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Solve the Differential equation: dy (x2 +xy + 4x + 2y + 4) ~y2 = 0,x > 0 dx...

Question

Solve the Differential equation: dy (x2 +xy + 4x + 2y + 4) ~y2 = 0,x > 0 dx

Solve the Differential equation: dy (x2 +xy + 4x + 2y + 4) ~y2 = 0,x > 0 dx



Answers

Solve the differential equation.
$y^{\prime \prime}+4 y^{\prime}+4 y=0$

In this video, we need to solve the following differential equation, X squared y double prime plus xy prime Plus four, y equals zero. And we know that the form of a koshi oil or differential equation is a X squared Y double prime plus B, X Y. Prime plus C. Y equals zero. So in this case if we match coefficients, we can see the A equals one, B equals one and C equals four. And we know that when we have a Koshiro their differential equation, the corresponding characteristic equation is A. R times ar minus one plus B, R or C equals zero. Look in that case for our differential equation, we've been given. That means the characteristic equation will be with a cold one. R times R minus one of course are Plus four since vehicles one and see equals form. Now we just need to go about solving this quadratic equation. So if we distribute we get r squared minus R plus our it's four equals zero minus R. And r cancel are quadratic equation we need to solve is R squared plus four equals zero. And you can either factor or use the quadratic formula. I'll just go with factoring If we do that this factors is AR -2. I Times are Plus two i. And then setting each factory can zero. We get the R. One and R. Two. The real part of solution in both cases is zero. So what we have is uh just the case where Our roots are R. one, R 2 Plus or -2 i. And when this happens um the corresponding characteristic equation takes the following form Y. Of X equals C. One times X. To the R. Times co sign or ours actually lambda. Excellent. Lambda co sign of mu natural log of X plus C. Two. Excellent. Lambda sign of mu L n X. And so in this case lambda is the real part of our roots and new is the imaginary part of our roots. But λ zero in this case. So our general solution is going to be C1 times X-0, which is one co sign of mu, which is to the imaginary imaginary part of our solution. And we don't really care about the plus, remind us we just care about the number itself. So we always just take the positive one. Uh So we have two Ln of X plus C. Two lambda again zero. So X zero is one sign of view, which is again too Ln of X. And so that is general form of the solution. This course you'll or differential equation.

This problem we need to solve the differential equation or x squared why double prime plus four, X Y prime minus Y equals zero. And we can see that this is a koshi oiler equation because it fits the form of a kosher other differential equation, which is x squared Y double prime plus b x Y prime plus C Y Equals zero. So in this case we see the A's four That these four and that C is -1 and the corresponding characteristic equation of a kosher. Other differential equation is R times r minus one plus B. R. Policy equals zero. So using those coefficients that we just found for a B and C, we see that in this case the characteristic equation is going to be for our Times are -1 plus four are -1 equals zero since C equals minus one, B equals four and equals four. So given that we know that now we can distribute this equation as four, R squared minus four are Plus four AR -1 Well zero So -4 and four are cancel leaves us with four. R squared minus one equals zero. And then we could solve this equation by factoring or using the quadratic formula, I'm just going to factor so I'm going to divide both sides by four, which gives R squared minus 1/4 equals zero. And then that can be factored as our plus one half Times are -1 half equals zero, Which means that our one Will be -1 half and are too Will be positive 1/2. And in the case of distinct routes of the characteristic equation for a Koshiro other differential equation, the solution form will be C one times X to the R. One plus C two times X to the R. Two. And so now we just need to plug in the roots so we'll have that. The solution Y of X is C one times X to the minus one half, plus C, two times X to the positive one half. And whichever order you put them in isn't really that important, So don't worry too much about it. You can also write these is C one over square root of X, but this is the same solution regardless. And so this is the general answer.

And this problem we are asked to find the solution to the following koshi oil or differential equation or X squared Why double prime plus why equals zero mm. And in general koshi earlier equation has the following form A X squared why double prime plus B. X. Y. Prime plus C. Why equals zero. And so in this case we're going to have the equals four. B is going to be zero because we don't have a white pride term And C is going to be equal to one. And a kosher. Other equation has the corresponding characteristic equation of A. R. Times R minus one plus B. R Plus C equals zero. So using those values of A. B and C, we can solve this characteristic equation, find its roots and then that allow us to determine the general solution of the differential equation. So in this case the characteristic equation corresponding to this differential equation With a equals four is going to be for our Times are -1, B is equal to zero, so we won't have that second term & C is equal to one Last Plus 10. Then we can expand this quadratic equation. So distributing there, we'll have four. R squared minus four hour plus one equals zero. And then we can use the Quadratic formula to find the solutions of this quadratic equation. So our routes one and 2, We'll be pulling into the quadratic formula. We just have b equals -4. So we have negative B. Oscar minus the square root of B squared minus four squared -4 times a which is four Time, see which is one and then all of that Over two times a. Just four. So now we can simplify minus minus four, is four plus or minus square root of In the square root. Here we're going to have negative four square to 16 -4 times for 16. So this is actually zero. So we can actually just get rid of the square root because that's just square zero. And in the denominator We have two times 4 is eight which simplifies four of eight simplifies to one half. So route one and route to are both one half. So we're in the repeated route scenario of a kosher another equation. And the form of this general solution in the case of repeated roots is white of X equals uh C. One X. To the R plus C. Two X. To the R. L. N. Ivax. Um So in this case the solution will be Y. Of X equals C. One is just C. One and C two are just constant. Uh X times are which is 1/2 plus C. Two X to the R. Which is one half again but then times natural lodge natural log of X. And that's a general solution for this calcium early differential equation. You could of course write X to the one half. In both these cases as square root of X, but in this case it doesn't really matter, it's the same thing. And so this will be general solution.

So for this example we are asked to find a solution to the second order of linear differential equation. Y double prime plus four. Y. Prime plus four, Y equals zero. To start off, we're going to find the characteristic equation of characteristic equation for this is going to be r squared Plus four R us or terrible. Four Plus four equals 0. So the reason that it takes this form is because anything, you know that is the second derivative of the function is going to take the form of square. If it's the first derivative it will take the form of the first power. And if it's just the function by itself, it will be that to the zero power. Mm. So now we're gonna find our okay, Bernthal for our so in solving for our we want something that multiplies to four and asked for obviously it's going to be tuned to are plus two no squared. Okay, so whenever you have this this is going to be a repeated route. Right? So whenever you have a repeated route, your solution is going to take the form of C. One and then you have to add an extra T. And then E. to the R. one. Actually I shouldn't say are one because they're going to be the same way E. To the RT right? And then plus c. two. E. To the R. See. So now what we're going to do is since this factors out to our equals -2. So I'm gonna write that in there really quick. So are Well to -2. Now what we do is we're gonna say that why of T. Is equal to see you won T. E. To the negative to T. Plus c. two e. To the negative to T. And this can be rewritten as Y. Of T Equals c. one T plus C two. In that quantity times E. To the negative too. And that is the answer this equation.


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