00:01
In this question, we have to find eigenvalues and the eigenfunction of the given initial value problem y dash double dash plus y dash plus lambda y equals to 0 with the condition y0 equals to 0 and y3 equals to 0.
00:20
So, first of all, we can write down the corresponding auxiliary equation of this, which is going to be m square plus m equals to 0.
00:35
Correct.
00:40
This is the first case where we have taken lambda equals to 0.
00:45
That means eigenvalue we have taken as sorry, lambda is 0.
00:50
So, we get this m times m plus 1 equals to 0, two values we are getting 0 and minus 1.
01:00
So, we write down the general solution will be given by y of x equals to c1 e raised to 0 times x plus c2 e raised to minus x.
01:15
This is nothing but c1 plus c2 e raised to minus x.
01:20
Since we have the initial condition y0 is 0, this implies c1 plus c2 equals to 0 and y at x equals to 3 is also 0.
01:32
This is c1 plus c2 e raised to minus 3 is 0.
01:39
So, this is equation 1, this is equation 2.
01:42
I would say from equation 1 and equation 2, we get the value of c1 and c2 to be 0.
01:53
Correct.
01:53
So, we are getting for this eigenvalue lambda equals to 0, our corresponding eigenfunction that is the solution comes out to be y of x equals to 0.
02:08
That is the trivial solution.
02:10
Now, we consider the second case where we take lambda to be some positive number.
02:16
So, we assume lambda equals to k square.
02:22
So, our equation now becomes hence our equation becomes y double dash plus y dash plus k square y equals to 0.
02:35
Again, we write down the auxiliary equation in this case.
02:40
Auxiliary equation, it comes out to be m square plus m plus k square equals to 0.
02:49
In this case, we get m value as minus 1 plus minus under root 1 minus 4k square over 2.
02:59
Again, we get two cases.
03:01
So, i consider here case 1 and here i will write case 2.
03:06
So, when 1 minus 4k square is greater or equals to 0, that means k square less or equals to 1 over 4, that is k is between minus half to half.
03:19
Correct.
03:20
So, what we are going to get in this case? when k is of this type, we are going to get y of x equals to c1 e raised to first root is minus 1 plus under root 1 minus 4k square over 2 times x plus plus c2 e raised to minus 1 minus under root 1 minus 4k square over 2 times of x...