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Won the previous two lectures, we have solved the differential equation for the time decay of radioactive

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substances, and you have seen that is a declining exponential function.

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And now we are going to use our newly defined function called Euler O the E to solve a other differential

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equation.

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So this differential equation is given by the change of Y.

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So why that is equal to minus the constant times y so pretty similar to the exponential decay times

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T, which is something new here.

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So basically, I called this time amplified decay, but it's a bit abstract and it's just generic.

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So not really a real physical phenomenon because here to change of the amount of the substance is given

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not only by the substance itself, but also by that time.

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So I don't know.

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Maybe you could imagine that for some reason, the rate of the decay would increase with time because

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some of some temperature changes or something.

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But yeah, this is just a quick example to show you that our method works also for other differential

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equations.

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And the important thing or the important distance difference here is that we now do not only have a

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function y dot that depends on Y, but also it depends on T, which was something that we did not have

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previously.

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But in general, you see here f depends not only on Y but also on T, which is why I decided to present

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to you this example where we have y in T.

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So when you look at this function and don't know the solution yet, then it's much more difficult to

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get a solution.

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So after some thinking, you may guess that the solution is such a Gaussian function, so exponential

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function of minus T squared time so constant.

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But maybe it is quite difficult to solve this analytically.

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So in any case, you could just take our code now and calculate the function numerically.

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So we, of course, we don't need to redefine the code because it is written in a general manner.

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We just need to change the function.

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So I will basically copy this cell here and just adapt to our new function.

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So instead of having to write him minus y, I write minus eight times y times T and A is equal to some

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number.

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And here we should take a pretty small number, maybe the zero point zero one.

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So I take your small value because when we increase T, then the rate will increase.

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So it's good to start with a smaller eight to see some interesting physics here.

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So then we basically just have to provide the T zero, the Y zero, the N Max and the H, and then we

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can just go ahead and solve the all an ordinary differential equation with the same line as we had before.

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So I run this and now it's plotted.

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I will also copy this code here.

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And I think this one, we don't even have to change.

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I can just run it and look at the solution.

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So here we cannot really see that much because we have still our analytical results present.

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So yeah, first of all, I should maybe change the analytical results.

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So I right here y zero times.

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So I have to write this one year terms and it's P minus T square times eight over two.

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And now it looks pretty, pretty decent.

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So it's still decreasing and it's only at zero point nine eight yet.

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So maybe I should increase the total simulation time and I could do this, for example, by decreasing

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or so by increasing the value of H.

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And then I could also increase here the value of and max.

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And then you see, that's first of all, our numerical solution represents the analytical solution quite

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well, even though we have increased the step size.

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And I think this is due to the fact that the function here a y t is pretty small since we have two small

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constant here.

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So you see there is not really a unique or an appropriate best choice for H, which is able to solve

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all the problems equally well.

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So you always have to think about if your function changes fastly or not.

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For example, here it was changing quite briefly, going from one to 0.5 and under a second.

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So when we when we.

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Used equal to zero point one before the solution was not so good and we had to decrease to zero point

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zero one.

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However, here after one second, we are still very close to one, so the function doesn't change too

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briefly, so we get a good result.

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Also, with a larger value of h equal to zero point one.

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So we see we have this nice agreement and also we see this trend of this Gaussian function.

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And I think when you consider this differential equation and put it in the physical context, it makes

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sense.

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So the rate at which the Y value decreases is given by the product of white itself and the time T.

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So if we only consider the influence of T, then the decrease would just get faster and faster with

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increasing time.

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And this is what we see in the beginning.

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So here the value of Y doesn't change much and all the influence comes basically from T, so it decreases

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and it decreases faster and faster.

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But then at some point, the value of Y will be quite small already, so this will then compensate the

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effect of the increasing T.

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And so overall, the decay will slow down and it will still then be very close to zero and it will never

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go below zero.

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So this means we have now tested our function for a new differential equation, which was a bit more

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difficult because it does not only include Y but also T, but in the way we have programmed this function,

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it still worked.

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And it's a nice thing that we have such a function available now and we can now solve basically any

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ordinary differential equation of First Order.