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Find the particular solution of the differential equation that satisfies the initial condition: Zxy' Inx2 = 0, x > 0, y(1) = 2...

Question

Find the particular solution of the differential equation that satisfies the initial condition: Zxy' Inx2 = 0, x > 0, y(1) = 2

Find the particular solution of the differential equation that satisfies the initial condition: Zxy' Inx2 = 0, x > 0, y(1) = 2



Answers

Find the particular solution that satisfies the initial condition. Differential Equation
$$y(x+1)+y^{\prime}=0$$ Initial Condition $$y(-2)=1$$

Okay, So for this problem right here, I'm going to start off by moving the natural log of experts, the right hand side. So I'll have natural log X square. And I'm going to go ahead and divide by X. So I have to buy prime. It's a natural log of X square. Try to buy X already. And so with that I'm gonna go ahead and write why primers do you like the X. And I want to move to the X to the right hand side. So have to D. Y equals natural log of X squared divided by X. D X. Going to interrogate both sides. It's on the left hand side. This is gonna be to Y. And on the right hand side, I'm going to use an identity here. So the identities simplify this down to the natural log of natural log squared of acts. Mhm. And we're gonna have a plus C. So one divided by two. So I'm going to get natural law squared of X. two by by two Plus C. And if I plug in our initial value conditions, which is why if one equals two. Okay, so two equals to the natural locks squared? Uh blonde. I'd buy two Plus C. And this entire term right here Goes to zero because the natural log of one is 0. So we got two equals to see. And by that logic we can get our final answer. Which is going to be why Equals two natural log squared Of X, divided by two plus two.

Okay start with this problem I'm gonna go ahead and move to E. To the X. The left side. And I'm gonna rewrite why Prime S. D. Y. T. X. And I'm gonna move dx the other side as well. So why d why? Of course to either the X. Dx and now I can integrate both sides. So this is gonna be y squared divide by two. And on the right hand side is just gonna be to E. To the X. Policy. Now I can multiply by two. So I have four E. To the X. Plus C. And I'm going to go ahead and plug in my initial condition here which was why A zero equals three. It's gonna be three squared Equals to 480 power. Just one part. It's four times 1. So it's just before plus c. And so Have nine equals to four plus c. Subtract four C equals 25 So we can plug that in. So we have y squared equals to four E. To the X plus five. And if we wanted to simplify down a little bit further, we can take the square root, so have Y equals to the square root four E. The X plus five, and that'll be your answer.

Problem. Six of of section 7.1. Asked first to solve the following differential equations. So the first thing that we're going to do to solve this is we're going to rewrite. Why prime as D y DX, we're then going to separate the variables. So we're going to divide this right side by Why divide the opposite side by the same thing, and then we're going to multiply both sides by the X. So during this we eliminate the DX is on the left side and the wise on the right side. So that leaves us with the UAE over. Why is equal to negative one the X? The next thing we're going to do is we're going to integrate both sides. That leaves us with Ellen of the absolute value of why is equal to negative X plus e. Now we're going to eliminate the the Ellen and sell for just why. So we need to have a base e on both sides and to simplify this this look like why is equal to e to the negative x times e to the C. Now we can rewrite this as some other variable. We're just going to call it a and our equation will now look like this. The last thing we're going to do is we need to go back to this initial condition and we're going to satisfy it. So we're going to plug in to for why two is equal to a Times E and we're going to plug in 14 x. So we get E to the negative one. So as we know, each of the negative one is just one over e so this could be rewritten. Us two is equal to a times one over e. We're going to multiply both sides by E so we get to eat is equal to a so our final solution looks like this.

So for these several differential separable differential equations, what we need to do is um literally separate then we need to get everything that's in terms of why on one side and everything in terms of X. On the other side. So to start this um we we can just we can subtract Y from both sides and then we'll have the wise on the right side and the X. On the left side. So do that now and we get yes. Uh huh. Um Now it's helpful to rewrite the white promise do you buy by dx? Okay so I'll do that. And now of course we still need we do need to get to why what's in terms of why on the side that some on the side the same side as the derivative of what? So what we're gonna do is we're gonna switch flip these around. Um So we're gonna divide both sides by X. And divide both sides by white scrutinized Y. So basically we get um you get one over why squared minus Y? Do you have I. D. S is equal to one over X. And uh now we can sort of multiply both sides by dx. So we get this and then we can integrate both sides. Um The right side is pretty easy to integrate. We get sorry we get well of course that's gone on X. Um Mhm. No no absolutely personality trait constancy. Um And then uh the left side. Yeah well using the for the left side using the partial fractions method you have learned previously you'd find that this is equal to Yeah. Uh Well you can split this up into this yeah. In this right. Yeah. Yeah. Mhm. Okay. Um So now integrating that you get uh lawn of absolute value. Why am I this one minus uh line of why equal to this? Yeah. And uh what we can do right now is we can we can do the initial condition part taking um this value and something that in to find R. C. And then afterwards we can try to isolate what if possible. So basically we'll get um On of absolute value of 1 -1. is minus lawn of absolute value. Uh One No no sorry um should be the negative one because that's the y value but either either way Mhm. Be the same deal for the second part. Okay so and then that's gonna be equal to law and of the absolute value. One plus C. Um No uh that would be just too and this would be one, you know when you take the absolute values um and a lot of one is zero. So this is really nothing. And this then that would have been one long of one is zero. So basically uh she is equal to Uh one of 2. So we can go back to the equation we had here and in place of C. Uh we can put along too from here we can use log rules to combine these things on the left side, we're gonna have lawn of absolute value of Why AM -1 over why is equal to line of absolute value of well, a lot of two absolute value X. Um.


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