00:01
So in this question, we are asked to derive equation 15 .5 from equation 30 .4.
00:13
So equation 13 .4 is dn over d t equals r minus lambda n.
00:31
So this can be rewritten as dn or r minus lambda n equals dd.
00:45
Now saying that let's say r minus lambda n is z, then dn would be negative d z or lambda so plugging this in d n is negative d z over lambda and r minus lambda n is z this is equal to dd and integrating this this comes out to be locked to the base e z equals negative lambda c lambda t plus some constant c now plugging the value of z z is r minus lambda n minus lambda t plus constant.
01:58
So this is the solution to the differential equation.
02:02
Now at time t equals zero, the number of atoms n would be zero.
02:10
And hence, plugging these values in the equation, log of r would be equal to c.
02:21
So c is actually log of r...