00:02
Okay, so what we have here is an equation that defines the rate of change of the level of hormone in the blood on a 24 -hour cycle.
00:09
So we are given with the values of lambda, which is equal to beta, is equal to 1, k is equal to 2, and x not is equal to 10.
00:19
So we are required to get for the value of x of t.
00:26
So what we're going to do first is to substitute the given values to the equation.
00:36
So by doing so, we now have dx over dt is equal to 1 minus cosine of phi t over 12 minus 2x.
00:53
So rearranging the equation, we have dx over dt plus 2x is equal to 1 minus cosine of phi t over 12.
01:08
And as you can see, this is now a linear differential equation, wherein the dependent variable is x and the independent variable is t.
01:20
So now we want to find the integrating factor.
01:23
So our p of t is equal to 2 and solving for the integrating factor of mu of t is equal to e raise to the integral of 2dt is equal to e raise to 2t.
01:39
So this is now our integrating factor.
01:42
So multiplying both sides by the integrating factor, we have e raise to 2t, dx over dt plus 2x, erase to 2t is equal to e raise to 2t times 1 minus cosine of pi t over 12, which is also equal to e raise to 2t times 1 minus cosine of pi t over 12, which is also equal to e raise to 2t minus e raise to 2t cosine p of t over 12.
02:24
So we know that we can express the left side of the equation as d over dt e raise to 2t x is equal to e raise to 2t minus e raise to 2t cosine of pi t over 12.
02:50
So taking the integral of both sides, we have x, x, e raise to 2t on the left side, and one half, e raise to 2t minus the integral of e raise to 2t, cosine of pi t over 12 dt.
03:11
So this is a complex one, so you can solve this by doing double...