verify-that-yxex-sin-x-is-a-solution-to-the-differential-equation-2-y-cot-x-fracd2-yd-x20

Question

Verify that $y(x)=e^{x} \sin x$ is a solution to the differential equation
$$
2 y \cot x-\frac{d^{2} y}{d x^{2}}=0
$$

Michael Jacobsen · Accepted Answer

and this problem were provided with the second order differential equation on her job is to show that why I&#x27;ve X equals either the X sign X is a solution. Our first job is to take the derivative of this given function well, right that D y dx will be equal to by the product rule first, a derivative of even the X copy sine X plus copy either the ex itself and now take the derivative of sine X, which turns into co sign X. Now, when we take the derivative of this function, we obtain detail Why divide by DX to and we go term by term. When we take the derivative of this first term, However, we note that was the original work we did on the first step in finding the first derivative. So it&#x27;s just copy down. This result, the driven of even the X sign X is either the ex sine x plus. You know the X Times Co sign X. Now let&#x27;s go to the very next term. Either the ex co sign X when we take this derivative will add either the ex copy co sign X. Let me add a little bit of space, a copy of the X and take the derivative of co sign Cosan becomes negative sine X salts Insert the negative between these functions and right to minus either the ex sine X And that&#x27;s our second derivative. Let&#x27;s simplify the second derivative as much as we can their very first term either The ex I necks cancels with the last term. And so the result is two copies of either the ex Times Co sign X. Now we have a second or derivative. We&#x27;ve completed the second derivative. So the next phase of the problem is to determine if this left hand side does, in fact equals zero through substitution. Denote this LHs to stand for left hand, side and right that the left hand side is equal to two times the function. Why? But why is given to be, you know, the X sign X sold do two times either the ex sine x Times co tanja necks minus due to why DX to that was the last step that we calculated on the second derivative which came out to to either the ex co sign X. So we get to you know, the X co sign X, and we&#x27;ve completed all of our substitution so far. Recall our goal is to now simplify as much as possible obtaining zero. If we can get to that step on the right hand side, then we&#x27;ll know that we have a solution. So start off by baby first expressing co tangent into this components co sign of X and sign of X. We have two times you the X sine x co tension itself turns into co sign X divide by sine X, and we have minus two either the ex co sign X by copying another cancellation with the sign Exes leaves us with two of the X coastline X minus two to the Exco sine x So we do it in fact get zero. And this then shows that why of X equals the X sign X is a solution.

Julia Choi · Answer

question number eight assets. Whether this given function is a solution to why double Prime minus two. Why Prime Plus five y equals zero. So we want to find the derivative of why, uh, why prime is equal to to e to the x co sign of two x plus e to the x sign of two x and why double prime is equal to two e to the x co sign of two X minus four e to the x sign of two X plus two e to the x co sign of two x plus e to the X sign, uh, to x and we can simplify this to four pups six Thatch simplified to four e to the x co sign of two X Um minus three e to the x sign of two ex. Know that we are using the chain role here because the function inside of the trick is to ext. Uh, now that we know why a prime and why double crime? We just need to plug it back into the original equation that they gave us and we get four e to the x co sign of two X minus three e to the x side of two X minus two times two e to the x co sign of two X plus e to the X sign of two X plus. I&#x27;m gonna run out of room here, So five e to the x x sign of two ex all equal to zero Ah, we can see that this term here will cancel out with this term once the two is distributed over. And this term here will also cancel out with this term when the two is distributed with this five each x sign of two X So we do see that zero does in fact equals zero. And so the solution is verified.

Robert Daugherty · Answer

16 meant to problem number eight were asked to differentiate. Should need to solve this difference or equation. So it is a leading your difference equation in his first order of since only have one derivative. So what you gonna recognize? You can put it in standard form Or you could recognize that e to the two x times y How would you differentiate this with respect to X? If you differentiate this with respect to acts, that would just be white way you use the product rule. So that&#x27;s two e two the two x times y plus e to the two x do y d x. And so what you see there, this is the exact thing you see on the left side of this equation. OK, so this equation is the derivative with respect toe acts of whyy to the two X is equal to two X. Now I can integrate both sides of this equation. So when I integrate both sides of this equation, I get why e to the two X is equal to, and I&#x27;ve got to integrate what two x the X, which is just X squared plus constant of integration. So why E to the two X is equal to X squared plus C. Therefore, why is equal to X squared e to the minus two x plus c e to the minus two x.

Patrick Vaughan · Answer

Hello. So today we&#x27;re working on first order differentials, and ultimately, this is gonna boil down to a problem of separation of variables or B ah, split up energy to variable sets. And then we integrate with respect each variable. Um, and the reason that we know this is because we have a d y in the d X. All right, so what do we do? We given an expression that looks like this where we have an equation that set equal to zero. Well, our first thing that we want to do is sort of split up the parts so that we can reduce it down to two variables. So first, we&#x27;re gonna want to add this complex expression to both sides. We have a brake line just how we differentiate steps within the problem. So that gives us two x y you I equals X squared e to the negative y squared over X squared plus two y squared DX. So next we can, uh, reduced down the expression but the differentials on one side and the only variable the others will divide by two x y and then by whatever DX, we&#x27;re doing two things simultaneously here And that gives us Do I already X equals our x squared e It&#x27;s the negative y squared over x squared What two y squared, all divided by to why accer two x life. Okay, well, no. From our textbook, we can recall that one way Teoh early Use a y ver expression is to equal to a new variable R b which has an X component within it. So the equals y over X or the other way to write that is why equals the time, Zach. So we can take this. Why expression and plug in for all of our wise. And then we can also do that with our, um the expression and plug that in when we have live. Rex. So doing this, we get the partial of, um, be of Akst times about about Deac equals X squared. Yeah, to the negative b squared, because why over act, um, squared. All right, we&#x27;re just putting of the end for that expressions that becomes B squared, X squared over X squared. So I think to be squared, or if you just take it as an expression of, um, why over x squared, then that&#x27;s the same thing is to be of X squared. So either way, it&#x27;s the same thing if you just plug in for why or for the whole expression. Okay, so then we plug in for the second line. So we have vivax times act squared, all divided by two x squared and then put into that Why vivax times x. So just ah, short reminder that the reason why I did not as the vaccine because when we&#x27;re taking the differential and I&#x27;ll do this in green So the differential of this vivax times X we have to, uh, variables that are, um, that have x contained within them. So you have to dio of the differential breakdown by part Sophie left side equation times the derivative of the right side plus the derivative of the left side times the original with the right side. And that gives us, uh, well, now that we&#x27;ve done that, we don&#x27;t need a do the and remind ourselves that it is a subset of acts or has subset of acts. So be plus e prime, and then we can take that and plug it in for that expression. So that&#x27;s just a little reminder that we dio until we get to that point. So this is the same thing is ready TV over the x Times X plus de And then we can rewrite the expression above where we have X squared. Yeah, nothing b squared plus two b squared X squared Hold about it by two x squared the x The reason I&#x27;m not combining that denominators because we see that we have an X squared in both Art is the expression so I can cancel out and then we&#x27;re left with e and I&#x27;m gonna be squared. What? To be squared all over too. The X equals R X TV over the X plus de So now we can attract the from both sides of the equation. And Teoh make it so, um sorry. Ah, so backtracking. I made a tiny mistake, so that&#x27;s not squared. So it&#x27;s always good, Teoh. We&#x27;ll check yourself. You&#x27;re solving these equations. So two x squared be, um And then there&#x27;s no x there, So just sort of show you what? Where I made my mistake. So I have about my two y x here and then when I&#x27;m playing in for why it&#x27;s two acts times p X and then we have the two x squared b on. We&#x27;re getting that. So then I boil some of that. OK, so now that we&#x27;ve corrected that mistake, do you get that expression? Uh, the same have the same denominator so we can combine the two expressions to multiply that by to be over to be So then this boils down to X TV over DX. Um, And then when we combine these two expressions, we have to be squared in two b squared, and that&#x27;s to B squared minus to B squared. So that cancels. We have e So the negative b squared over to be so this very complex expressions getting much simpler. And now we can separate variables such that we have. We can, uh, multiply the whole expression by to the over eat the negative b squared. Um, and I just I won&#x27;t do this simultaneously this time. So we have that equals one. And then we have to be, um, he to the B squared. Because when you&#x27;re dividing by a negative beast square, that&#x27;s like saying one over one divided, divided by eat the V squared. So that&#x27;s just equal to eat. of the B squared. Um, and then times are over the ec. So that may want to multiply the whole thing by the X over X, and that will give us to the B to the B squared equals, uh, one over X, the X. So now we&#x27;ve successfully separated are variables. We have these on one side X on the other, and we can integrate. Now, which the whole goal. Um, and I apologize. I left out my TV. There we go. Okay. So he can integrate. So the integration of the right side would be Oh, are only on the vax plus Ah, see some constant and then the adoration of the left side while we have two sets of the variables. So what do we do when we have that? Well, we can use a substitution or a u substitution. So you more equal b squared. The derivative of you would be to be bebe. And then this plugs in very nicely. We have are two vdv to be TV and then we have our b squared. So student Matt in, we have yeah to the u Do you you do you to you is are integral Well, sorry, just et you. Because the derivative of each of you is Are you dio so plugging in for you. So now we&#x27;ve successfully done both for immigration, but we want to put it back to our original format. So with our wives and our axes Sorry. So we wanna put it back into our original format with our wives and our axes. So first we have to reverse, solve and plug. Are you back in? So this becomes e. B squared equals natural. Lagerback&#x27;s what I see. And then we can plug our, uh, wild racks squared back in, Harvey. And that&#x27;s natural log max plus C. And, um, that&#x27;s our final answer. Because we have successfully reversed integrated or river solved back and for what we substituted in for, um and ultimately, what you could dio is go back in and, uh, suffer why? And sort of winning. I realize that, Um, but you don&#x27;t have to. It wasn&#x27;t asked for in the problem. So this is indeed the final answer, and I&#x27;ll just seem out so you can see everything we did to get to this point

Ernest Castorena · Answer

They asked us to verify that it satisfies the central vision for Let&#x27;s Find the first year all of negatives he wanted for the negative to be for the X. We take our second riveted and we&#x27;ll have C one e to the negative. We&#x27;ll see to be to the X notices. The same thing someone has attracted him. One from me of it well, under 10 does satisfy the differential equation.

Robert Daugherty · Answer

Section six to problem number seven. So I&#x27;m dealing with first order, ordinary, linear, different gel equations. So I&#x27;m gonna get this in standard form. So standard form is going to be why prime plus a function of x times y equal another function of X. And then I need to bolt by both sides of this equation by this expression in order to be able to different shapes the different equation. So first, get this into standard form. So this is going to be why prime minus 1/2 why is equal to 1/2 e to the X over two. My integrating factor will be e to the integral of minus 1/2 DX, which is you to the minus X over two. So you multiply this entire equation by E to the minus X over two. And when you do that on the left side, you&#x27;ve got the derivative of a product which is e to the minus X over two. Time is why is equal to when I multiply the either minus X over two on the right and simply get 1/2. And so now I just take this equation, integrate both sides, So I&#x27;ve got the derivative you to the minus. X over to why is equal to 1/2 and I integrate both sides of this equation and I get why e to the minus X over two is equal to 1/2 X plus c. Therefore, why is equal to 1/2 X? Multiply everything by E to the X over two and there is my final answer. So why is equal to 1/2 x e to the X over two, plus a constant integration Times E to the X over two.

Problem

By writing Equation (1.1.7) in the form $$ \frac{…

Problem 7 Medium Difficulty

Verify that $y(x)=e^{x} \sin x$ is a solution to the differential equation
$$
2 y \cot x-\frac{d^{2} y}{d x^{2}}=0
$$

Answer

$y(x)=e^{x} \sin (x)$ is a solution

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$y(x)=e^{x} \sin (x)$ is a solution

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