00:01
In question 34, we have two linear waves, and we represent them by d1 equals f1, xt, and d2 equals f2, xt.
00:15
And both satisfies the basic definition of equation, that is, d x2, dx squared, equals 1 over b squared of directive, second directive over two.
00:33
So if both d1 and d2 satisfies this equation, we want to show that any construction d equals c1 plus c2 d2 also satisfies this equation.
00:48
So in order to solve this problem, all we need to do is to plug everything inside.
00:54
So over d, x squared.
00:58
Is the derivative of the derivative.
01:10
So this, since derivative is derivative of d with respect to x equals c times derivative of d1 with respect to x plus d2 times derivative of d2 with respect to x.
01:27
And then when we take the second derivative because c2 and c2 are constants, we can just take the derivative of c2.
01:35
So that gives us c1, the second derivative of d1 plus c2, the second derivative of b2.
01:47
And we can do the same to the second c2.
01:55
When i will be square, and we have ddt over dx ddt, which is equal to c1 derivative of d1 with respect to t plus c2, derivative of d2 with respect to t.
02:21
And again, because c1 and ct are constants, we can take the derivative of the derivative of d1, d2...