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Point)A first order Iinear equation in the form y' + p(z)y f(z) can be solved by finding an integrating factor /(z) = exp p(z) dx(1) Given the equation 3y'...

Question

Point)A first order Iinear equation in the form y' + p(z)y f(z) can be solved by finding an integrating factor /(z) = exp p(z) dx(1) Given the equation 3y' 12y = 6 find p(z) = e (2x)Then find an explicit general solution with arbitrary constant C .Then solve the initial value problem with y(0) = 1

point) A first order Iinear equation in the form y' + p(z)y f(z) can be solved by finding an integrating factor /(z) = exp p(z) dx (1) Given the equation 3y' 12y = 6 find p(z) = e (2x) Then find an explicit general solution with arbitrary constant C . Then solve the initial value problem with y(0) = 1



Answers

Find the general solution to the given differential equation. $$z^{\prime \prime}+2 z=0$$

Who has to find the general solution to the differential equation here is the double prime plus two. Z equals zero. Um We can see that in this case B equals two and z must equal zero, C equals zero because there's no Z term. So that means that B squared minus four C. Is for which is greater than zero now. Um We can then find, you can then find the roots of the characteristic equation And those are zero and -2. So we plug those in here and we get A. E. To the zero to you, plus B even minus two. T. Zero T. Is just one. So we have that. The other thing we could have done here is, you know, just done an integration integrated at once equals some constant, which I'm in this case would have been let's see here. I think it would basically be yeah. And so then we could uh could it actually be a over to um no, is that B to a? It would just be another constant, Some other constant here. So we could have integrated at first and then, you know, had a first order equation. Or we could have gone through this process here, and I just realized that we have a constant, which is clearly a solution to this and then an exponential.

All right. So let's go ahead and start this problem by substituting out the vital prime Y. Prime and the white term. And so we replace it with r squared minus r -6 equals zero. And then we can do a little bit of factoring so we'll have AR -3 Times are plus two equals to zero. And from that we can derive that are most equal to negative two and three. And so with these routes, what we can do is we can go ahead and build our solution. So our solution is going to be Y equals two K, one e to three x Plus K two E to the -2 X. So that's our final answer.

Now we've got D. Z over D. T. Is t squared Z squared putting all the Z's on the same side. So that would be Z to the negative to power. DZ is t squared D. T. Um Taking the integral of both sides. I would get Z to the negative one. Power over negative one is gonna be t cubed over three plus C. And so or rather I'm going to put a plus D. Here because then I'm going to just change it a little bit. Um One over Z is going to be I'm just multiplying by negative one on both sides, negative T. To the third power over three minus D. Um So then I need to add these together negative T. To the third power minus three D. Over three. Then I can invert both sides. Three over a negative T. To the third power plus see. All right. Because I can write see in replacement of any constant because we don't know what the constant is yet. Okay now we know that Z equals one third win T equals one, so one third is going to be three over negative one plus C, so negative one plus C is nine, so C is 10. And so it's going to be the particular solution is going to be Z equals three over negative T cubed plus 10. Mhm.

We were given nine Z double prime minus equals zero. Um Well, first thing we need to do is get this into the form that um well basically the form that they use, you don't really need to use that for him. They just make sure that a this coefficient here is one. And so you can always do that by dividing throughput whatever this constant coefficient is. Because these all have constant coefficient. So we get these double prime minus nine minus 19 Z equals zero. So now he can identify B0 and see his -1 9th. Which means b squared minus four C. Is four nights, which is greater than zero. Now we can get our characteristic the roots of the characteristic polynomial. Those are well plus or minus the square to this over to and so that's plus or -1 3rd. So a general solution then is easy, some constant A. E. To the T over three plus some other constant E. To the minus T over three. So these two routes here and then. So we have these two solutions here, and then we take a linear combination of those, and that's a general solution to this differential equation, and obviously you can plug it back in and you can see that that's indeed the case.


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