, which gives a solution y=mx+b. So it's just for this particular DE that I'm using y=mx+b. Hope that helps!
how can we find its solution in matlab ??