00:01
Here is the initial value problem that we are going to solve.
00:06
So we first write the differential equation as y prime plus 1 over x plus 1, x minus 2, y equal 0.
00:20
Then we look at this function, the coefficient of y.
00:27
So we write the star as partial fraction.
00:32
The reason we write it as partial fraction is that we want to integrate this.
00:45
So let's find the integration factor.
00:49
So it is a over x plus 1 plus b over x minus 2.
01:03
So by comparing coefficients we have a plus b equal 0 and negative 2 a plus b plus b is so this implies that a is equal to negative 1 over 3, b is equal to 1 over 3.
01:32
So this is 1 third 1 over x minus 2 minus 1 over x plus 1.
01:48
Next we integrate star, so we integrate 1 over x plus 1.
01:53
1 x minus 2 d x this is 1 over 3 log absolute value x minus 2 minus 1.
02:16
We can write it as 1 3 log absolute value x minus 1...
