00:04
Okay, folks, today we're looking at problem number 43 in which we're going to solve the initial value problem.
00:12
I guess the third order initial value problem as given.
00:19
And so the steps are pretty much the same as in the second order, initial value problem.
00:28
But in this case, of course, we have to solve a third order.
00:36
And this is pretty much, you know, i've done this a few times now, but i'll just sort of summarize what you have to do.
00:44
You look at the auxiliary function, auxiliary equation, and in this case you're going to get r cubed minus r.
00:55
Not equal zero, sorry, just r cubed minus r is the auxiliary expression, auxiliary equation.
01:04
And then you find the roots of the auxiliary equation, and in this case, it's just going to give you, you get three distinct roots, 0, 1, and negative 1.
01:15
And now you want to show that the three functions, y11, y1, sorry, y1 equals 1, y2 equals e and negative t, and y3, equals e to the t are both linearly independent and solutions to 41, which you can in fact do.
01:44
And then once you've done that, you can see that the general solution to 41 has this form.
01:56
Okay, so now once we get here, now we have to look at the initial value problem.
02:02
And that's the next step.
02:05
So you just plug in the initial values into your general solution.
02:15
And you'll notice that here we have three constants because it's a third order.
02:23
And the three constants mean you have to have three equations.
02:29
And that is, in fact, what you get right here...
