00:01
Okay, so here we have x minus a times x minus b times y prime minus y minus c is equal to zero.
00:09
Okay, and well, the objective here is to solve this given differential equation using the variable separation method.
00:18
Okay, so we can rewrite this as, well, the integral of dy over y minus c equals, well, so i can rewrite the given equation, write as dy over y minus c equals dx over x minus a times x minus b.
00:37
And then once i separate my variables, i might dys with my ys, my dxs with my x's, i can then integrate both sides of the equation.
00:45
Okay.
00:46
That's where we're at here.
00:47
And then, well, we can then write the sum, you can write this as a sum of partial fractions and get the integral of 1 over y minus c, d ,y, equals the integral of 1 over a minus b times, while 1 over x minus a times 1 over x minus b dx.
01:09
Okay.
01:10
Now, since, right, 1 over x minus a times x minus b is equal to a over x minus a plus b over x minus b, then while we can have 1 is equal to a times x minus b plus b times x minus a.
01:31
Now, if we put x is equal to a, so if we let x be equal to a, okay, so let, you know, or put x be equal to a, then what happens is, well, we get 1 is equal to a times a minus b plus b times, well, a minus a because x is equal to a.
02:00
And well that implies, right, that, well, this term is gone...