00:03
Okay, so for this problem, we are asked to solve the initial value problem.
00:06
We have x times d, y, over dx minus y equals x squared, with an initial value of y of 1 equals negative 1.
00:16
We will come back to that initial value later.
00:19
The first thing i want to do is i want to divide by x to every term.
00:24
So i have d, y, over dx, minus 1 over x, y, equals x.
00:30
So i did this in order to isolate d, y, over dx.
00:34
Then what i want to do is i want to find my integrating factor.
00:39
So i'm left with e to the derivative of negative 1 over x dx.
00:43
And yes, that negative is very, very important.
00:46
So then the next thing i want to do is it's going to be, so if i look at this, i can do negative ln of x, which is also going to be the same thing as e to the ln of 1 over x, which is just going to give us 1 over x.
01:06
So the next thing i want to do is i want to go ahead and multiply the whole equation.
01:10
So 1 over x, d .y over dx, minus 1 over x times 1 over x, y equals x over x.
01:25
So what i'm going to be left with is 1 over x, d, y over dx minus 1 over x squared y equals 1...