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The partial differential equation $c^{2} u_{x x}-u_{t t}=0$ where $c$ is a constant is called the wave equation. Show that $u(x, t)$ satisfies the wave equation for the given value of $c$.$$c=2, u(x, t)=4 e^{2(x-2 t)}$$

Calculus 3

Chapter 6

An Introduction to Functions of Several Variables

Section 2

Partial Derivatives

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12:15

In calculus, partial derivatives are derivatives of a function with respect to one or more of its arguments, where the other arguments are treated as constants. Partial derivatives contrast with total derivatives, which are derivatives of the total function with respect to all of its arguments.

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The partial differential e…

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The wave equation $c^{2} \…

So in order for us to show that this satisfies the wave equation, we're first going to need to go ahead and find the partial book Perspective X or the second with perspectives as well as the second partial derivative Perspective team. And then just plug those into that equation of top there. And before we do that, I'm gonna actually rewrite this a little bit. Because remember, when we have, um, something to the power and they're being, like, added or subtracted. We could actually rewrite this as the multiplication of the base of those. So all right, this is for e to the two x times e to the negative to t like that. And the reason why I want to do this, then I don't have to really think about like, actually, this is a four here. Not too, so yeah, again. The reason why I want to do this, because it would just make the derivatives a little bit easier. Um, toe work with. Okay, so let's go ahead. And first you the partial, but respect X. So I'll do that over here. So the e to the negative 14, which is a constant. So would be four e to the negative 40 and then we take the derivative E to the two X so that would be e to the two X and then chain rules of times too. And if we were to kind of multiply that together, that would be eight e to the negative 14 times Eat two X And now we can go ahead and take the partial of this again. So we get Actually, these aren't FC's. We're supposed to be used here, so u sub x use of double X and then we just do the same thing. Think upfront is just a constants. We pulled out. Take the derivative E to the two x b e to the to exchange rule says multiplied by two. And so that is going to give us 16 e to the negative 40 e to the two X Right now, we could go ahead and do the same thing, but with the partial but respect Why? So I'm just go ahead and change this out. Cordell Beidle Why are not y t getting all the letters mixed up today? It seems s o b u t is equal to So now we treat The two X is a constant would be four times E to the two X and then over here e to the negative 40. So again, chain rule so derivative of the inside would just be negative for multiply those together negative 16 e to the two x e to the negative 14. Now, we could go ahead and take the partial of this again So we get u T T is equal to so negative 16 e to the two x times the derivative this so e to the negative. 14 driven on inside negative four And then we multiply all that together and that should give us 60 force. We have 64 um, e to the two x e to the negative 14. That will be our second partial derivative there. So now let me actually go ahead and write out what that was again. So we had c squared you x x minus U t t. And now let's go ahead and plug. These s o. R C was too. So we'll go ahead and plug that in. So this is gonna be equal to so two squared times 16 e to the negative 40 e to the two X and then minus 64 e to the two x e to the negative 14. And, well, that would give us 64. Here we have negative 64 with the same thing. So those just cancel out and so we end up getting this is zero. So it does look like this satisfies the wave equation with a sea of being too.

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