00:01
For this question, we're asked to solve this initial value problem, and i'm going to use the integrating factor method to do that.
00:06
So first, let's calculate the integrating factor, i of x.
00:11
So this is equal to e to the integral of whatever function is being multiplied by y.
00:17
And right now, nothing is, so we just have 1 times dx.
00:23
And note that we're only able to do that because we just have y prime here.
00:28
For instance, if there was an x out in front, then we would have to divide this whole equation by an x, so that the coefficient of y prime is just 1.
00:36
And in our case it is, so this is the integrating factor.
00:40
It's just e to the x.
00:42
So then we take this equation and multiply each term by e to the x.
00:47
So we have e to the x times y, and this is all equal to 4e to the 2x.
00:56
And the reason why we did that is because now this left -hand side, this is the derivative with respect to x of e to the x times y.
01:07
And this is equal to 4e to the 2x.
01:11
And also it's a good mental exercise to verify that this quantity is actually equal to this...