verify-that-yta-cos-omega-t-phi-is-a-solution-to-the-differential-equation-11-21-where-a-and-omega-a

Question

Verify that $y(t)=A \cos (\omega t-\phi)$ is a solution to the differential equation $(1.1 .21),$ where $A$ and $\omega$ are nonzero constants. Determine the constants $A$ and $\phi$ (with $|\phi|<\pi$ radians) in the particular case when the initial conditions are
$$
y(0)=a, \quad \frac{d y}{d t}(0)=0
$$

Mike Gaerlan · Accepted Answer

Okay, so just start this problem. We first are going to take a look at the roots of the exhilarate equation here, which is going to be p of our is equal to r squared. Plus, Omega, not squared, is equal to zero. So we&#x27;re gonna have that r is equal to plus or minus I Omega. Uh, Omega, Not now. This part here has the route b is equal to omega eso if b is able to omega, Uh, but we need to take a look at the or compare the roots here so we&#x27;ll have plus or minus i omega. So if omega is not equal to Omega, not so let&#x27;s do that case first. Omega is not equal to Omega. Not so. If this is not Omega, not then our zp of tea is going to be equal to than just a not e to the Omega or Iomega like so. So this is going to have the derivative c p of t is equal to I unless should be Iomega t. So we&#x27;ll have omega, uh, I or Iomega that a not e to the i omega t the men zP double prime of T is going to be able to the negative omega squared a not e to the I omega t like So let&#x27;s plug these into our ah differential equation, which is going to solve our plus omega, not squared, is equal to is equal to. And then we&#x27;re gonna have, um, on the right hand side, this is equal to f not. And then then e to the I omega t like so So Ah, oops, sorry. And then there should also be a zp here. So if we have negative omega squared Ah, so omega squared, um, a not e to the i Omega t and then plus omega not squared a not e to the I Omega T is equal to f not e to the I omega t Then we have omega not squared minus omega squared. A not is equal to f Not so a not is going to be able to f not over omega not squared minus omega squared. So we&#x27;re gonna have our particular are complex valued particular solutions. E p of t is equal to f not over. Oh, Megan. Not square in minus omega squared. And then times e to the I Omega t so either the I omega t remember, we&#x27;re gonna replace that, actually with co sign Omega T plus I sine omega t. Since we have the co sign part, we want the real part. So that&#x27;s going to just be this here. So r y a p of tea is going to be equal to f not mining or divided by omega not squared minus omega squared cosigned omega t And this is for Omega not equal to Omega. Not so. That&#x27;s for that scenario. Now we need toe Take a look at the case where omega is equal to Omega. Not so for Omega, equal to Omega. Not then That means our particular solution CP of tea. Then it&#x27;s going to become a not tea e to the I Omega t like So So then zp prime t. Okay, so let&#x27;s do that. So we&#x27;re going tohave first times derivative. First time second is gonna be, uh, that and then plus, now we&#x27;re going to have Iomega tea or so Iomega Times a not ah t e to the I Omega t. So, um, that is here. We took derivative of first time Second, then here we have derivative of second, which is the Iomega E to the Omega T times first, which is Iomega I not or a not tea. So the C double prime of tea this is equal to And then we have a not, uh, or sorry, i omega a not e to the i Omega t and then here we&#x27;re gonna Ah, the derivative of this is just this. And then we multiply by Iomega. So we have plus Iomega a not e to the I Omega T and then plus I omega squared. That&#x27;s going to become negative. Omega squared. Sorry. Negative. Omega squared. Okay, negative. Omega squared times a not and then ah t e to the I Omega t. Okay. Ah, we can now combine these two terms into just to be to Iomega a not e to the I Omega t and then minus We have our omega squared a not t e to the i Omega T. Next, we plug it into the complex value difference equation, which is E p uh, double prime plus. Sorry. Plus, and then Omega not is able to omega, so I&#x27;m just gonna write it as omega squared, and then zp is equal to R F not and then e to the i Omega t. So now plugging these into our ah, to this here we have to Iomega a not e to the i omega T minus omega squared a not t e to the i Omega T and then plus omega squared a not t e to the i omega T This is evil to f not e to the i omega t the these tube can now cancel out and we just have to Iomega a not it&#x27;s gonna be able to f not when we also cancel out these terms here. So a not is kind of eagle two f not divided by two Iomega or this is also equal to F Um, sorry, I ah, negative. I f not over to omega. So zp of t is equal to, um negative I f not over to omega times t eats the Iomega t So this e to the Iomega&#x27;s he I can replace with cosigned omega t plus I sine omega t And again, Since we&#x27;re still taking the coastline part, we want the real part of this. So the real part of this is when we multiply this by this here. So this negative I becomes ice negative. I squared, which is positive. So why P of tea is gonna end up becoming okay. We have f not divided by to omega and then tea sign Omega t like so. And this is for Omega equals omega, not It&#x27;s our final solution here.

M S · Answer

first, I want to verify that this has a initial value of while pipe will zero. So we&#x27;re just gonna plug in pie for every t value. So negative pine times coast on a pie minus pi. I don&#x27;t want this equal zero. So cousin pie is the quadrants of angle pies. Located here does zero. I have high and three pie, perhaps, and pies Here this coordinate is negative 10 and that&#x27;s co sign sign. So comes on in these cases, negative one. Still have zero equals negative pi times negative one because co Santa pies negative points all minus pi. So zero equals in a negative times negative gives us a positive pot and anything might it still is zero. So this initial condition checks out, we also need to verify that it is so It is a solution of this equation here. So what we have to do is we have to play in our derivative and we have to plug in our original Why equation? So our suspect, the derivative of our y equations. So we have d y B c equals the first times the derivative of the second, which is negative, scientific, close the second times, the derivative of the first, which is negative point. And then the river of, uh, minus T that&#x27;s just minus one. So we simplify. This will have the positive t Sorrenti minus a co sign T minus one. Just most applied everything together and we simplified. So this is far as we can simplify the situation. Now we need to plug in our were C Y d T equation in our Y collision. So have I t times t sai Inti minus co sign T minus one. Now, when this starts equals, we&#x27;re gonna play guitar y equation. Our original migration here. So we had negative t co sign t minus t and we bring down our plus t squared sine ti. So now we distribute owner left side will have t squared sign T minus t co sci fi minus E on left side, on the right side. We don&#x27;t have anything to distribute, So we&#x27;re just gonna bring everything down. If you look on the left hand side, it matches our right hand side. We have everything. We have a t squared Zion T we have a minus. Seiko Cy Inti and we have a minus t. So since our looks inside, it was our right hand side. We have verified that these solutions

Smita Praharaj · Answer

So this question were asked to find a particular solution off the differential equation to you over tt equals to turn you that passes through the 0.1 comma by two. Given that the general solution is you equals to sign in bursts off e to the C plus t. So basically, we need toe put in this point into the general solution toa obtain a particular solution that actually passes through this point. So we&#x27;re not a substitute 1 40 on by right to for you into the general solution to solve for C and then were used the value of C dropped in the particular solution. So we have by over two equals sine inverse off e to the scene Last one. Now sign in verse off a number. We know that sign off my right to equals to one. So if signed in verse or something equals two pi by two so that something needs to be one, so we&#x27;ll have e to the C plus one equals +21 Now we&#x27;re to take Ellen on both sides so we get Ellen off. E to the C plus one equals Alan off one now. Ellen off E to the C plus one simplifies to see plus one. And that tickles Ellen one, which is zero. So that means C equals negative one. So our particular solution you b equals sine inverse e to the T minus one.

Ernest Castorena · Answer

buying that this differential equation has this function as a solution. Let&#x27;s take our derivatives to test that well, have you? Prime is equal to you. Let&#x27;s see. See one into the tea. Estate is the same. That Iranian product world for the second. So well, I have seen too e to the T plus c Teoh t need to the team. Okay, We can group some coefficients together. We like the C to the t t in the team now we can take our second derivative. Then we&#x27;ll just have one. Well, it is gonna be to see to see to the teat. Why is that? Well, when I do the product world for the next one and one of them, I&#x27;ll leave the tea alone, take the dirt of either the t get itself, and the other one will take the derivative of tea. Yeah. Now, let&#x27;s see if this satisfies the different Jill equation. I&#x27;ll have one plus two C two, you know, T No, let&#x27;s see, Plus into t unity. Then we&#x27;re gonna subtract two times the first year. That so too. See one CTO eternity. Track C two to C two. See you to the team, and then we&#x27;re gonna add the original function back and you didn t see to t It is a team. Okay, let&#x27;s see what we get. Well, these two C two utilities, they&#x27;re gonna cancel out with these negative to see two deputies. So let&#x27;s get rid of what about this. See one into that he Well, there&#x27;s two negative ceiling, you tease. So one of these will cancel out one of these. We&#x27;ll have one remaining, but there&#x27;s another scene. One either that he here. But somehow this cancels out. What else do we have there, too? Negative. C two t either The tees. Well, there&#x27;s one here. There&#x27;s one here. This is these two. Cancel out I get now. This one was cancelled out earlier, and we got to appreciate it out on rhythm.

Smita Praharaj · Answer

we&#x27;re asked to show that the function by equals to buy times e to the negative co sign X solves the differential equation by prime equals y times sine x to resolve of my prime and simplify it too. Ah, show that it equals white times sine X. Now let&#x27;s ah, uh, find what? Why Prime ms No white prime. So why prime equals spies a constant. That&#x27;s honest as it is, and we&#x27;ll get the debatable e to the negative coastline. Ex recall that the derivative Opie to the f of X is each of their four x times of prime off. It&#x27;s so the derivative off E to the negative co sign X will be e to the negative close eye Next times the derivative or negative ghost I next. Now that becomes five times E to the negative. Kasai next. Now the derivative of negative ghost I next minus one would be a constant. So that would stay as it is debatable. Go sign. Next is negative. Sign X. Now this simplifies to negative times negative becomes a positive and then we have by e to the negative ghost I next times sine x. Now if you notice by times e to the negative because I next. That&#x27;s precisely what why ISS simplifies to white times sine x on. That&#x27;s what the right inside of the creationists. So we have shown that why prime equals two y times. I next for why to be dysfunction. So what this way satisfies the differential equation?

Problem

Verify that $$ y(t)=c_{1} \cos \omega t+c_{2} \si…

Problem 30 Easy Difficulty

Answer

$yp(t)=\frac{F_{o}}{2_2o} t \sin (w t), w=w_{0}$

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$yp(t)=\frac{F_{o}}{2_2o} t \sin (w t), w=w_{0}$

Differential Equations and Linear Algebra

Anna Marie V.

Caleb E.

Samuel H.

Joseph L.

Integration Review - Intro

Definite vs Indefinite

Stephen W. Goode, Scott A. AnninDifferential Equations and Linear Algebra

Anna Marie V.

Caleb E.

Samuel H.

Joseph L.

Stephen W. Goode, Scott A. Annin

Differential Equations and Linear Algebra