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Point) Find the solution ofy" +4y +4y 288with y(0)and y (0) = 6.Preview My AnswersSubmit Answers...

Question

Point) Find the solution ofy" +4y +4y 288with y(0)and y (0) = 6.Preview My AnswersSubmit Answers

point) Find the solution of y" +4y +4y 288 with y(0) and y (0) = 6. Preview My Answers Submit Answers



Answers

solve the given differential equations. $$D^{3} y-6 D^{2} y+12 D y-8 y=0$$

Welcome back to another different location problem. And this one we have a wide double prime plus six. Wide prime plus 13 Y equals zero. How to solve this. We have to take the characters took equation to be r squared plus six are plus 13 equals zero. Great. Now let's complete the square to get our into one term. This will be our plus three, 20 squared. I did this by just dividing the middle term by two. Now we have to find out what this next part is. Since we have a nine here in the 13, you know that the leftover part is a four. They will be equal to zero. Great. If we subject for it and take the square roots, We get our plus three is equal to plus or minus to. I like that. And if we subtract the three we'll get our equals negative three plus or minus two. I All right now we have two solutions the positive to I and the negative too. I. So we'll have to take a linear combination of these uh solutions to get our total solution. I'll start off with the positives with the positive solution. The characteristic equation. And remember this gives us um results of the form Y equals E. To the R. T. C. In the front historically very constant. But let's uh Let's write this up at the top as why one equals C. One times. What is this negative? Three plus two I. E. To the T thomas negative three plus to I. All right. And if we simplify this, what we get, we take the real part out here And that will be easy to -3 T. And if we take this complex part that will turn into signs and co signs. So this result will get us why one is equal to. So you still use green here Because why one equals C. one E. to the T. Times cosign of two T. Plus I signed up to T. Right. There we go. Uh now there will be very similar case for the 2nd 1 where will have a negative two I. Inside here. Uh Up here. But remember we're going to have a different constant out front. So this will not change the uh the value of this uh total linear equation. Um Well the only difference will be there will be a negative I sign here. But since uh when when we combine them we will have two different constants playing out. So we'll have a C one minus C two on uh next to this uh sign term. So what really matters is that there is a constant in front of this sign term in the general solution. So let's write that as why is our final solution And this is going to be E to the T times C one co sign of two. T plus C two. Sign of two T. Now, if you really want to talk about the exact values of these constants, they're not the same as the ones here. Um In fact, this one should be C one minus C two, but we're just going to combine all the constants into new ones because it doesn't really matter what their value is um since we don't have a precise uh the point quite yet, but that is our solution to this general to this difference equation.

So here we are given the second order differential equation. Why prime prime Plus two of Y Is equal to zero. And writing the characteristic equation here we have arab squared Plus 12. A constant is equal to zero. So remember here that We don't put an hour when there is no derivative of why? So here since we just have a Y. It's just 12, not 12 are now going forward serving for the roots of this equation. With the characteristic equation. We have r squared, our squared being equal to minus growth. And taking to the rules both sides, we observed that we are going to have complex roots and they will be plus or minus For the root of -12 which is plus or minus. There were 12 4 x three x -1 which is minus proof. In here. We have plus or -2 Who drove three i. and these our roots. And now our general solution will be in the form of why T. Is equal to C. One E. To the power of lambda T. Of course new T plus C. Two E. To the power of lambda T. Sign, new T. And we're lambda zero here. And our mule is 203. Now substituting this into our general solution. From here we have whitey Is equal to C1 and since we have E. To the power zero T. The E to the Power zero becomes 1. So we don't have an e term here. So of course our new is to root of three T. Right? Plus C. Two. Sign To write off three T, and this is our general solution and our final answer.

First Dickerson is X minus y. It's a close to six. Second is four x plus three way is it was too minus food putting point four comma minus two. We get in fast. A crescent four minus off minus two equals 26 This implies four plus two, is it? Question six. This implies six is a close to six, which is do in second Question four, multiplied by four plus treat My departed I minus two is the question minus four in place 16 minus six is equal to minus for this in place, then it a close to minus four. Well, we just falls, so the point is not a solution. Second point is six. Coma zero. So let's put in first equation six minus zero is a close to six. This implied six is a close to six, which is true. Next is four multiplied by six plus three. Multiplied by zero is the question minus food. This implies 24. It was to minus food that is false. So again, this is not a solution. Third point is two comma full. So let's put it in the given in crescents In first it Chris and we're open to minus four. It's a close to six employees, minus two, because 26 which is false. Let's put it in secondly question. We often for multiplied by two plus three. My deployed by four ID equals two minus four. This implies it plus 12 is a close to minus four. This implies twenties because to minus food, which is falls so neither six comma zero nor to calm a four is a solution. Do given system off in questions.

I wouldn't decide if this given 0.132 is a solution to the given system. So for a point to be solution, it needs to work in every equation in the system. So 132 would have to be true for all three of these equations for it to be solution. Now it goes in alphabetical order for this coordinate. So it's X y Z, and then I want to fill into each one. So, like if I fell into equation one with X, Y and Z, I feel in one plus three plus two and then I'm trying to think, Does that actually equal six? And in this case, yes, A desk. So then I want to check it in equation too. So I'm gonna fill in one minus three and then minus two billion in accident. Why? And then Z and I want to know. Does that equal negative for Well, I guess one minus three is negative. Two minus two is negative for So it does work for that third for sorry second equation. And I want to check the third equation so two times x plus, why minus Z and I want to know does that equal to three. So that's gonna be two plus three, which is five minus two. Makes three. So then, for my final answer, I would say yes. This is a solution for the system.


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