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Write down an explicit formula for the solution of the wave equation:u(1,0) = sinI, "(1,0) COsI. u(0.t) = 0 1 > 0. +> 0The solution is periodic function ...

Question

Write down an explicit formula for the solution of the wave equation:u(1,0) = sinI, "(1,0) COsI. u(0.t) = 0 1 > 0. +> 0The solution is periodic function of t, true O1 false? Now solve the forced problem40Ir coSwt.u(r.0) sin I, "(r.0) CUR Ia "(0.t) = 0 r > 0, t > 0.Consider the cases W= 1 and L ` In what essential aspect do the two solutions differ from each other. and from the solution in part (a)?4471

Write down an explicit formula for the solution of the wave equation: u(1,0) = sinI, "(1,0) COsI. u(0.t) = 0 1 > 0. +> 0 The solution is periodic function of t, true O1 false? Now solve the forced problem 40Ir coSwt. u(r.0) sin I, "(r.0) CUR Ia "(0.t) = 0 r > 0, t > 0. Consider the cases W= 1 and L ` In what essential aspect do the two solutions differ from each other. and from the solution in part (a)? 4471



Answers

Show that the wave function $\psi=A e^{i(k x-\omega t)}$ is a solution to the Schrödinger equation (Eq. 41.15), where $k=2 \pi / \lambda$ and $U=0 .$

So in this question, we have to wear functions. Why one? And why two. And both of them are solutions for the wave equation. Now we want to check if the superposition of these two waves is also a solution to the wave equation. So why of ex Ante? This is the superposition between them. Why? One off accent T plus why two of ex ante Now the wave equation is the second partial derivative off. Why would respect to space equals one over the square? The second partial derivative off the wave function. With respect to time now we're going to plug in the wave function. I can write it as follow. So partial partial eggs. Partial partial X. Why one plus Why too equal to partial partial t partial partial T Why one plus Why, too multiplied by one over the square. Now we can get from this partial partial X partial y one partial x unless partial y two partial eggs equal to one over V square par Shit. Partial t partial. Why one partial t plus partial y two partial t. And here what we did is we took the first derivative and now we can take derive it again so we'll end up with the second derivative. So we'll have partial. Why one partial X Close The second derivative off the second wave function partial X equal one over V Square. And here I'll open a square bracket, the second derivative of the first one with respect to time, plus the second derivative off the second one with respect to time. Now, if I distribute one over the square here, it's easy to see that we're going to have two terms. So I'd write them, like as follow So partial worry one partial X Square, plus partial to let me write the second one in blue, still far shell to partial X Square, and this will be equal to one over the square. Partial square. Why one partial T Square plus one over the square? Partial Why to Partial T Square. And as we can see, the terms in blue are the wave equation for the second wave function, and the terms in black are the wave equation for the first were function and since both of them are solution to the wave equation, superposition of them will be also a solution for the wave equation as we saw in this term here and therefore from this term, we conclude this term is yes, it's

So in this problem we share to obey functions. Why wanted by two on? So this problem we need Teoh, understand that everyone function no matter what the form is, must satisfy the general wave equation which is given here. So in order to actually get is two solutions on show that their linear combination is also where you function We need to address a certain property, solve the differentiation or functions So when we show, for example, if off eggs this will hold true for the partial derivatives off multi variable functions this is for the single variable function. For example, he could NGO facts plus a low off eggs than the partial derivative when the function half off eggs is going to be able to the linear combination off these two personal religious for these two functions that our own day on the side here So this property allows us to the date is a standing wave function, for example, or generally, when we have ah, they hear way functions linear combination of two way functions you can call it a standing wave function but anyway, this combination or two we'll do away functions and say the combination equals y one. Plus, why do now from the property here we can say that further. We want to partnering with you or ext. It's going to be the derivative. The first functional Rex also edition secondary with no brakes. Now what happens when we try to make A to create the second only the differentiation on this side. So this will be the second order partial derivative over X which will be equal. And now we will apply standard those of depreciation. So this is Barschel derivatives. Why line over X and so for the white too. And now we will apply these derivative here on the first and on second tearing. So the Sequels second burst derivative Rx, or the white one function plus the second partial derivative or the white to function over the horizontal position coordinates. So this is very important because this relationship you could see here shows that the first term of the general way function, which is the turning off the differentiation over the X coordinate. So this is the turn will be leaner if the wise addition off two functions So the whole the river there will be a linear combination off the relatives of the two functions or the spatial coordinate. No need to brew sending for the dive. And also we don't really to prove because this since this holds true for the X coordinate, which is one of the variables in the function, why axity the same the whole 40. So both the second partial derivative over X and the second part of the river do over t behold the same the same property off linearity. So in that case, we need to actually see what is the final conclusion of this. Well, we can now substitute into the baby question directed, so start from partial derivative of their Y function. Second brush of the remedy over X is one over the lost squared far from the religion for the first one over tea. In addition part for the relative why do over tea? Then we can see that this term right here from the Santa disproven before is is equal to one over been lost two squared and then we will will move the differentiation outside on the brackets of the linear combination. So it will be the secondary with you over tea and inside little camp. Why one plus y two and notice one thing since linear combination. Basically, this this deserve here is also it will do over axle. Why want plus who I do. And this means that that the common in since this is a solution, this is the solution is equal to this. So it is the solution for the wave equation. Therefore, this proves that linear combination. Why one plus one? Why? To really be the actual solution as well to the wave equation, the proper solution wave equation. And they're using the same property in the teeter. The second order derivative over time do they in this form? Why? Well, because now we can just say these equals do why and we finally obtain it right? Partial derivative off the wife function, which is the function it is obtained by addition of two functions is in fact equal to we lost two squared and again second. Why should they really be over time off? Do I function so on both the time and the horizontal position aspect this fashion Why one plus one? That's something by additional y one and y two indeed satisfies the veil equation and therefore it is A. It is a proper solution. Fourth of eight equation. After this, Siri's off Bruce and substitution. So this is the end of the problem. I hope you find this helpful, and they hope to see you in another lesson.

Hello. So today we're asked to show that each solution sat given below a three D satisfies or in other words, is of the wave equation which is given right here. U subscript e t equals a squared with u subscript x X. So the first thing to do is to sort of identify what this exactly means. Well, you t t is basically saying that the left hand side of this equation, you has to be equal to the second partial derivative of T squared that the right hand side should be a squared times, the second partial derivative of you with respect to X two times. So what this is saying is that essentially, what we want to approve that the left hand side is equal to the right hand side of the equation and were given four different solution sites that were there proposing are off this whip equation. So we have to check that each one of these a through D functions of you are indeed wave equations. So they have to satisfy both the left hand side condition as well as the right hand side condition. All right, so now that we know this, let's color code, so Read will be designated of a blue of B green of C and saying of D Okay, so the first one U equals sine of CAC's times the sign of a Katie So focusing on the right hand side first will take the first partial derivative of you with respect to X. So we see that we only have one X component in this equation in this function. So that will take the sign of a Katie and pull it out because I just constant in these conditions and then we have sign of K X. So what is the derivative of the sign of a function while the narrative of sign is co sign so co sign of K axe times the derivative of what is inside the parentheses? So the derivative with respect to acts would just be K. Therefore, we can re simplify this point all of our Constance once more. So we have k sign of a K T as a constant times the co sign Que axe, and that's the partial derivative of you with respect to X. Well, now let's take the partial derivative of this new function again with respect to acts which gives us the second partial derivative of this function. You with respect to acts twice. So again we retain our constant K times a sign of a K t. And then what's the derivative co sign? Well, that's negative. Sign of our function times the derivative of the prince sees which is again Okay, so making a little more space. Our final derivative here with respect to X twice is case negative K squared times the sign of a k t times the sign Que axe. And that's the partial derivative The second partial derivative of you with respect attacks twice, which can then be written, as you expects. So going back to remind ourselves that we just solved for you xx And then we also have this a square term. So we have to plug in dysfunction or you xx times are a squared and that's our right hand side equation. So now we have to do is see if the left hand side is equal to that. So the partial derivative of you with respect to t do this for the first one. So we have only won t very also again will pull out constant of Sign K axe and while with the derivative of Sign once more, that's co sign of a k t times the derivative with Respect T, which is R a k so we can pull out her Constance once more a k times sine que axe times are variable interest, which is coast of a k t. So that is our first partial derivative with respect to t of that function. So now we can move on to the second partial derivative with respect to T for the second time, and so are constants are retained and then the derivative sign her coastline. Sorry, derivative co sign is negative. Sign times a k t. Times the derivative of the inside. So again a k therefore the final form of this partial derivative second partial derivative, you would expect t twice or U T equals well, we have our negative sign. We have a twice So we have a squared. We have Kate toys. So we have k squared times the sign of K axe, that constant times a sign of our changing variable Katie, And that is our left hand side. So let's see if he's in the same Well, we have our a squared or negative a squared. Both. We have our case where both and then we have our to sign components and both. So the right hand side equals the left hand side. So scrolling back up, we can say yes. This, um, equation is indeed a wave equation because it satisfies the condition where we have the left hand side equal to the right hand side. And now we can move on to the next problem and blue. So you equals t all over b squared, t squared, minus x squared. And then we just have to do the same thing again. So let's take the partial derivative of you with respect Teoh X first. So again, working on that right hand side and we'll rewrite this equation was that it's our constant tee times are part of interest. A square T squared minus x squared to the negative one. So what is the first derivative of this? Well are constancy is the same. And then we have our, uh Then we have our derivative of this function right here. So we have negative one times a a squared T squared, minus X squared to the negative two times the derivative of the inside, which is just negative two x So let's rewrite this. So we have a negative one times native to so a positive too. Um so you're a positive two x t all over a squared T squared minus x squared. And then that whole function is squared so we can rewrite that as to t r constant part times us times a squared T squared, minus x squared to the night of to So this is our first derivative with respect to X. So the partial derivative of the U with respect to X. So now let's take the second partial derivative of this function you with respect to X again. And while we have our constant, that for me is the same our to tea and then we're gonna have to use that product differentiation. So as a reminder, that is our left side times the derivative of our right side, plus the derivative of our left side times the right side retained. So this gives us X times negative too. A square T squared minus X squared to the negative three times again that two eps and then this is plus the derivative of that left side axe. So that's the one times the rights entertains are a squared T squared, minus x squared to the night of to And then let's simplify this right all out. Nice. So that we get, um, negative four. Sorry. We have a native two times a two way. Go think it a four x t uh, x squared. T b r x interacts are negative to one or two in her tea. Um, times heard divided by come on a squared T squared minus x squared to the third power and then distribute that to t And again we get plus to t all over He squared t squared minus x squared, squared. And now Teoh combined those two together we have to multiply that right side by hey, squared T squared minus x squared all over a squared B squared, minus x squared. And then this will give us negative four x squared T plus two. Hey, squared T cubed minus two x squared t all over a squared T squared, minus x squared. Cute. All right, so now we can further simplify because we spirit here we have a two X squared t and we have a get a four x squared piece of those can be combined and moving over into space. That gives us than a negative 60 x squared T plus two squared, cubed all over He squared t squared minus X squared Cubed, which is a second partial derivative of hue with respect to X for the second time. Or are you act fax. So as a reminder, Ari and side has to be times that a squared and that is our great hand side equation. So let's see a little bit and then go into space and you might a little more where we dio Okay, so now we want to find the left hand side of the equation. So let's take the partial derivative of this given function you with respect to t. And as a reminder will rewrite our function. You up here, which is, uh, t times a squared T squared, minus X squared to the negative one. So right off the bat, we have two functions a t. So we have to again use that product Rule of differentiation and then actives list tee times e squared t squared minus R. X squared times that negative one in that city, negative second power and then times the derivative of the inside with respect to T. So we have our A squared to T plus the derivative of tea, which is one times that right side retain so B squared T squared, minus X squared to the negative one. Now let's simplify this so it makes your life easier moving forward. So we have a negative, too Times a a squared, T squared times this, let's put it divided by and then we'll put it back later. So divided by a squared T squared minus X squared, squared plus one over a squared T squared minus X squared. And again to get those to be the same denominator, we're gonna multiply by a squared T squared, minus X squared over the same thing. So that's basically multiplying that rate side by one. So it's not fundamentally changing the value, and then that gives us a negative to a square T squared. Plus, he squared t squared minus X squared. All over are a squared T squared minus X squared, squared, and we'll see that we have in a square cheese scored on top and a negative to a square cheese squared so we can simplify that even more. So are partial of u with respect to t can Finally, simple high down Teoh Negative X squared minus a squared t squared, Um, and then I will put it back into the easy form. So he squared t squared, minus x squared to the night of to just cause I don't like using quotient rule. Doing it like this always forces it back into product role. Um, I forgot my parentheses there to distinguish it. Okay, so now we can take the second derivative of this function with respect to you again on and then just as a reminder, in case you're not aware, you forgot, um, the second partial derivative of you with respected t twice is basically the same thing of saying the partial derivative of you with respect to t and the partial derivative you with respected T. Um, and that's why you could do this. Um, step by step. Like this. Okay. So I know that we have that out of the way. Um, let's take it. The derivative with tricked you again. So we have that product full but with two larger components. So we have that lefts entertained. You have X squared minus B squared, B squared times negative too. A squared T squared, minus X squared to the negative Three, uh, times our he squared times to t And then plus the derivative of that left side. So that becomes a negative. A square to t times I writes I retained b squared t squared minus x squared to the negative two. And then, uh, getting rid of what we just that little explanation we wrote that gives us a little more space, and then we can rewrite this out so that we have group. Sorry, little technical difficulty. We're gonna rewrite this out so that we could try and simplify and see what it looks like. So we have our we'll break this down slowly. So we have negative two times this whole expression. So we have positive two X squared minus plus negative times negative. Plus two a squared T squared times Teoh a squared t all over a squared T squared minus x squared, cubed plus, uh, negative to a squared t all over a squared T squared minus X squared squared. So again We can dio this multiplying by one using a square T squared, minus x squared or the same thing. This allows us to get the same denominator. And then we have so simple Find this right side That gives us a four x squared a squared T plus for eight of the fourth t cubed and then we have minus for eight of fourth two cubed uh, plus Sorry this No, before that's a to so plus two in the fourth t cubed and then plus two b squared x squared t all divided by a squared B squared minus X squared, cubed. Okay, so now we have next verse or an X square, a square tea and an X squared a square t so this could be combined Don't put that circle are noble And then we have a to the fourth t cubed so those two can combine. So then we have, uh, before polls to six x squared a squared tea and then four minus two. So plus two eight of the fourth t cubed all over He squared t squared minus X squared, cubed And then to better compare it to that right hand side will pull off the A squared. So that gives us six X squared t plus to a square t cubed all over that denominator. And that's the second partial derivative of you with respect to t squared. Or that you t t And now let's check out the left hand side equals very hands tied. So does your mind a little bit. Okay, so we have six x squared t plus to a square t cubed all over that. So it looks like we I might have missed a negative sign somewhere. So let's go back in check. Because these air very similar. Um, let's check this right hand side. So, yep, there's my mistakes. So we have a negative two and a negative, too, which will make it positive for and then so that would be positive. OK, so besides, that looks like I made two mistakes, So there's a positive four. Uh, well, that's becomes eight, actually, because we have 22 in two. So two times two is 44 times two is eight. And then So that positive eight minus two, then that becomes the positive six. Sorry about that, folks. Okay, so then, yeah, we see that the left hand side does equal the right hand side. So once more zooming all the way out, going back to the top. So once more we have, Yes, the left hand side is equal or it inside. Okay, Now let's go into some space and start on the third problem. So for the third part, we're going to do it in green, and we have you equals X minus 82 6th plus X plus 80 to the six. So pretty similar. But we have a negative sign in a positive sign in those functions. All right, so let's start with that Ray inside doing it with respect to X. Well, I give the six times X minus 82 the fifth times, uh, one plus six times x plus e t to the fifth again. Times one. And then, um, the second partial derivative of you with respect to X for the second time equals 30 times x minus 80 to the four plus 30 x plus e t to the four. And then this congest be simplified down by pulling the 30 out and multiplying it by the X minus 80 to the four plus X plus a teach in the fourth. So this is the second partial derivative of you with respect to X twice or you xx so recall that we have a squared times. That and that is our right hand side. So moving into more space, let's look at that rate inside. So the partial derivative of With of You with respect to t so we have again six x minus e t. To the fifth times a negative a plus six times X plus 82 the fifth on and then just times a day. And then if we take the partial derivative of you with respect City again, we get 30 X minus 80 to the fourth and that negative eight times another negative a plus three times x plus a t to the fourth times a times another. So that gives us, um, well, you have a negative and or negative base that becomes positive so we can have a 30 a squared pulled out Times X minus 80 to the fourth plus X plus a t to the fourth. And that's our partial squared of you with respect to t twice or U T. T. So that's her left hand side. Now we can look at it and compare and see that. Yes, the right hand side does indeed equal the left hand side, and we can go back to our original problem function set. So we see once more that the left hand side does indeed equal the right hand side of the equation so that this is also a wave equation. Now let's response to that last problem set in science going into open space. It's getting hard to find at this point, um, and we'll rewrite it down here as a reminder. So you equals the sign of X minus 80 plus the natural log of X plus 80. So the partial derivative of you with respect to X for the first one while the dirt of sign is co sign and that's X minus 80 times that one plus one over X plus 80 times one and and we can really write this again. So we have the co sign of X minus 80 plus, um X plus 82 the negative first power so that we can use that exponential derivative, and then we have the partial derivative of you for the second time with respect acts and, well, what's the derivative of co sign? That's X minus 80 times one plus a negative one times X plus 82 d native to their form. The final solution for that partial derivative is negative. Sign of X minus B T minus one over X plus 80 squared. So that's the right hand side. Okay, um and then Oh, yeah, Don't forget. We need to multiply all of that by a squared. And I'm looking at the left hand side. Take the partial derivative of you with respect T while the derivative of sign is again co sign of X minus 80 and then times negative A Because that's the derivative of negative 80 plus one over X plus 80 times A. Because the derivative of that inside of the law algorithm is just a Okay, so let's rewrite this look. So we have the partial derivative of you. The first partial derivative with respect to t is negative. A co sign X minus 80 plus a Times X plus 80 to the negative first power. Well, now let's take the second partial derivative of you with respect city again, and we get negative a times. Well, what's the sign of her? What's the derivative of co sign? That's negative. Sign Princey their silly We don't pick it. Subtraction. So negative sign of X minus 80. And then we have another negative A for that derivative of the inside of co sign and then we have plus are a times negative one times X plus 80 to the negative, too. And then times the derivative of that inside of that, which is just a so simplifying this down. We have a negative and negative, a positive a squared. And then we have that negative from the sign. So again we have a negative a squared sine of X minus 80. And then we have minus he squared all over X plus 80 squared and then pulling out a squared. We don't have the same equation as on the right hand side of negative sign of X minus 80 minus one over X plus 80 square. And that's the second partial derivative of you with respect to t twice or you're so going back. So going back to original statements, we can also say that yes, the left hand side is equal to the right hand side. So all of these, I guess we'll do it in gold. So all of these are with equations because they meet the criteria outlined. And then I'll just try to zoom out everything so that you can see everything we did to prove this. All right, so there's everything we did.

But a we are U T. T equals a squad. You xxx. So we have to prove this statement. So first of all we have you equal sign kicks sign a day. Now here we have these equals U. D. T equal a squad you X X. This is the answer for part of the problem. Now we have further part B. Here you equal T. Upon Esquire Esquire -6 sq. So further therefore we have U. T. T. Equals this and now we're going to find he works therefore works equal. So we have you X. X. As this this can be written as two D. Into esquire T squared plus three X square. But then esquire T square minus X squared to the power three. This equals esquire. Sorry yeah further you can right this as this is our new X X. No U D t equal twice a quality into yes quite a square Last three x Squire farm esquire two squared minus X squared to the power three. Therefore it is written, esquire into U. X. X. Hence we proved your DT equal. It's got you X. X. Now we have part C. Where we have to find you do T. Equals this and we have to prove you D. D. Equals a square you of xx. Now we have to find out you D. T. We'll be finding further you X. So You x equal six into X minus 82. The power five and 2 one plus six into X plus 80. The bar five into one. No therefore we have U. T. T. Equal esquire you X. X. This is proof now in part B of the problem we have to find out you equal sign X minus 80 plus Ellen X plus 80 here. Also we have to find out you D. T. Equals yes where you xx Now, funder we have so here we have find out U. T. T. Now we will be finding You X. X. So we have two x equal. They're finally, if we have U. D. T equal inquired you xx, this is the answer.


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The autocorrelation function Is maximum at 0 None ofthe mentioned Origin and Infinity Infinity Origin...
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This question is designed to be answered with a calculatorUse the four functions shown:1.y =5-XIL: y = I5-xlIlI: v =35-x IV: y = (5-x)-1 Which functions are non-differentiable at x = 5?Il and IIl only IIL and IV only 1, Il, and IIl only II, III, and IV only
This question is designed to be answered with a calculator Use the four functions shown: 1. y =5-X IL: y = I5-xl IlI: v =35-x IV: y = (5-x)-1 Which functions are non-differentiable at x = 5? Il and IIl only IIL and IV only 1, Il, and IIl only II, III, and IV only...
5 answers
Chapter 3, Sectlon 3.5, Questlon 18EJ Incorrect.Find the sensitivity of y = f (x) a x = 0 and use It to calculate Ay for Ax = -0.02y = X+1Enter the exact answer.Lyexact number; no toleranceClick if you would like to Show Work for this question: Qpen_Show_WorkShOW HINTLINK To TEXT
chapter 3, Sectlon 3.5, Questlon 18 EJ Incorrect. Find the sensitivity of y = f (x) a x = 0 and use It to calculate Ay for Ax = -0.02 y = X+1 Enter the exact answer. Ly exact number; no tolerance Click if you would like to Show Work for this question: Qpen_Show_Work ShOW HINT LINK To TEXT...

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