In this video, I want to then solve the differential equation which will be a linear first-order equation. This is the equation DY, DX plus 2Y equals E to the minus X. We want to find Y as a function of X and we have the initial condition, Y of zero equals three quarters. So, let's remember what the formula is for solving a linear first order equation. So, the standard equation is DY, DX plus P of XY equals G of X. Our example equations is in that form. The first thing you do is calculate the integrating factor. So mu of X is equal to the exponential of the integral from X naught to X of P of X, DX and X naught is the initial value of X that you know the value of Y, and then the solution is Y of X equals one over the integrating factor mu of X times the initial value of Y, Y at X and zero, plus the integral from X naught to X of the integrating factor, times the right hand side G of X, DX. So, that's the formula we derived last time for solving a linear first order equation. So, we just apply this formula here. So, our P is two, so we compute the integrating factor. So, mu of X is equal to E to the integral from zero to X of 2DX. So, that's just equal to E to the 2X. So, that's our integrating factor. Then we can construct the solution. So, Y of X is equal to one over the integrating factor, which is E to the minus 2X, times Y naught, Y of zero is three quarters, plus the integral from zero to X of the integrating factor E to the 2X times the right-hand side, E to the minus X, DX. Then we just have to do this integral. So, this is E to the X. The integral of E to the X is quiet easy. So, this is E to the minus 2X, three quarters. The integral is E to the X from zero to X. So that becomes plus E to the X minus one from the lower limit of integration. This is the solution but we can simplify it. Three quarters minus one is minus a quarter. So, this is E to the minus 2X times E to the X minus one quarter, but if you want to write this in a very more transparent way, you really don't want a growing exponential here multiplied by a decaying exponential. So, you would like to just have, this thing is going to be decaying. So to make this cleaner, you can multiply through by E to the minus X. So we get E to the minus X, and then we have one minus one quarter E to the minus X. I think that's the cleanest way to write this solution. So, let me review this example. If you have a first-order differential equation to solve and you recognize that it is a linear equation, then we have a very nice formula to solve this. So, to solve, you need to compute the integrating factor, E to the integral from X naught to XP of XD, X. You compute the integrating factor associated with that equation. Then you just write down the solution. So, you have to do two integrals, one for the integrating factor and then one for the integral of mu of X, G of X, DX. If you can do those two integrals, then you can get a very nice close form for your solution. I'm Jeff Chasnov. Thanks for watching, and I'll see you in the next video.