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5.Which of the equations given below is the auxiliary function (characteristic equation) of the following differential equation? (5 Puan) 2/ Sy 6 sin 2x + 7cos 2rm&...

Question

5.Which of the equations given below is the auxiliary function (characteristic equation) of the following differential equation? (5 Puan) 2/ Sy 6 sin 2x + 7cos 2rm" + Jm + $ = 0nI2m? + Sm6sin 2 + 7cos 28Zm + 2 = 0Asin 2Rcos 2 = 0

5.Which of the equations given below is the auxiliary function (characteristic equation) of the following differential equation? (5 Puan) 2/ Sy 6 sin 2x + 7cos 2r m" + Jm + $ = 0 nI 2m? + Sm 6sin 2 + 7cos 28 Zm + 2 = 0 Asin 2 Rcos 2 = 0



Answers

Find values of $m$ so that the function $y=e^{m x}$ is a solution of the given differential equation. $$2 y^{\prime \prime}+7 y^{\prime}-4 y=0$$

All right, let's start with this problem by solving for the homogeneous solution. So we'll have our squared plus four equals to zero. If we move forward to the other side we'll have the R squared equals two negative four. Now, if we take the square root we'll have the R equals to plus or minus the square root of negative four. And we can simplify that down to plus or -2. I. And so now we can build our homogeneous solution. So homogeneous solution in this case is going to see one coach signed two X plus C to sign two X. And we can also take a stab at our guest for the particular station. So our guests for the particular solution is going to have the form A co signer two x plus B signed T. Rex. Now note that their terms that match in the particular solution and the homogeneous solution. And so we don't want that. So what we end up doing is we end up multiplying particular solution by a factor of X. And so that gives our guests To be AX. co signed two x. Was B X. Sign two x. And now we need to take the river this twice. So for the first derivative YP prime, We're going to have a product rule. We're going to have a co signed two x minus two. AX. sign two x Plus be signed two x plus two. The X Co signed two x. We have taken the areas again. So have wiped double prime equals two negative to a signed two X minus for a X. Co signed two x minus two. A signed two X plus Be to be signed two x. Sorry that's co signed two x mm. And minus rather plus To be co sign two x minus four B. X. Sign two X. And so now let's find this down a little bit. So we'll have and so we'll have um these terms can be condensed. Um let's see what else going contents. We can condense think these terms right here. That's about it. So we'll have this condensing too negative for a sign two X minus four A. X. co sign two x. This condenses Plus four B. You can assign two x minus four B. Ex signed two x. Okay so with that we want to plug these into our original equation which is Y double prime. So whatever this is right here is a negative for a sign to your ex minus four A. X. Co sign two X. Plus four B. Co santo X minus four B. X. Sign two X. And then plus four. Why? So why in this case was That's so plus four a. x. Co signed two x. Plus four B. X. Sign two X. And this all equals to two coz I two X. Okay. So now let's see if we can simplify this down a little further and let's see we have these two terms right here that cancel each other out. We also have this term right here that cancels out with this term and that's it. So were we end up with negative for a sign two X. Plus four, Vico, Santa X equals two to co sign two X. And now we can try and make a system of equations here. So let's take all the like terms together. So whatever has co sign in it. So these two have co sign whatever has signed it. It's just that and nothing on the right hand side. So we'll have negative for A equals to zero and we'll have that for B equals it too. So we can translate this to being A equals to zero and B equals to two divided by four, which is one half. And so with that we can actually build our particular solution in this case or a particular solution is going to be um one half ex scientifics and we can add this onto our head mantra solution to get the total solution for this differential equation. So we'll have one half X. Signed two X Plus a homogeneous solution. In this case we set our homage a solution of C1 co sign two x plus C two. Sign two X. That's her answer.

Okay we are going to find are empty and we're going to do that by integrating our prime of T. And then using our zero to find our constant of integration. So in order to integrate these notice that both are Sine and Cosine have a two T inside. So when we do go to integrate this, if we did use substitution extra half is going to appear in front so we can take that into consideration as we go to integrate. So we'll have a one half of sign of two T. Now you know your um derivatives better. So sometimes after you do your integration you save yourself. Okay. If I took the derivative of sine of two T, I would get co signer to T and then multiply by two that too and that one half becomes one. So it's a good way to check your integration. So the same thing happens here but we also get a negative because of the integral of sine to that negative. Co sign. Now also notice that as I'm integrating, I'm adding my Constance of integration so my two becomes a two T plus C. Three in the K direction. Okay so now that I have my integration with my constant of integration I'm going to use the idea that when I put a zero in I should get a one in the I. Direction. So one equals now the sine of zero is zero. So it ends up that C one equals one. Okay now looking at my J. Components, you don't see a J component. So that means that it is zero. So it should be a zero when I put a zero in. Now remember the co sign of zero is one? So it ends up that we do have that negative half there. So when we added over we get that seat two equals 1 half. Now our last one if we put a zero in we should get a K. Equals one out. So one will equal that zero plus C. Three so C three is one. So now I can go ahead and rewrite and I'll just be adding in those constants of integration. So I get a one half signed to T plus one in my eye direction. I will have a negative one half co sign of two T plus a half in my J. Direction. And then finally we will have a two T plus one in the K direction.

Okay, so start off this problem by determining the order of this differential equation. And so right off the back, the highest power that I see on the left hand side that contains something of the form of D. To the N. U. Divided by D. To the E. R. N. To the end is D squared to you. D. R. Squared. And so what this, well we can determine from this is that we have a second order differential equation. And to determine whether it's actually a linear or nonlinear, let's go ahead. And first look at the right hand side. So we know that the right hand side has to be a function of our. And what that means is we cannot have any other variables except are in there. We can only have constants and the variable are now in this case note that there's a U. Turn here and you is in fact a variable. And so what we have to conclude here is that it's a non linear equation. So our final answer is that it's a second order differential nonlinear equation or second order nonlinear differential equation rather. So that's your answer.

Come to this lesson in this lesson, we'll find a particular ems like that. Why is it called to eat? The party? Mx becomes the explicit solution for the differential equations. So he will need my prime from here. We have em E M X. We also need by prime prime that is M squared E M X. So we're pleased that in here there. Yeah, so we have M squared E MX -5. M E M X. The last six EMX got zero. So we can go through by M. S. And E M S. Then we have m squared -5 M. The last six years go to zero. Okay. The last thing is that Instead of writing five and we're right at us negative three M minus two. Um the last six call to zero So that the two collectively become -5. So you would group them into tooth? Yes, so there's a negative year. Make the six. Mhm. A positive. Okay, So the next step is to factor them out from here. So am out the M -3. And here we can factor out too so that FM -3. So so am -2 Than M -3 spoke to zero. Which implies that M -2 scored zero. That implies that I am escort to. Uh huh. And minus thirties quote 20 Which lies Embassy called 2 3. So why he calls to E. Two X. Mhm. Or why it cost to eat? Sorry. X. Is the solution the explicit solution for the differential equation? Okay thanks for your time. This is the end of the lesson.


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