00:01
Number 39 from chapter 2 .6.
00:03
So we're asked to solve the differential equation given in problem number seven, which is written here.
00:13
Okay, so let's just rewrite this in terms of dydx.
00:26
So all we're doing is dividing both sides by cos x plus y and we get sine x plus y over we're cos x plus y, which is tan of x plus y.
00:41
Okay, so this is dying out for the substitution, or calling out for the substitution, z is equal to x plus y, in which case, dz, dx is going to be equal to 1 plus dy, dx.
01:15
So we can rewrite this difference of the equation as dz, dx, minus 1 which is equal to dydx which is tan of z.
01:51
So this is a separable differential equation.
01:57
So we can write it as you can see clearly that we can put this one on this side and we'll get dz, dx is equal to 1 plus tan z.
02:11
We can then integrate both sides.
02:14
So we're going to get 1 over 1 plus tan z, dz, and that's equal to the integral of dx plus a constant of integration.
02:41
Okay, to do the integral on the left, if we let u equal tan z, then du d z is equal to the integral of tan z, which is sex squared z, and there's a common identity that sex squared z is equal to 1 plus tan squared z, which is just 1 plus u squared.
03:34
Okay, so we can write this left -hand integral as 1 over 1 plus u times 1 over 1 plus u squared d u because this equation here has told us that d z is equal to one over u squared the u.
04:08
The right side's dead easy that's just going to integrate to x plus a constant c.
04:16
Okay, left -hand side.
04:21
I won't go through the details here, but we can use partial fractions to show that 1 over 1 plus u times 1 plus u squared is equal to a half times 1 minus u over 1 plus u squared plus 1 plus u d u that's equal to x plus c we'll take the half out the front and split this up into three fractions okay so now let's integrate each of these fractions separately each of these rational functions separately.
05:36
Okay, the first one is just a standard integral that integrates at 1 over 1 plus u squared integrates to arc tan or the inverse of tan times u.
05:51
The second one if we write two here and divide by two here then this thing that's in the red brackets you can see that the numerator is the derivative of the denominator.
06:12
So we can write that as a half natural log of the denominator.
06:23
Similarly, the third function in the integrand just integrates to the natural log of the denominator.
06:36
Okay, so that's equal to x plus c.
06:48
Okay, so let's times both sides by two...