Find the general solution of the following differentialequations. (please use undertermined coeff. and clearly find yp,y'p, y''p)y''+y'...

Find the general solution of the differential equation. $$x y^{\prime}=y$$

So for these first up in this problem, I went ahead in rewrote ah Wai prime as D. Y T X And I did this because it's going to allow me to multiply both sides of my equation by the X and to start a separation of variables. So on the left side, we have d y. And on the right side, this is equal to Y d. X. So to finish separating our variables, I'm just going to divide both sides of the equation by why? And when I do that, I get one over. Why do you Why is equal to D X and from here? Because we have a separation of variables, we can go ahead and integrate each side of my equation independently. I'm so on the left side. We know that the integral of one over why is just the natural log of y and the integral of D X is going to be X and we'll write plus C on this side of the equation. And so finally we know that the natural log is telling us that why is equal to e to the X plus c power, Um and so we can go ahead and rewrite this in our final form of our general solution as wise equal to a E to the

Yeah, we want to solve a given differential equation which is dy dx is equal to see can't why for the initial conditions Y of zero equals zero. This question is challenging our ability to solve an initial value problem in particular is challenging us to solve this IVP with separation of variables. There are four steps to execute first. We're gonna isolate are X and Y terms on either side of the equation. So why? In the last Excellent. Right. In step two we integrate both sides because each side has a differential dy dx step three we saw the integration techniques and in step four we saw the IVP plugging in our initial conditions. So in step one, isolating X and Y. I guess coast Y Dy equals the X. That's because Dy oversee can't Y equals coast Y. Dy that we have integral coach. Why do I equals integral? The X or solving sign Y equals X plus constant integration. See in step four we saw percy using our initial conditions, we plug in Y X equals zero. The sine of zero equals zero equals zero plus C, so C equals zero. And our final solution is sign of Y is equal to X.

In discussion there, Astra. Find the general institution off the differential equation. Why Prime equals to y O operates. No. What? Why is considered as a function of next? So white crime would be d y over dx. And that's why you're Rex If we bring y to the bottom here and take the d x to the tough So we get d y o ver y equals two d X over Rex. Now integrate both sides of this previous equation. So we get Ellen absolute value off. Why calls to Ellen Absolute value of X plus C where C is any constant. Now I can write. See as Ellen off a for some constant A So we get Ellen. Absolute value of y equals two Ellen Absolute value backs plus Ellen, eh? Now, by using laws of logarithms, this boils down to Ellen Absolute value of a X. So Ellen, absolute value off by equals to Ellen, absolute value of the X or Y equals to a X. So this is the general solution to this differential equation. Here. We can also check if the solution is correct or not. If why is they extend the Y over DX equals to a which is same as eggs over eggs. Now a X we solved for is why. So that's why Rex and that's precisely what under potential equations.

I have d Y over the X equals e to the y squared all over why this is a different Joe equation, because I have that d y over DX. And if I want to get it back to a general solution that I need to do the integral. But first, I would want to separate the variables so I would send the d X over to the right side because that gets it off the denominator. But I noticed that really have the wrong variable there with it. So I'm gonna need to move this entire rational over to the left side. An easy way to do that is to use the reciprocal. So if I use the reciprocal, that's me flipping that fraction so that this will cancel out so that reciprocal to keep things balanced goes over to the left side. So my new line of work is gonna be why over e to the y squared times d y equals just DX or could write one DX Now that my variables air separated, I'm ready to integrate integrating on the left hand side. I want to do one important step first. There's not like a great rule for having an e on the denominator. So I actually can bring that e up to the numerator. If I write, the exponent is negative, right? That will reciprocate it for me. So I'm gonna write that exploded as a negative y squared to bring it up, and everything else is going to stay the same. And now becomes more obvious to me that I can use a substitution here to make it just e to the U, which its anti derivative states the same. It's still eating the you. So we're going to replace this exponents negative y squared, and I need to also check its derivative. Well, the derivative would be negative to why d y and I have some of that already here notice I have. Why do you Why? So if I solve for what I already have, then I would need to divide that negative to to the other shy. So as we're place, everything we have each of the U. And when I seven for this, Why d y I get do you over negative too. Okay. So again, I saw for what I have currently in my integral, which is why in D Y, which meant I had to move that negative to over that negative two is out of place and then replace the you that I saw it for and replace this. Why do you? Why now it's ready to integrate this negative two on the denominator is just a coefficient, and the anti derivative eat of the U is eat of the U. But let's change it back to our current variable or the correct variable. And so we're gonna call it negative 1/2 e to the negative Y squared on the left hand side. I'm sorry. On the right hand side, there's not a whole lot of work cause it's one DX wine goes Toe X. When I do the anti derivative, I need to have a plus C. I put plus C because of the fact that there's could be some constant there that I don't see when I took the derivative, because any concert number has a derivative zero. So just be careful when we have these indefinite integral to always input plus C plus C plus c. So just to make this a little bit cleaner for my final answer, I would get rid of the fraction in front. I would multiply by negative too. But notice I keep the see the same because, see, it's just some number I don't know and they have two times Some number I don't know is actually still some number, I don't know. So I just leave it as a C there at the end, and this is a more simplified version of the general solution.