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 $y(6) = 0$, which is equivalent to the following BVP $y'' + lambda y = 0$, $y'(0) = 0$, $y'(6) = 0$. (a) Find all eigenvalues $lambda_n$ as function of a positive integer $n ge 1$. $lambda_n = $ (b) Find the eigenfunctions $y_n$ corresponding to the eigenvalues $lambda_n$ found in part (a). $y_n(x) = $

Transcript

00:01 Hi, so we are given that the differential l of y is equals to negative of y double dash with boundary condition y dash of and y dash of 6 is equals to 0 which is equivalent to the bvp y double dash plus lambda of y is equals to 0 with initial condition y of 0 is equals to 0 and y dash of 6 is equals to 0.

00:28 So in the a part we need to find the eigenvalue lambda n as a function of n greater than equals to 1.

00:35 So let it started with the a part.

00:38 So in the a part if you consider the first case that is if lambda is equals to the k square which is greater than to 0.

00:46 So in that case y double dash plus k square of y this should be equals to 0.

00:51 So let's make an auxiliary equation we get m square plus k square is equals to 0 this will give me m is equals to plus minus iota of k.

01:01 So by using this roots i can write y of x this should be equals to c1 cosine of kx plus c2 sine of kx as the solution of it.

01:13 Similarly this y of x should be equals to c1 cosine of k would becomes under root of lambda because we compute that lambda is equals to k square.

01:23 So it will remains only x plus c2 of sine of under root lambda of x.

01:33 Now let's differentiate this we get y dash of x is equals to minus of c1 cosine will become negative of sine.

01:42 So it will be sine of under root lambda of x plus c2 into derivative of lambda of x.

01:51 So here it should comes under root over lambda.

01:55 C2 under root lambda sine will become cosine lambda under root lambda of x into derivative of lambda x will becomes under root of lambda...

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