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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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Lectures

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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04:08

The partial differential e…

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The wave equation $c^{2} \…

So if we want to show this is a solution to the wave equation, Um, all we need to do is go ahead and find the second partials with prospective ex ante and plug it into here. Onda show what's equals zero. So let's go ahead and do that to start. So if I were to first take the partial of this with respect to X, we get use of X. Actually, before we do that, I'm going to rewrite this little bit. Use a little bit of algebra, so would be four times E to the two x times e to the four t. And I just want to do this because we'll make it a little bit easier. Thio kind of see what these derivatives should be. So now remember, going to treat this eats the 40 is a constant, so I could just kind of pull that out front so four times each of the 14 and then taking the derivative of E to the two X So we get to the two X and then change rules is take the derivative inside, which is to So we get eight e to the fourty e to the two X and now we would go ahead, take the parcel of this to get our second full partial. So it'll be the same thing again. Um, we ignore the stuff out front and it would be easy to the two x times through it on the inside, which would just be, too, which is going to give us 16 e to the fourty e to the two X. So we have our personal there. Now let's come over here and find our partial with respecto or not. Why t so we ignore the four e to the two X since we're assuming that's going to be a constant when we take the partial and we just do derivative each of the 14 b e to the 14 times derivative the inside, which is four. So it be 16 e to the two x e to the 14, and now we go ahead and take the partial of this with respect to T. So that is a second full partial and then same derivative, essentially, except we have a 16 out front. We'll try all that together 64 e two x E to the 14. So now we just need to come up here and plug all that in, and we hope we get zero. So let's write that out again. So C squared u x x minus U T t is equal to zero. And actually, that's assuming what we want. You plug these and so C is too so to square this four. So four times 16 e to the fourty e to the two x minus 64 82 the two x e to the 14 and then four times 16. 64. We have the same terms there. So these actually just cancel out with each other and we end up getting zero, which checks out for being a solution of the wave equation.

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