00:02
All right.
00:04
So in this question, we're asked to check if certain functions are solutions to the wave equation.
00:11
Because it's easier, let me just start directly with the second example.
00:18
So a bit of a recap.
00:20
So the wave equation very briefly is a second degree equation.
00:25
So two derivatives in time of a certain function you must be equal to some parameter times the second derivative in time with respect to the variable x.
00:33
Some function u where c is a positive parameter.
00:37
So let's start with the first, the second example, which asks us to check if u of x t given by x squared times t is a solution to the equation.
00:48
So just put a box for it to be clearer.
00:54
And here we just really have to make the calculation.
00:56
So in this case, the first derivative of you with respect to x, we have an x squared.
01:02
So to xt.
01:04
The second derivative with respect to x of the function u 2 times t times x so it's simply 2t and now let's just check the derivatives in time which in this case are very easy first derivative with respect to time is x squared times t so we simply get that x squared now there's no t's in here so the second derivative with respect to time of the function u is simply zero and these are not the same so the conclusion is it not and this is not the solution to the wave equation.
01:37
Now let's do the first example.
01:39
The first example is not difficult per se, but the calculation, these are not only different, but there is no positive constant c such that these are equal, right? so it's not a solution.
01:51
So now the first example is a bit more involved.
01:54
Our function is defined as tangent of x minus c t plus x minus c t to the three over two.
02:07
Okay.
02:10
So let's go and make the calculation.
02:15
Let's start with maybe the derivatives in time.
02:17
Let me maybe use a color green so that we know what we're working with.
02:21
So first derivative in time of this function u.
02:24
Let's just recall the rules.
02:26
Derivative of the tangent as we know.
02:29
Oh, sorry, here there was a plus.
02:33
T tangent of x plus c.
02:35
So derivative of the tangent term, as you know, is 1 over the cosine squared.
02:42
But we need to multiply by the derivative of what's inside with respect to t, which in this case is simply c.
02:49
So let me put a c in there.
02:52
And then for the second term, this is just a power.
02:54
So the rule, you remember, the power comes down.
02:57
3 over 2, x minus c t.
03:00
And we reduce the power by 1 .3 over 2 minus 1 is 1 over 2.
03:04
And we'll multiply by the derivative of what's inside with respect to c to t, which in this case is just minus c.
03:13
So let's make the calculation now so that this becomes c over cost squared of x plus c t.
03:28
Here, just in a more friendly way, let me write it minus 3c over 2, x minus ct, the power of 1 half.
03:37
So this is the first derivative.
03:39
Let's do the calculation for the second derivative with respect to t.
03:42
So now this challenging bit, that's just to recall.
03:45
So it's 1 over a function u.
03:47
So, you know, derivative of 1 over u is minus u prime over u squared, right? so this is going to become a minus c.
03:58
Now we need to differentiate cosine squared.
04:01
So derivative of the cosine squared is 2, cosine of x plus ct.
04:06
And now we need to differentiate cosine of x plus ct, which gives us a sign of x plus ct...