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The equation (zy+y? +z?) dr z? dy = 0 is & homogeneous equation. Verify that it is and then use substitution to golvc it_ (Hint: y(z) = ru(z))_...

Question

The equation (zy+y? +z?) dr z? dy = 0 is & homogeneous equation. Verify that it is and then use substitution to golvc it_ (Hint: y(z) = ru(z))_

The equation (zy+y? +z?) dr z? dy = 0 is & homogeneous equation. Verify that it is and then use substitution to golvc it_ (Hint: y(z) = ru(z))_



Answers

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

In the problem, we have indications dessert upon the equals integration Y. DY always Ellen injured equal. Why is square upon to class C? So we have a giraffe while equal use part was quite upon to plus C. Or giraffe zero equal 8.0 plus C. One equal parsi or C equal zero. So we have jennifer Y equal power. Why is quite upon to? This is the answer.

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.

For this problem, we are asked to solve the homogeneous differential equation. Why prime equals X cubed plus Y cubed over X Y squared. So our first step is to make the substitution why equals X times V. And to note that we have that, why prime then is going to equal the plus X times V. Prime. And that before making our substitution we can rewrite our right hand side up there as being X squared over Y squared plus Y over X. Where we note that the equals Y over X. So we can then write the plus X. V prime equals one over the squared plus V. Which in turn helps. Which in turn means that X times V prime equals one over V squared, subtracting the from both sides. Which then means that we have V squared V prime equals one over X. Or that the square the we'd have then that the integral v square D. V equals the integral of one over X Dx. Which means that we have v cubed over three. It's going to equal lawn of X plus C. So then V is going to equal uh the cubed root of lawn of X plus C. But we know that V itself equals Y over acts from up above. So we have then that why is going to equal X times the cubed root of one of x plus C.

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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