In this video we'll go through how we can use this green functions formalism to solve a second order differential equation with non constant coefficients. So previously saw how to solve these steps of equations when a to a one and a not for constants. This then we're not going to assume that now they can be functions of X and will change variables from our previous description. We previously used time now going to use X but obviously the same idea holds regardless of what you call your variable. So the idea to solving this type of differential equation with a greens function is your first have to find the greens function which satisfies this different function. This differential equation. This has to be because the greens function is an impulse response function which change the non homogenous term to a delta function. If we're able to solve this equation to find our greens function. Then the solution to our original differential equation of interest will have a solution of this form. So once you find the greens function you just have to integrate it with the original non homogenous term and you get your final solution to your differential equation. We're not going to assume a specific form of f of X we're going to solve this in general mostly by going through how we can solve this differential equation up here. And this is as you mentioned that this is of course subject to the boundary or initial conditions of your problem you're trying to solve. So we can come up with an algorithm of sorts for finding the greens functions. This isn't to say that this is just something you should memorize you should understand the process but then eventually it's just a matter of following those steps. So we first have to look at the case when X is not equal to X prime because then we can get rid of this delta function so we're to put on the perspective of the previous motivating example we're looking at what happens when the force is not being applied. So again when I say to X prime the delta function is just equal to zero and we're left with a homogeneous differential equation for the greens function. And the key condition for greens function is to keep in mind that this is a physical property this represents the response of a physical system to an external impulse. And being a physical property. Or a physical quantity if you will. It must be continuous. So whatever greens functions you find it must be continuous everywhere in particular. What this means is when you look at X is equal to X prime. Then the left hand side has to hold. It has to be the same thing. So this equation still has to be satisfied for whatever greens function satisfies the homogeneous differential equation. So what we're going to do then is integrate both sides of this equation under this condition that G X X prime has to be continuous and build up our solution that way. And we're only going to integrate about X is equal to X prime. And so over a very very small interval about this point and what that means. It is a one of X a not affects a one of X and a two of X are approximately constant over this very very small interval we're going to consider. So integrating both sides and treating our coefficients as constants. Integrating about this point means taking a little bit to the left of X prime so X prime minus epsilon and a little bit to the right of X prime so X prime plus epsilon. And again, we were able to take out the coefficients because we're assuming that they're about constant in this small interval. Recall that when you integrate the delta function, if your domain of integration includes the point I was able to X prime, then this is just equal to one by definition. Just to write it out, this is what we have. This is just a side remark. So carrying out our integration under the condition that G has to be continuous. If your integrator continues function over a very small interval and then take the difference. Then this term over here is approximately equal to zero. Because you're whatever this integral is, you're evaluating at two points that are very close to one another, which means that you're basically subtracting the same number. So epsilon tends to see our books were only looking in the neighborhood of the point X is equal to X prime. This combined with this leaves us with the following expression. Thank you.