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Solution by eigenval4) Write the initial value problem vector form b) Solve by using eigenvalues/eigcnvectors. Write the solution in z(t), U(t) form;1 I +2y I(0) = ...

Question

Solution by eigenval4) Write the initial value problem vector form b) Solve by using eigenvalues/eigcnvectors. Write the solution in z(t), U(t) form;1 I +2y I(0) = 4r + 3y" v(o) =I(0) v(o) =1I(0) JI + Sy' v(0) = 11 2(0) = 1 41 - 3y" y(o) = -21=81 4 3(0) = T(0) =I+2yI +24 21 _ 3y' v(o) = 2131 - 5y I(0) 1 3(0) =I - I(0) = 21 - y' "(o) = 21=_T T(0) = 2 9 31 - y ' v(0) = -38-2-y I(0) = 3 1 = 2r 2y' v(o) = 51 61 = %y I(0) = 1 v(o) = -

Solution by eigenval 4) Write the initial value problem vector form b) Solve by using eigenvalues/eigcnvectors. Write the solution in z(t), U(t) form; 1 I +2y I(0) = 4r + 3y" v(o) = I(0) v(o) = 1 I(0) JI + Sy' v(0) = 1 1 2(0) = 1 41 - 3y" y(o) = -2 1 =81 4 3(0) = T(0) =I+2y I +24 21 _ 3y' v(o) = 2 1 31 - 5y I(0) 1 3(0) = I - I(0) = 21 - y' "(o) = 2 1 =_T T(0) = 2 9 31 - y ' v(0) = -3 8-2-y I(0) = 3 1 = 2r 2y' v(o) = 5 1 61 = %y I(0) = 1 v(o) = -



Answers

Solve the vector initial-value problem for $\mathbf{y}(t)$ by integrating and using the initial conditions to find the constants of integration. $$ \mathbf{y}^{\prime}(t)=2 t \mathbf{i}+3 t^{2} \mathbf{j}, \quad \mathbf{y}(0)=\mathbf{i}-\mathbf{j} $$

So you have this one. And straight away we're gonna integrate this one and this in a VTR DT. And this is gonna be eat a t I. Plus he's a negative. T j uh, plus two e two to TK We'll see right eso Once you have this immigration, uh, you're gonna use this initial condition, right? So where every c t here gonna put zero and equated the result to negative iPods for J? So what is eating? Zero is one. I in fact, them just make it. I write. So one eyes just I What is t 20 again is just j. So this is gonna be J, right? And then what is gonna be this one? This is gonna be to Kate, right? Because e t zero is always gonna be one, and then just see, that is gonna be negative. I plus four chain. So what is he gonna be then? Well, see is gonna be negative to I write this. When I subject I from either side, you're gonna have negative to I when I subtract j for me to society, gonna have three. J um, I subtract two came from the other side is gonna have native to Kate. Right? So our equation now becomes minus two. Oh, right. Because this is E T I. And this is negative two I. So I combined them straight away. Right? This is gonna B e t plus three o. J. Because you have he to the negativity and have positive, uh, three j So I combined them right in the same way. Finally, you have to e to the t. They have negative, too. So I'm gonna pull out, too. They have e they have minus one que right. So you have this one, too, to integrate, right? So when you do that now, you can sit separated That variable straightaway raisel gonna have the r equals, um, e to the T minus two. Right, So this is gonna be e to the T minus two. I the 2nd 1 is gonna be I was eating my to negative Teoh plus three j. And the final one is gonna be to out e t. He ain't any U two's incident to t minus one k. Right. And then we separated it here. Then we're going to do now is in a great, uh, if each term raise. It is really our of tea. Now, what is the integration of this one is gonna be e t u minus two t i plus negative e to the negative t right. Plus three t j then Plus, what is this one gonna be? This one is gonna be just, uh, e t two t, right? Uh, minus two t k races. So that is what we have. And then we'll see. They're gonna use the second initial condition. Here, Teoh find, uh ah. See? Right. So ever received t here, will it put zero? So this zero is gonna be one. So and the zero is gonna is gonna be zero right to negative. Two times zero is zero. So this is gonna be I She was gonna be the same thing here. Is gonna be zero here is gonna be negative one. Right? So this is gonna be negative, J This is gonna be zero now, this is gonna be one, because heated as heroes one. And this is gonna be zero. So it's just gonna be positive, K right? Positive K plus e. And we're gonna equate that This result Teoh three i plus J Plus two k. Right. So this is gonna be, uh, three I plus J plus two K. Right? Says three. I plus James was two k. Now we're gonna so c is gonna be, uh, When I subtract I from either side, you're gonna have to I wanna add J to set aside, I'm gonna have to j When I subtract k from inside, you're gonna have plus K. So that is our see Right. So see, here in this equation is all of that. So in place of C, I'm gonna put this one right When I do, That's what am I gonna get? Well, this is I This is I am gonna combine them. So this is gonna be a E to a T minus two T minus two. Oh, I This is Jay, this is Jay. I'm gonna combine them, so I'm gonna have ah, plus two, uh, minus eat a negative. T plus three t o. J. And this is K. In this case, I'm gonna combine them. Eso I'm just gonna have plus one plus either to t minus two team. Okay? And this becomes the long answer and not pretty

May I have this one and still gonna be are simple Differential Equation s said this is same thing that we've been doing before. It's still gonna be the same thing that we're going to continue. So you have. Ah, this is Plus here. Uh, so this is first t. I was t j plus to t Square K DT and integrating, uh uh, either side of the equation. So you're gonna have a function of tea, right? This is gonna be t 2 to 4/4, plus duty squared, great duties. Where'd I then? This one is gonna be t squared over two j plus, uh, two or three t cubic k. Right plus laboratory, Constant C. And then we're gonna use this initial condition to find RC way. Have t We're gonna put zero there, right? Cause it's t here. I'm gonna put zero there, So to find our initial conditions. So I suppose you're here. This is gonna be zero. This is gonna be zero right concealer time to you. Number zero. This is gonna be zero. This is gonna be zero. So everything is zero. You just have see? And then you equating the result to this one, right? That is the initial condition. So quit in this result to I plus Jay So see it as you're gonna be. I opposed a because everything here is zero. So let's just see being equal to I plus J, right? So see, is I A plus J Therefore, whenever I see see, I'm gonna put I I plus J there. Right? So after the initial condition, we know that sees I plus J Therefore, we're gonna have t to the +44 close duties where I plus t squared over two J plus two with three t's way. Okay. Plus I plus J. Right, So that's what you have. And then what we're gonna do is just group like terms, right? So this is gonna be t to the 4/4, plus duty squared, plus one. Right, cause this is just one I, sil close one, and then all is gonna have I attached to it. And then this is also one right one j. So this is just gonna be t squared over two plus one, and then you just attached J K. Does not have anything, so it's just gonna be the same thing here

We have. Why? Double prime of T is equal to 12. T squared I minus two t j with initial conditions. Why of zero equals two I minus four j and why prime of zero equals zero. Let's go ahead and take and into grow of Why double prime, um and that's going to give us 40 cubed plus C one in the first component and negative T squared plus C two in the second component. Now we have initial condition and we can calculate why. Prime of zero a C one c two. Using this initial condition, we see C one C two must be zero. So we have y prime is equal to 40 cubes the T squared. Let's go ahead and integrate one more time to figure out what why is that's t to the fourth plus D one Now notice. I've switched over to using D's and Nazis to prevent any confusion, and we have negative t cubed divided by three plus D to if we evaluate this guy at zero, we get d one de to, and we have initial condition up here. That tells us the D one equals two. The D two equals negative four, and we're done

We have. Why? Double prime of tea is equal to I plus e to the T. J. We have initial condition that why of zero is equal to two I and that why prime of zero is equal to J. Let's go ahead and cake and figure out what? Why? Prime of tea is just by integrating the components. So if t plus c one for I and we're gonna have e to the T plus C to for Jay now, we're going to evaluate this at zero. That tells us C one I and one plus C two j now using our initial condition. That tells us that C one equals zero and that C two equals zero. Okay, so we have why Prime of T is equal to t. I have read it in brackets here to the T, and now we can go ahead and take another integral. So we get t t squared, divided by two plus D one and e to the T plus D to now notice that I'm using D one and D two because and the previous step already is C one and C two. So I didn't want it to be any confusion about that. Now, let's go ahead and use our initial condition here. Um, and also evaluate why at zero. So when we evaluated zero, we get d one one plus D to and we know from the initial condition this he used equal to I. Okay, so that tells us the D one equals two and the D two equals negative one. And that's it. We're done.


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