There's an integral, every one of those equations we can actually integrate and come up with these conserved quantities. You can think I'm calling him K here, this is not kinetic energy, it's something different. But it's like energy, so it's an integral of motion basically. So it leaves us a waypoint to go to a set of slides. So let's say we have our classic mx double dot = -kx. Anybody know what the integral of motion of this one is? What's the preserved quantity of this differential equation? Spring mass system. What's constant? You did it in the homework with two particles in the spring mass. Part B, what did you have to use to get that relationship? Energy, right. Basically. We know out of this, this is nothing but a spring mass system. So we know that that kinetic energy and the spring potential energy summed up will give us a constant, right. But we know that. Can we mathematically prove it? So I'm going to show you quickly how to integrate these system. And this is a simple linear system. With this other term there, the cubic stiffness has an x cubed instead of an x. But even x cubed is trivial to integrate. I expect you guys to know how to integrate x cubed. In the next homework you will have other terms in there. There was one homework where you end up with a sine theta, but even sine theta isn't that evil. I know you guys know how to integrate sine theta, hopefully. So you can come up with these integrals of motions, all right. So let's see how we can do this. And there's some little tricks. You'd have to write them down to remember all the little steps. So I've got the second ordered derivative of x equals to minus kx. This works nicely. Now I'm going to do a substitution where I'm saying v is simply x dot, right? It's commonly done, like you do in second order system, we put in first order forms and integrate it. That's typically what we do. So this side can be rewritten as dv/dt = -kx. And you can can also say, well, let me multiply this dt Over to the other side. So mass times differential velocities equal to -kx times differential time. This phone, this side is easy to integrate. That's just mdv. The integral of that is just going to a give you v plus the constant. But the right hand side, x depends on time and we don't have an answer for it in this form yet. Is not as easy, so once you get to the stage breaking these dts and dvs up, the trick is you want to multiply both sides times v which is of course dx/dt, right that's what .x means. So if I multiply both sides over with this. Here on the left hand side, I'm going to have mvdv, that's easy. So, right hand side -kx times v and v is dx/dt.dt. These dt's cancel and you end up with nothing but -kxdx, that's it. Now we've got a set of equations we can easily integrate and I'm not doing in between two specific points in time I'm just going to get the indefinite integral. So, if you do the indefinite integral of this, the integral v with respect to v is just going to be v squared over 2 right. So you end up with mass velocity squared over two which is nothing but kinetic energy. The right hand side, the integral of x with respect to x is just going to be -kx squared over 2. And then out of this whole thing you have to do plus a constant. What's this constant physically going to be in this problem? If you look at it the constant mv squared over 2 plus kx squared over 2, what is that? Total energy exactly that's it. That's because it has a physical thing but in a more abstract sense when you have differential equations we can often do this step but you don't get something that's actually in terms of energy it's just an equation constant of whatever these other things are right? So here happens to be total energy but if you go back to these differential equations and do the same steps bring this to the right hand side. Instead of just having a minus A omega I also Also have minus B omega cubed, I did the integrals of those in the end then I'll have an x to the 4th over 4 and an x squared over 2 with the right constants, and an integration constant, right? And that's what we found here, so if you do that step, and I encourage you to practice that once because in the homework you have to use it anyway for one of the solve the homework for, you can get this. So this is the integration constant, you can see this is not mass over velocity squared the units are wrong this is actually velocity rates, this is an acceleration squared, this can't be energy but it's something energy like in that term and this is a preserve quantity. If I get my resulting tumbling motion out of your integration, and every integration times that if you're computing on mega .2 integrated and you have your current omega and you go through this math, you should always get the same k. All right? So this becomes the great integration check for talk free motion as well. And I think one of the homeworks, in the next homeworks said has you numerically stimulate this and validate that this actually, you stimulate this numerically, you compute this analytically with the T's and the H's. And this math should always give you back that same value that everything make sense. That's a good numerical check of what's going on. So this k1, 2's and 3's are now written out in terms of the energies, these your integration constants for the system. So I'm giving you the complete answer here at this stage.