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The particular solution ofym _ y = sin tis of the form:8 Y = A sin t + B cos tOY = (At + B) sin tOY = At sin tY=t(A cos t + B sin t)...

Question

The particular solution ofym _ y = sin tis of the form:8 Y = A sin t + B cos tOY = (At + B) sin tOY = At sin tY=t(A cos t + B sin t)

The particular solution of ym _ y = sin t is of the form: 8 Y = A sin t + B cos t OY = (At + B) sin t OY = At sin t Y=t(A cos t + B sin t)



Answers

Substitute $ y=a \cos (2 t)+b \sin (2 t)$ $y^{\prime}+y=4 \sin (2 t)$ to find a particular solution.

All right, So for question number four, we're going to solve this non Homewood unions differential or second differential equation. And to do this, we're gonna have to find the solution to the homogeneous differential equation and add that to a particular solution for the non homogeneous solution. So let's go ahead and start by making this language in use by just putting his ear on the right side and then solving for it. So we're gonna guess e to the Lambda T and will be able to factor the Lambda t out for something like this here. So this is gonna be lame. Two squared plus for it's the people zero. And we're gonna factor the e to the lambda t out to sell for Lambda. Okay, you can already see the solution that I'll just write the solution. I know. So lame. No one is gonna be to wife. We plug to I and here we're gonna get that or two y squared is gonna give us negative four. And negative four plus four is going to give a zero. If you don't know the solution right away, you can just do the quadratic formula. Use your calculator in Lane 22 is gonna be negative to I before on imaginary solution. We don't actually need this second value of Lambda. We just need one of them. So we're just gonna be using the positive value since that makes things easier down the road. Okay, so let's put this into our guests. So we guessed that second homogenous general equal to E to the negative or positive to t I Let's apply. You lose formula. And when we do a play, you lose for him. When we get that, this is gonna be equal to co sign of two t plus I time sign to t. And since we want a real solution, we can actually just get rid of the I. And now that we have to win early independent solutions, we can put the C one and C two variables over here. And this is going to be our final, homogeneous general solution. And to get the non homogeneous general solution so right as just general to we just have to add the humble home Virginia solution plus a particular solution. So copy and paste that over there and then add the particular solution. So too I'm signed of tea. And this is our general solution. I think I wrote the particular solution wrong. I think that particular solution should be to t yeah, duty.

Ladies and gentlemen, today we're looking at section 4.5, number one and now, just in the case where we're starting to look at not homogeneous equations and how to solve them. So in this case, way have ah, we have the following. We have to We have a solution, um, to this equation and this equation. So we're given one solution to this equation and one solution to this equation and were asked to apply what is called the superposition principle too. So the following questions First off, the question is you could What is the superposition principle? While the super present position principal says If I have a solution to this to an equation of this form, If I have one solution to an equation of this warm and I have and I have a one solution to an equation of this form, then for any real number C one and C to, uh this solution here for this function solves this linear r OD. So I mean, it's pretty straightforward. It looks pretty straightforward that, you know, you have a solution here. You have a solution there. Then a linear combination sells any, uh, OD. That's equal to linear combination of these two. So I mean, it looks pretty straightforward. And, you know, in general it is. It's just good to know that you know this hold, because you'd like It's a very simple thing that you'd like to hold. So with that in mind now, we're really each of these three problems A, B and C is really just about applying it and more or less the applications. Durant, Um, so what if we look at the first case? We have five signed T. And of course, um, so we have this difference. Your equation equals five. Signed t, and we go back to the original one. We see that CO signed t is a solution. So, um, by the homeless or by the superposition principle, we would have five co signed t solves A and that's that's all we need here. We're not looking for general solutions or anything like that. We're just The question is just what does the superposition tell us about, um, solution to here? And the next question, of course, is the same type of thing. But in this case, we have this sign t plus, I guess negative. Free uh, me too. The two tea. And of course, again, that's of this warm here books. Um, so what do we do? Well, we apply the same thing, and we get this for answer. Now, um, you'll notice that in my answer, I leave it in this form. I actually don't encourage. Um, students to cancer are to simplify in general. So I mean, obviously what I'm saying is is that you could simplify this further and say, Well, it's just equal to co sign t uh, my receipt of it to tea. And of course it is. Um, you know that this is true. Um, you know, and that's fine. But I'll tell you, a lot of times when students do something like this, they make mistakes. So obviously in this case is pretty obvious. But what I mean is, for instance, like maybe you do something like this instead of the negative, right? You canceled out or you do something like this are you know, many different possibilities. Um, Then this answer here technically is wrong. You know, the second answer would technically be wrong. Whereas this answer here is correct. And and for some people, um, your solution would be wrong, or you would lose at least some points. Because what you did going to hear, you know, going to the end. If you say this is my answer here. Well, your answer technically is wrong. So I often, you know, tell students these from my personal perspective is that leaving it leaving? Um, it was See if I can get rid of this here. Um, leaving it the answer. Sorry. Um, oops. Okay. I guess it didn't work, but leaving the answer here at this point is better. I say leaving it here is better than trying to go further. Um, and I avoid simple, fine and see. And partly because it's easier to mark. It's also easier to mark when you have it like that. It's easier to mark because I could just say Yeah, this is the 1st 1 This is the 2nd 1 I can see where you got it from. Bang. But when you go further than that, uh, you know, it can lead to confusion. So I'm just mentioning that here, you know, pulling on a sort of a violent if you Well, um, because I think it's important that Ah, one should pay attention to what repressed professor teacher suggests. And in general, I would always tell students not to simplify beyond, um, just the answer. Once you get the answer, leave it as simplifying room lead to mistakes. Um, in the next one, get it? It's straightforward because, you know, it's just a linear combination. Um, and so again, I you know, the answer is just, uh, doing the super present super position principle. Um, says that four. You know, the constant four times co sign T plus 18 times this and see, this is a little clear toe when I was referring to, because now I can see where the 18 came from, You know, today are, um and you know this. I know where this came from. I know where this came from and as clear to me. Whereas if you wrote, this is six each of the two tea. Um, you know, four coastline to you, plus six to the two t. Of course, you would be correct, but, um, technically, six, each of the two tea is not, um, you know is not the C two times. Why two, if you get what I mean, because we go back to the superposition principle it says, see to why, too. And, um, so 18 times why to cause this is why to here is the correct answer. Um, and four times why one plus 18 hands way too. Uh, would be the correct answer. And I would suggest the students just stop here. Don't go any further. Okay? So, uh, just a little note. I do think, um, students get in the habit of simplifying things, and that often just leads to mistakes. And it makes it harder for a greater for a marker to mark about my opinion. Thank you very much.

Thank you, Songer. Not much is equations with you. I'm lost four games. You equal to do you tend t close thing two key He squared sine hamachi portion off this V. You don't for you because you're so h which is a Hamas portion of the yuan per sign u T C two sign a team u P key which is not homogenous portion of the equation. First a Time T plus B multiple over t since there's multiple over co starring to t since since the very close trying to tedx t there. So we've been ad c times t t plus d multiple greatly sign t and then afterward me dio this represents for the G Q Tempting coastline kooky. But now for the minus t score Trance signing key which is simply gonna be e times t squared plus f times t sign a he plus eight times t squared. Plus I James t plus que co sign you have keen being called to you new r h teeth. Plus you pee off teen

So for this problem, Although we're given a differential equation, we don't actually need to use it at all because of the fact that we're given a general solution in the point I'm always to do is plug our point into our general solution insult for our constancy. And so when we do that, we're going to plug in X or a pie for X and pie for teeth. So we get five equal to see minus 1/8 time's a sign of two ply minus one over 32 times the sign of four pi. And so we know that this is going to be equal to C minus 18 times the sign of two pi is zero minus one over 32 times. The sign of four pie is also zero. So that's going to just tell us that pi is equal to see. So our particular solution is going to be the same with our general solution. Um, the only difference is we're plugging in pie for RC, so our particular solution is going to be pie minus 1/8 Time's a sign of two teeth, minus one over 32 times, a sign of 40


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