00:02
Okay, so for one dimensional heat conduction, we have t -cube t over t x to the v plus i over f equal to zero.
00:12
Now, the second, the p -e using second note, we have t i plus one minus two times t i.
00:22
T i inverse over t x to the v which is equal to i over k here we have four modes different values of i equal to two two one two three two three four at i equal to one t two minus two times three plus t zero is equal to negative seventeen point eight six and two minus two times t one plus t zero is equal to negative 17 .86.
00:53
Let this be the equation first.
00:55
Similarly, at node 2, we have 3 equal to negative 17 .86.
01:00
Let this be the equation 2.
01:03
And at node 3, we have d4 equal to negative 17 .86.
01:10
Let's be the equation 3.
01:14
Now, at node 4, that is, i equal to 4, we have d5 minus 2 times d4 plus t3 equal to negative 17 .86.
01:26
Let this be equation 4.
01:28
Now use concave by context at node 4...