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(2y^2 x^2 Jdx xy dy = 0 Find the general solution: Write down the methed You use_...

Question

(2y^2 x^2 Jdx xy dy = 0 Find the general solution: Write down the methed You use_

(2y^2 x^2 Jdx xy dy = 0 Find the general solution: Write down the methed You use_



Answers

Solve the differential equation.
$ xy' - 2y = x^2, x > 0 $

All right. So start off by replacing these white terms with our terms. So we'll have our squared minus two ar minus two equals to zero. And I don't see any obvious factoring here. So I'm gonna use the quadratic formula So I have r equals two um negative B rather to plus or minus the square root of be square -4 times a times c. All divided by two times a. Mhm. So that simplifies down to two plus or minus the square root off -4 divided by two Which is one plus or -1. So with these routes we can go ahead and build our solution. So solution is going to be in the form see one co sign of X plus C. Two sine of X. Mhm. All times E. To the X. And so that's your answer.

In this problem, we have to solve the differential equation. Why Double prime minus six y prime plus nine y is equal to zero so forthis kind of differential equation. We can always associative characteristic a creation, which is going to be D squared minus six times D plus nine years ago to zero. Now this implies D minus three. Whole squared is equal to zero thirties. They called to three. In this case, we see that we have only one solution. So the fundamental solution it's going to be easy to the power three x and x Times E to depart three extra. Remember anybody on the same route? Then the fundamental solutions becomes X times E to the poor that route time. Six. So then that general solution it's going to be some constant c. One times e to depart three Ex Placido times X times It'll be poor three X

Hello and welcome to another differential equation problem. And this one we have Y double prime. When it's too wide prime plus two Y equals zero. To solve for this, we have to use the characteristic equation and from section 3.1 we learn that if we want solutions of the form E. To the R. T. We can actually simplify this using the characteristic equation of R squared minus two are plus two. Where each power of our is equivalent to a derivative of the differential patient of above. All right now let's solve this. Uh quadratic by completing the square and we'll get our minus one quantity squared Plus one equals 0. All right, let's subtract one and take the score. You get AR -1 equals plus or minus I. And lastly we will add one to both sides. So we are equals one plus or minus. I. Now due to this plus or minus term here, we'll have two solutions to this. Ah The differential equation and the optimal solution will just be a linear combination of those two solutions. Um So if we remember this characteristic equation will get us results of the form by equals C. E. To the RT. So let's substitute RN for our discovered value from this characteristic equation. So why someone look right with green Is going to be C1 E to the one plus I times t. Now, it doesn't matter which one we do first. I'm just gonna do plus uh for this case. Now, the other solution, all right in blue Will be white to let's see two E to the one minus I team. Great. Now let's uh let's use oilers pharma. This is what we learned earlier in this chapter. And that's if we have the uh if we have E to the I theater, we can actually rewrite this as kusenov theta plus I sine of theta. All right. So I'm going to uh do that here. Start off with the left side. So if we note that we can take out this E. To the T. Because there's a there's a real component. So why one equals c. one eat to the T. But times is uh E to the I. T. So that means in this case uh theater will just be T. Because we'll have I times T. Because I hear the data in this case must be T. This will be times co sign of T Plus I Sine of two. Great. Now for the second case we're gonna we're gonna have a very similar situation because the the real part will factor out. But the only difference is instead of tea it will be negative T. Because we have this negative sign right here. So I'm gonna write that out now. We'll have y sub two the C. Two E. Two T. Times. Cosine of T. Plus I signed up co sign of negative T. Plus I sign of negative T. Great. Uh Now what's the next step? Well the next step is actually to relate these two values. Um We we know that uh they're the real solution. The most broad solution will just be a linear combination of these two solutions. So what we can do is we can uh take all of those linear combinations and just put it into uh these variables here, these constants there. Uh those will absorb all the factors. So the next real step is to resolve this intersection here. So we can really add these two together and get why one plus Y. Two. So let's first think of what is co sign of negative T. Well if we look at the graph of co sign of negativity, we see I'm going to write it up here. Co sign of negativity will look something like this where this is the Y axis and this is the exactness. And if we take the negative, if we substitute any value of X. Access for the negative X. X. For its negation it won't change value. So co sign of negative T. Is just co sign of T. So I can erase this negative there. On a similar note, we note that uh the negation of sign will actually be it's negative because the sine function looks like this for its initial values X. And that's why. Um So when we have a negative inside the sign will be negative sign. And on the outside so we can do negative on the outside and raise this intersection. Great. Now let's take the some of these two uh components. Why one plus Y. two. This is the uh we'll get the solution to our differential equation. Uh And this one will be E. To the T. Because that's coming to both um Plus two co Santee. Because we have co sign of tea on both both of them. So call it a C. One co sot of T. Plus. Well this uh sign I sign of two will have a C. One value attached to it. Um So we can't uh we don't know what this value is compared to this negative uh term over here. So we are still going to have to include it uh in our final solution. But we'll have to see to sign to T. They're a sign of two and that concludes our problem.

In this problem, we have to solve the differential equation. Why? Double crime minus y prime minus six. Why is he called to zero so forthis kind of differential equation? We can a specific characteristic in creation. So the characteristic decoration it's gonna look like the square minus T minus six. Is it called a zero? Now let's try to solve this equation. I can see that there is a nice factory ization. This can be factored as t plus two dines. T minus three is a call to zero so that in place either tease a call to negative to Ortiz is a call to positive. Three. Certain The fundamental solution are going to be Ito the poor. Negative two times six on each of the power, three times six. So then, general solution, it's going to look like Why is he going to see one? For some constant? It'll depart negative to x plus C two you to the poor three x So that's going to be might general solution


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