00:01
We're going to solve the initial value problem, differential of y with respect to x, or derivative of y with respect to x, equal y squared minus 1 over x squared minus 1, and y at 2 equal 2.
00:16
So in this case, we can separate variables, that is, from the expression or the differential equation given here.
00:27
We can separate this way.
00:30
We can divide by y squared minus 1.
00:33
We get differential of y over y squared minus 1 equal differential of x divided by x squared minus 1.
00:42
This is informally treated because we know the differential of y respect to x is not a fraction.
00:48
But if we do this, we are currently separating the variables.
00:52
And then the integral of differential of y over y squared minus 1 is equal to the integral of differential of x over x squared minus 1.
01:03
So we've got to calculate the integral on the left with respect to y and the right with respect to x, and we see the form of the expression where the integral is exactly the same.
01:12
So we calculate this one, and we will have this readily calculated.
01:16
It.
01:18
So let's say we're going to calculate the integral of differential of x over x squared plus 1.
01:31
That is, we are integrating the function 1 over x squared minus 1 respect to x...