Let's go on and solve that equation. Now this is using a, standard technique in solving differential equation and let's just walk through this. It's really not a complicated thing. So here's the differential equation I need to solve. Q is a function of time and that's the function that I'm looking for. R and C and V0 are all constants of course. Now the initial condition like we said is that Qt equals 0 is 0. Now to solve this what I need to do is write this as two different equations. The first one. I'm going to just write the same equation again but I'm just dividing through by R. And so this is what we have. this is what's called the in homogeneous equation. The in homogeneous differential equation. It has the function, derivatives of the function and the function itself that we're looking for on the left hand side. And there is some kind of a constant term here on the right-hand side. Now there's an associated so-called homogeneous equation to that where the right-hand side is just 0. Now there's kind of a trick to solving a differential equation like this, which if you give it some thought really is a logical thing. And the first time through may seem a bit puzzling to you. But what you have to do is following. The first thing you do is find any particular solution to equation one. So I just have to find A function Q that will solve this equation. And then what I'm going to do, well let's do that here. So if I were to say okay let's assume that Q particular is just going to be a constant. Because then I know that dQ dt will be 0 and I can ignore the first term. And so I'll say okay, I'll just in the second term, I have 1 over RC Q particular is V0 over R. So a solution to the inhomogeneous equation, the original equation is QP is CV0. So I just need to find any old solution. Now what I need to do is go back to the homogeneous equation and find the general solution to that equation. Now that easier because I don't have a terminal on the right hand side. And so the solution to that is going to be that exponential function. Source at e to the minus t over RC. Now I know when I take the derivative of this I get a minus 1 over RC times the same thing and if I plug that in the minus 1 over RC. And the plus 1 over RC will cancel out and this equation will be satisfied. Now I can also multiply by some constant A because that would be carried along in both terms and I need to put that in. This is so called integration constant. So here is the general solution. Now if you, this is going too fast and you didn't follow it, just take a minute now and plug this in to this equation and then prove that that equation really does add up to 0. Now what you do, here is the trick is the solution of this full equation here, up here is going to be the particular solution that you found plus the general solution. So here it is. So here's the particular solution, here's the general solution and that's it. So now there's one thing left to do. A, the so called integration constant is has to be determined and the way we find that is we use the initial condition and the initial condition is that q, this function q at time 0 has to be 0. So if I take and I say okay , I'll plug t equals 0 in here. So let me write that down. So if I plug t equals 0 right there then this whole term becomes a 1, e to the 0 is 1. So then I have CV 0 plus a has to be q of t at t equals 0 is 0. So that means that a then has to be minus CV0. So this whole equation then is I can factor the CV0 out of the first term. And then the second term has a minus CV0. So there is the general solution of that equation. Now the voltage as a function of time is just the charge divided by the capacitance and so here it is. So what happens in this circuit is the, the I start off at t equals 0. So this is time along this axis and I'm dividing it by RC. So RC is this number that appears and it's called the RC time constant of this circuit. It has the units of time. And so I'm going to divide time by RC. I'm just normalizing it. And what I'm plotting on the vertical axis is the capacitance voltage divided by V sub 0. So it's like taking dividing by V sub 0 on both sides putting this down here. So I'm just plotting this factor. So this is the factor that's being plotted over here. Now that starts off at 0, of course and eventually it reaches 0. This factor goes to 1. When time goes to infinity when t becomes a very large number, I have e to the minus something large that e to the minus infinity goes to 0. And so this whole term vanishes when t gets large and VC just goes to 0 so CV over V0 because 1. So this shows how the compacitor charges up over time and it's this 1 minus this exponential function. Now and as I just said the final stage of course is V0. Now the at t equals RC so the ratio of t over RC is 1. At that point if I plug that in here t equals RC, I have e to the minus 1 and 1 minus e to the minus 1 is about 0.63. And so if I write this down here, is 1 minus e to the minus 1. That's what happens when t equals RC and this factor becomes about 0.63. So after one RC time constant, the capacitor is charged up to about 63% of the final voltage that it's going to reach. By five time constants, it's pretty much charged all the way. In fact I'll give you a little problem to figure out what is this factor when t is 5 times RC.