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Coursework 2.1 Give examples ( at Ieast three) of Difference equations considering the three cases Real and Di stinct, Real and Equal, and Com plex conjugate Roots)...

Question

Coursework 2.1 Give examples ( at Ieast three) of Difference equations considering the three cases Real and Di stinct, Real and Equal, and Com plex conjugate Roots) with any of forms below ~ } ''~T't : ~ :' the following i1 nput [ ` " ` . . 7 ". . . .' x[n] = 3(0.2)". Ypln] P(0.2)". [n] = r Yp[n] = P + Pzn + Pzn? + Pan? x[n] cos n. Ypln] P1 cos Zn + Pz Sin 2n. ~ " ` ` ` ` Show the solutions for the total response using the Classical Method,

Coursework 2.1 Give examples ( at Ieast three) of Difference equations considering the three cases Real and Di stinct, Real and Equal, and Com plex conjugate Roots) with any of forms below ~ } ''~T't : ~ :' the following i1 nput [ ` " ` . . 7 ". . . .' x[n] = 3(0.2)". Ypln] P(0.2)". [n] = r Yp[n] = P + Pzn + Pzn? + Pan? x[n] cos n. Ypln] P1 cos Zn + Pz Sin 2n. ~ " ` ` ` ` Show the solutions for the total response using the Classical Method, consider Initial conditions other thanO. `' : ' :'. . `~ Submit the scanned solutions in pdf .



Answers

We have shown that in the case of complex conjugate roots, $r=a \pm i b, b \neq 0,$ of the indicial equation, the general solution to the Cauchy-Euler equation $$ y^{\prime \prime}+a_{1} y^{\prime}+a_{2} y=0, \quad x>0 $$ is $y(x)=x^{a}\left[c_{1} \cos (b \ln x)+c_{2} \sin (b \ln x)\right]$ (a) Show that $(8.8 .26)$ can be written in the form $y(x)=A x^{a} \cos (b \ln x-\phi)$ for appropriate constants $A$ and $\phi$ (b) Determine all zeros of $y(x),$ and the distance between successive zeros. What happens to this distance as $x \rightarrow \infty,$ and as $x \rightarrow 0^{+} ?$ (c) Describe the behavior of the solution as $x \rightarrow \infty$ and as $x \rightarrow 0^{+}$ in each of the three cases $a>0, a<0,$ and $a=0 .$ In each case, give a general sketch of a generic solution curve.

Yeah. We asked to sketch what this the solutions that this differential equation kind of looks like what we would expect it to look like, why prime equals 2/3 Y -3. So we'll call this thing Z. And we can plot that. It's obviously just a line here. So Z is a function of why it crosses zero at let's see 4.5. So what that says is when y equals 4.5, why primary zero? So why is it changing? So there's this equilibrium point here. But that equilibrium point? Well, in in future instability analysis you would say that is an unstable equilibrium point. And why? Why is that? Well, you can see here, say we started five, just a little bit above that. started five. We can see that the derivative is positive. So we're gonna go to higher values of why. So we're going to move out here where the derivatives positive, we're going to move out here and then we're going to keep moving and moving out and moving out. So where if we start, if we think about our trajectory along this line here, basically parameter. Rised by T. We start here and we're gonna go off this way, we start over here, notice that we have the slope is negative, why prime is negative over here. So we're gonna go to smaller values and it's still negative, we're gonna go to smaller values and go to smaller values and we're just gonna keep going off in that direction. So unless we start exactly right there, we're going to basically move away from here. And so what happens is we can see the first case, we have three, so we're gonna move off that way. So this plot is gonna go, you know, it's gonna cover around here, but then eventually it's gonna actually three is down here in the blue, it's gonna shoot off to infinity. Um as he goes to his t coast infinity. So it's just gonna keep growing exponentially. Obviously you can solve this differential equation exactly, and so you can see but the exponential growth Now for six, that case we start over here and so we actually grow exponentially and why it gets um large in the positive direction exponentially. So we have a curve that looks like this. And so obviously they started them for values here, but because I've plotted out, you know, I've used a big scale here, we can kind of everything gets smeared together.

Original equation is a solution. We could go ahead and take the derivative, and this would use the quotient rule. So we have low de high negative sine X minus high do you glow and then squaring the bottom? All of that is equal to y prime. And as you can see, it's going to get multiplied by an X here. And so we're first going to just substitute this in. Or why prime? Let's move by all of this, my ex, which would then just cancel one of the exes. Hello and then Plus, we'll add why, which is the original equation, those sine X over X, And we want to determine if all of this is equal to negative sine X, so we'll just leave the right side along as negative sine x, and then we'll rearrange the left side. From here, we have common denominator of X says you could see the positive and negative co signs cancel. And then we do in fact, have negative sine X. Once we divide both these exes out, and then we're just left with negative sine X on the left side. Negative sine X on the right side. And so we could go ahead and, you know, again I have, which would then end up being the same thing on both sides here as well. So, in either case, we do, in fact, have, um, the satisfaction of this differential equation, given this solution.

Given a little bit. Dash X is three x plus sign by X on effort to go to three. We need to finally, and today we're still on for the crops. So we're just find the ability to do because a photo truly eggs this same by X the eggs. So this is the goto three x squared, divided by two minus caused by Exteberria. But by blessed see So let us find effort toe It is equal to three into pulled by two minus cause to buy by five city and it is given ist three. So make it see best minus three plus one way they were kopecks is a well too three x squared, divided by two minus course by x q eight. Their thing minus we lose one way, right? The DUI it wishes formed, uh, this on this And the graph for the questions are even as this so orange Line is representing. There is no question with the costume and Red Line is representing the function of things


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