00:01
Heat equation such as del u by del t equals to 1 by 4 del square u by del x square and the boundary condition as given as ux of 0, t equals to 0, ux of 5, t equals to 0 for t greater than 0 and initial condition as u of x, 0 equals to x square for x varying from 0 to 5.
00:29
Now the most general solution for the heat equation is given as u of x, t equals to c1 cos of mx plus c2 sin of mx into here a to the power minus c square square t.
00:48
Now we are given the value of c square as 1 by 4.
00:53
So the solution for this now differentiating it with respect to x we have ux of x, t equals to minus c1 m sin mx plus m c2 cos mx into a to the power minus 1 by 4 m square t.
01:13
Now we have ux of 0, t equals to 0 the boundary condition.
01:19
So substituting the value we can have 0 equals to 0 plus m c2 a to the power minus 1 by 4 m square t.
01:27
So from here we can have the value of c2 as 0.
01:30
As for the non -trivial solution m cannot be equal to 0 and a to the power minus 1 by 4 m square t can also not be equal to 0.
01:40
Now we get the solution as u of x, t equals to c1 cos of mx into a to the power minus 1 by 4 m square t.
01:53
Now differentiating it we can have ux of x, t equals to here minus m c1 sin mx a to the power minus 1 by 4 m square t.
02:08
Now using the boundary condition that if ux of 5, t equals to 0 we can have ux of 5, t equals to minus m c2 sin 5m into a to the power minus 1 by 4 m square t equals to 0.
02:28
So from here we can have the value of here it is c1, the value of c1 cannot be equal to 0 and m can also not be equal to 0.
02:41
So equating sin 5m equals to 0 we can have here sin is 0 for sin n pi so we can have 5m equals to n pi.
02:54
So from here we can have the value of m as n pi by 5.
02:59
Now we get the value of m...