5

6 h~3) - 0[(~pvr)^ ~ %]...

Question

6 h~3) - 0[(~pvr)^ ~ %]

6 h~3) - 0 [(~pvr)^ ~ %]



Answers

$$ 6^{m} \cdot 6^{n}=_______ $$

Yeah, to find a lab. Les Waas Parents. This is that waas off six. Bon. Aged minus. Wanted About four sequenced you the power t into due to the power three. So this is the answer.

Here we have six times zero, so six times zero is zero.

What is negative six times zero. The thing to remember here is that any real number times zero is zero and negative. Six is a real number, so negative six times zero is zero.

You know everyone, uh, today we're looking at problem number seven here, which is to find a solution to equation number seven and a couple notes about it. Um, to make that are important. Um, here is that this is a linear second border home. A genius constant coefficient. Um, in your equation are a, um, ordinary, different chill equation. I should say, um, and again, each of these terms linear, linear, second order home a genius and constant coefficient. They all mean something. And it's important. Um, when you're studying this course that you should know what each of these terms mean. Contrary to the common held opinion, um, you know, the vocabulary of mathematics is is, you know, is very important. And so you really should know what these terms mean and understand them in in the case, you can apply them. And really, it will tell you a lot about what their solutions you look like. And in doing so, it will probably help you. I'm no, um or at least give you a good good information about how to solve it. So, in a more general, like on an exam or something, if you know these kind of things. Well, it will tell you, um, a message. Solve it if you if you know what these things are. So they're good done know and understand. So the first step in solving dis equated this. This equation is we find the auxiliary. It's equation. So to do this, you just assume that your solution y is on the form y equals need to the rt year. And when you plug that in a while, you get this expression here, and, um, once we have that well, we, um since eat it, Artie is never zero. We just end up with this guy here, which is six course, you can see the expression. So now this is a polynomial on. This polynomial, in fact, has a special name. It's ah, auxiliary equation. So the auxiliary equation in this case is just the polynomial here, and it's a straightforward quadratic, and the next step is to now solve the auxiliary equation. So in doing that, we just substance, we just plug it. Are we Look at the expression, um, sorry, uh, zero equals to this and we solve for our and just using, you know, straightforward quadratic. Um uh quadratic formula we get our is negative to service. And ours 1/2 on DDE. That tells us our general form dissolution then is like this. And now, in this case, we have two distinct roots Are auxiliary equation meaning the general form is just see one times e to the, you know, to the first room, sometimes tea and plus C to each of the second room time. And that's what we have right here. Okay, um, now, the final stop, of course, is to check that this is, in fact, a solution, and that's what I do. But I just focus on, um, you know, this one here. So I just focus on first, um, solution. And you could do the same thing with the the second Lynn early independence solution. Um, And when you plug that in, you just end up with this expression which, um which, uh, you know, tells you then that why of tea is a solution. Um, and one thing you could one thing that's interesting about this and I think you'll see this later on is when you go back here, you might kind of quibble with the idea of why am I using this? Why you, Artie? Um and just saying, Well, I'm just saying, Well, I'm just assuming. Okay, Why, um is of this form and wise a solution, and then I'm plugging it in here, you know, in doing all this on so you can say Well, but why is why does that mean that the general form of solution is guaranteed to be this? In other words, how do I know there's not Maur possibilities? Well, it turns out, and this is really part of the theory of differential creations, which, um, you know, it's not really my place to get into, but I still think it's interesting too. At least mention, um if you go back to here, we said it's a second order, um, differential equation. So second order here, it means that since it's an ordinary different equation, well, you're gonna have to linearly independent solutions. And that's sort of the key to everything we do is that, um, we should have to literally independent solutions if it was ordered three and we should have three so on so forth. And you can shock that the e to the negative to t and t t t over to aren't factoring really independent. So these are the solution we want hand the key. And part of the key is is that, um you know that since we have two of them, uh, and this method works, it's really all used in other words, long as you confined to linearly independent solutions. Those are the solutions. There is no other. So this is kind of one of those meat. Things about ordinary difference equations is that if you confined to literally independent solutions and its order to those are the that is the general solution. So, um, just keep in mind any I just wanted to mention that because it's kind of one of those neat little tricks about ordinary, different situations that is.


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