00:01
Hi, here for the given question we are given that we need to consider a road of length l and insulated side is given an initial temperature distribution of fx for 0 less than x less than or equal to l.
00:15
So, here now further we need to find the final distribution of the subsequent times where t greater than 0 both the end of the road are kept at 0 degree celsius that the initial temperature distribution can be defined as f of x we need to find this value.
00:33
So, here in our case we know that here to solve the heat conduction equation for the one dimension can be written as del t upon del t is equal to alpha multiplied with del square t upon del x square.
00:49
Here t of x t is equal to temperature distribution at a position x and at time and alpha is the value of thermal diffusive constant.
01:20
So, here in our case by using variation of separation variables we can say that the solution will be of the form t of x t is equal to x of x multiplied with t of t.
01:32
So, here by substituting the value in our heat equation we can say that this equation can be written as x multiplied with x multiplied with dt by dt is equal to alpha multiplied with t of t multiplied with d square x upon dx square.
01:51
So, here now further on simplifying this we can say that here we have 1 by alpha t of t multiplied with dt by dt is equal to 1 by x of x multiplied with d square x upon dx square.
02:11
Now, here further we can observe that here the left hand side is dependent on t and the right hand side is dependent on x.
02:18
So, here we can say that there exist in lambda square as a constant value and here further this ordinary differential equation can be written as 1 by alpha t of t multiplied with dt by dt is equal to lambda square minus which is equal to 1 upon x of x multiplied with d square x upon dx square.
02:43
Now, here further we need to solve this equation and by solving both of the equations one by one here we can say that for lhs of the equation by integrating we have value of 1 by t dt is equal to alpha square integration of alpha dt.
03:05
So, here by solving this integral we have t of t is equal to c2 e to the power lambda square multiplied with alpha t.
03:15
This is negative value...