Find the eigenvalues and eigenfunctions for the differential...

Find the eigenvalues and eigenfunctions for the differential operator L(y) = -y'' with boundary conditions y'(0) = 0 and y(5) = 0, which is equivalent to the following BVP y'' + ? y = 0, y'(0) = 0, y(5) = 0. (a) Find all eigenvalues ?n as function of a positive integer n ? 1. ?n = (b) Find the eigenfunctions yn corresponding to the eigenvalues ?n found in part (a). yn(x) =

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00:01 Hello friends in this question we need to find the eigenvalues and eigen function for the differential operators where the given the boundary value problem is y double dash plus lambda y is equal to zero so let this be star so now consider case one where if lambda is equal to zero then star becomes that y double dash is equal to zero then star becomes that y double dash is equal to therefore y is equal to the auxiliary equation is c1 plus c2 x and y dash is equal to c2 so consider the initial conditions that y dash of zero is equal to zero implies c2 equal to zero and y of y of y equal to zero implies c1 plus five c2 equal to zero which gives c1 as equal to 0.

01:05 Therefore, y equal to 0 is the trivial solution and therefore lambda equal to 0 is not an eigenvalue of star.

01:22 Now consider case 2 where if lambda less than 0 but lambda is equal to minus mu square where mu is not equal to 0 then star becomes y double dash minus mu square y as equal to 0 so we write the auxiliary equation as y is equal to c1 e power minus mu x plus c2 e power minus mu x so y dash is equal to minus mu x so y dash is equal to minus mu x so y dash is equal to minus mu c1 e power minus mu x plus c2 mu c2 e power mu x so sorry here it is plus so now y of 5 5 equal to 0 implies that c1 e power minus 5 mu plus c2 e power 5 mu is equal to 0 and y dash of equal 0 is 0 is equal to 0 implies c1 minus c2 equal to 0 so this is because mu is not equal to 0 so this implies c1 equal to c2 is equal to 0 and therefore y equal to 0 is the trivial solution of x and therefore lambda less than 0 is not an eigen value of star.

03:15 Next we consider case 3 where if lambda greater than 0 we put lambda as equal to mu square where mu is not equal to 0 then star becomes y double dash plus mu square y is equal to 0 therefore the auxiliary equation y of x is equal to c1 cos mu x plus c2 sine mu x and then y dash of x is equal to minus c1 mu sine mu x plus c2 mu cos mu x and and so y dash of 0 is equal to 0 implies c2 mu equal to 0 which implies that c2 is equal to 0 and then y of 5 equal to 0 implies c1 cos 5 mu is equal to 0 now for non -trivial solution, assume that c1 is not equal to 0, then cos 5 mu is equal to 0 implies that mu is equal to 2n minus 1 into 5 by 10.

04:58 Therefore, we find the value for lambda n as equal to 2n minus 1, the whole square in...

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