00:01
So in this problem, we are given an initial value problem, that is y times dx plus x parenthesis, natural log x, negative natural log y, negative 1, parenthesis close, day y is equal to 0.
00:20
And the initial value condition is y of 1 is equal to exponential.
00:26
Fine.
00:27
So we have to solve it.
00:30
Now if you look over here at this given initial value problem, you can see that it is homogeneous.
00:37
So it means that we can make the substitution that x is equal to vx.
00:45
Fine.
00:46
So when we'll take it derivative with respect to y, sorry, it's the vy, fine.
00:59
So we are going to take its derivative with respect to y.
01:03
V plus y t v y d y y d y d y d y fine and next i'm going to shift this d y over here so that i get t x is equal to v d y plus y d v fine next what i'm going to do over here is to put this and this equation over here fine so i'll get y square brackets v dyy plus y dv plus vy plus vyy parenthesis natural log vyyy negative natural log y negative 1, t y is equal to 0.
02:10
The next step is to do some algebra manipulation, so we can get this equation.
02:17
D .y is equal to 1 over v...