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Question

Solve the given problems.
Show that $y=e^{-x} \sin x$ satisfies the equation $\frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}+2 y=0$.

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.

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.

Smita Praharaj · Answer

so we want to show that the function y equals 2 to 8 to the negative. X plus X minus. One solves this differential equation by prime equals X minus y. So basically, what we want to do A software. Why prime and then verify the situation here. To find my prime, we need to find the derivative of dysfunction to find the disreputable e to the negative X recall that the derivative off key to the F of X is each other f of x times f prime of x. So that means the derivative of all me to the negative X It would be e to the negative x times the derivative off the exponents, which is negative one. So that would be negative. E to the negative x So let&#x27;s find out what why Prime Miss so white Brian would be derivative of dysfunction derivative, off to eat of the negative. X two is a constant, so that stays as it is to reboot of o. P. To the negative. X and speeches found is negative e to the negative X plus derivative of X is one minus debatable one is zero. So this simplifies to negative to the E to the negative X blessed more So that&#x27;s my wife prime. Now let&#x27;s find out what the right inside of the equation simplifies to sew the right inside is X minus Y. So we have X minus for why we have to eat to the negative x plus X minus one. So we get X minus two e to the negative X I distribute the negative negative times positive becomes a negative x negative times negative becomes a plus one So the ex and negative X cancels out and we&#x27;re left with negative too Lead to the negative eggs lost one. So as we just saw the left inside of decoration and the right inside of the equation simplified to the same thing. So this function, why here indeed does soul the differential equation by prime equals two x minus Y

Taylor Shimono · Answer

So for this problem, because our differential equation involves a Y prime. We&#x27;re going to want to start by finding the first derivative of why. And so we&#x27;re going to do that on the left side over here. So why Prime is going to be equal to a first a derivative of each of the ex, which is just e to the X power plus the derivative of sine X over too, which we know is just going to be co sign X over, too. Because the sign the derivative of sign, is just positive coastline and that coefficient won&#x27;t do anything. And finally, it&#x27;s going to be minus the derivative of CO Sign X will be minus negative Sign X over too. So when we simplify this, we get why prime is equal to each of the ex. That looks plus go sign X over, too. And we&#x27;ll distribute this negative here. So we get plus sign X over, too. And finally, we just want to plug this into our differential equations on the right side of the screen here. So why prime, we said, was e to the X plus co sign X over too, plus sign X over too. And this should be equal to co sign X plus. Why? So that&#x27;ll be plus e to the X plus sign X over too. Minus co sign X over too. I mean, all I did for this plus y portion was just plug in. Um, are why equation that we have that were given in the problems. And so we can see that since we have an e to the X on both sides of our equation will cancel out. Um, since we have a signed two x on both sides of the equation, they&#x27;ll cancel out, and, uh, we&#x27;re going to be left with co sign X. Over, too, is equal to co sign X minus co sign X over, too. And when we simplify, this is going to leave us with co sign X. Over, too is equal to co sign X over too. So since these two equations, these two sides of the equation are equal, that tells us that our equation for why is a solution to our door

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

Solve the given problems. For what values of $m$ …

Problem 44 Easy Difficulty

Solve the given problems.
Show that $y=e^{-x} \sin x$ satisfies the equation $\frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}+2 y=0$.

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Allyn J. Washington, Richard S. EvansBasic Technical Mathematics with Calculus

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Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus