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So we have seen that it's pretty easy to use the command solve underscore IVP to solve differential

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equations.

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So in this case, we didn't have to define any function ourself.

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We did not have to define the Arla function.

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And we could even choose a method AKI 45, which seems to be superior compared to the Euler methods.

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And in fact, it really is much better because you need fewer points.

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The point size will rtd you judge to step with step size will be chosen accordingly, and it adapts

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to the change of the function to get you always the good result, with only a few points for the calculation.

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So it's really the best of both worlds.

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And what we haven't done yet is looking at higher order differential equations.

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So I want to show you here we can also solve second order differential equations like the free fold.

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So I copy this and let's change it to our needs.

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So what we need is we need to go to zero from zero to 10, and we need to provide the starting values

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for why zero and zero.

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So all rides, why zero is 10.

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I think this is what we chose previously and V0 is 50 and now we only need to provide the function.

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And for this, I will scroll up to our example where we have considered the free fall because I want

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to show you the very important difference.

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So of course, I could have just written the function myself, but I want to show you really what's

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important to make sure that you changed this.

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So I copy this from our previous example and go back down to our where we just left.

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Um, so let me paste it here.

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So the thing is, our function can be called f only e.

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This is good.

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And previously we had provided two arguments why zero and why?

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One?

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Because this is how we have programmed our Euler methods.

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We had two separate variables y zero and y one.

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And our output was only the value of the function.

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However, in order to fulfill the syntax of solve IVP, we must provide the arguments in a different

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way and we must provide the output in a different way because here y is a list like two starting values.

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So this was the starting value.

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So like the starting values, also, the arguments themselves are lists.

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So it's just a single is called Y, and it will have two elements in this case.

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And this we have to make sure that we respect here for the return function.

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So we have to write the minus g.

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This is good.

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But then we have to also provide the other variable and this will just be the velocity.

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And this is actually pretty similar to our trick that we have used before.

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Let me briefly scroll up to the theory part.

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So in your notebook, you can stay where you are.

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Um, let me just go here.

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But we have introduced a new variable.

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For example, Z one is equal to the first derivative of Y.

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And this is exactly what we are going to provide here.

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The one variable will be the first derivative of Z.

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One is equal to the function, which in our case is minus G.

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And then the other variable will just be equal for the derivative will just be equal to see one.

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This is what we right here see one or in this nomenclature that is used here, it's just y one.

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So to be honest, it's not 100 percent necessary that you understand really all the details that go

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on here.

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Just remember that when you try to use T solve IVP methods to solve a second order differential equation,

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you must provide a function that has to return arguments.

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And so this one would be the actual function.

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And then you just write y off one.

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And if it's higher order function, then you use more terms and you write before you're y to y three

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and so on and so on.

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So I think now we have a good solution here and we can now go ahead and plot this.

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So I will use the same command as before, but without an analytical solution.

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This time I will just right here y.

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And now, of course, I have to provide which of the two ys, I want to have zero or one.

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And I want to have to coordinate, so I want to have zero.

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So this is this.

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Solution, and to be honest, it looks pretty confusing, so let's check if it is OK and we will do

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this by copying the analytical solution from before, so I would just go up here.

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Here it is.

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Copy this.

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Go back down and copy it here.

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And let's see if it's correct.

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Yes.

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It seems to be correct, but for some reason we went way out of range here, so let's just go to 10

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and then in steps of zero point one.

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Yeah, exactly like this.

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So you see, in fact, the solution from during a quota once again got it correct.

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But a lot of points are missing here, which is not not not a problem at all.

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If you just want to have the ending value here, but if you want to have to hold your victory, then

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of course, it's pretty bad that there are no points here.

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But still, overall pretty remarkable that this method is able to solve this problem by only using five

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points for the propagation.

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So what you can do to at this point is just to right, as we did before this one here to evaluate.

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And then just make sure to get here to correct ranges.

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So we want to go to 10 and and Max was one on one.

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So you see, now we have a smooth curve and we have all the points that we wanted to have.