00:01
R dot is equal to 0, r dot is equal to 0 that is first derivative dr divided by dt is equal to 0, dr divided by dt equal to 0 which implies r to be a constant say lambda, r to be a constant say lambda.
00:25
Hence our j dot juliet will be equal to ar plus bg which gives you dj divided by dt, dj divided by dt is equal to bj plus a lambda, bj plus a lambda which can be written as dj divided by dt minus minus bj is equal to a lambda.
01:04
Consider this to be equation number 1.
01:07
So, our integrating factor if is equal to e power minus bt, e power minus bt.
01:17
Hence the solution of equation 1 will be in terms of j multiplied by e power minus bt that is y multiplied by integrating factor is equal to integral of e power minus bt that is q multiplied by a lambda multiplied by dt plus the integral constant c1.
01:48
Now, on integrating this we will be obtaining j multiplied by e power bt is equal to minus a lambda divided by b multiplied by e power minus bt plus c1 plus c1 which implies our j is equal to c1 e power bt minus a lambda divided by b.
02:19
Consider this to be equation number 2...