Solve the following IBVPU - Uxx = 0 ux(t,0) = 1 Ux(t,1) = [ (uco,x) = 1...

Solve by any method. $$ \frac{2}{u}=\frac{3}{u^{2}}+1 $$

Hello there. So for this exercise we have a set of two vectors and we need to apply the grants need or organization to obtain an Ortho normal set. So here is the graphical representation of these two vectors. So the victory You want is equal to 1 -3 and the vector YouTube is equals 2-2. So you're going to kind of visualize what is happening either in the process of the Grand National procedure. So let's start. So the first step is to choose victory one that here, I'm going to the note with the hat here, the unitary vector. So in this case this is not unitary. So we are we assigned to the we want the vector you want. That means that there's equals to one minus three. So basically we're calling this vector here the one. Then we need to change with respect to this victory. We want the victory you to, to have, uh, we need to make This veteran YouTube or phenomenal to be one and that is to assign the video. That is not normal yet. It's not unitary yet. There's going to be what? Well, we're going to take the orthogonal component of you to with respect to be one and that is taking you to minus the projection off you too On the Vector V one. So geometric, that means that the projection of U two on the one is something like this. Okay, so this projection is going to put in a different color here minus The projection of two. Everyone. So that corresponds to this vector here. This vector here corresponds to the projection of the two Overview 1. So what happens if we take you to and we subtract this, This is like the what is the component of Youtube that is aligned to be one. So when we see obstruct them, we will obtain a victor here. A component 32 That is going to be orthogonal to be born. So that's exactly what we're doing here. So this is equal to take him to to minus the inner product of you two would be one B one over The norm of B one square. This result in a bit that is equal 2, 2 to -1 over the square root of 10 times Actually 1 5th time, one -3. And when we saw these two vectors together We obtain the actual solution. That this 1/5 times well stupid Here. So we too would be one is equal two to minus six and the norm the square of the norm of B one is equal to to 10 times The Victory one that is 1 -3. And this is equal to victory to to minus which is minus is plus to fifth, one minus three. And then the result is equal to one 50 12 4. Okay, so that corresponds to these two vectors the two. And in this case we want will be this factor is the same as you want. And then what we need to do is uh normalize the factors that means that we will obtain in a small factor here, the two that is unitary and a small victor over here. That's going to be the one unit terms. Just to take the directions basically. So that means that we need to take V one. We need to normalize me one And that corresponds to taking everyone divided the norm of B one and that is equal two, one over the square root of 10, one minus three. And for me too, to normalize it, we need to take me to and divided the norm and this is equal 2, 1 over the square root of 10, 31 And this is the Yeah, the final result. The final transformation of the applying the garnishment procedures. So at the end We started with two vectors this summer ISIS, you won and you too And we obtain two vectors that are Ortho, no, you the two and the one. We transform these vectors in an Ortho normal set following the grand procedure. So just to re capital Ating we just need to pick the first vector will be the same vector and for the second vector then we start to see obstructing the projections of the second factor with respect to the first vector that we choose. So basically this is a kind of recursive algorithm. And after obtaining these two vectors that are Ortho no, then we normalize them and we obtain the final orphan and all set.

He's got Yeah I guess were given two vectors and see to uh you with components one plus I two minus four, I and V with components one minus side two plus three. I now in part A were has to find the infinity norms of U. And V. We'll just as in the real case the infinity norm of you. This is the maximum of the ma july of the components. So this is going to be the maximum of the module lists of one plus I and the modules of two minus four I, which is the maximum of the square root of one plus one. In the square root of four plus 16, which is the maximum of Route two and Route 20 which is clearly the square root of 20 hungry masses. We should get Likewise the infinity norm of the Well find this, we want to find a maximum with the module I. Of the components. So the module it's of the first component one minus I is the square root of one plus one, which is route two. And the modulates of the second, component two plus three I. Is the square root of four plus nine which is 13 and therefore the paranormal V. Is the maximum of these two. Route 13. Thank you everyone. Thanks for listening to show in part B. Bye. We're asked to find the one norms of U. And G. Well similar to the real case, the one Norman a complex space of a vector. This is the sum of the module I. Of the component. So this is The module s of one plus I Plus The Module Lists of 2 -4 i. Hitting them, which we found this was route to and route 20 in part A. This is route two plus route 20. Okay, Likewise, the one norm of v. This is the module lists of the component one minus I. Plus the modules of the second, component two plus three. I We found these were route to And through 13. This is route two plus route 13 but constantly sort of like something. Yeah, but just there's got to be racist and part C rest to find the two norms of U. And V. It also like it'll start growing hair on the inside. Sometimes this is a norm that were kind of familiar with already with complex numbers. It's the square root of the sum of the squares of the module I. So this is the square root of the Module lists of one plus I Squared plus the module lists of 2 -4 I squared, which is the square root of one plus one plus four plus 16, which is the square root of 22. I feel like I'm Likewise the two norm of V is the square root of the modulates of one minus I squared plus the module lists of two plus three I squared, which is the square root of two plus 13, which is route 15. Okay. Finally, in part D, we're asked to find the infinity distance from unity, the one distance from unity and the two distance from utility. While by definition the infinity distance from U to V, this is the infinity norm of u minus B. So to find this, let's find what U minus V is. Well, this is one plus I minus one minus I, which is uh to I and the second component is to minus four, I minus two plus three I, which is a negative seven. I I'm typing away. And as we saw before, the infinite normal of the vector is the maximum of the module I. Of the components of the vector. Now the module lists of two I is simply too. And the module lists of negative seven is simply seven. So the max the infinity norm of U minus B is to sorry seven find the one distance from U to V. This is the one norm of U minus V to the some of the module of the components. So this is the module lists of to I plus the module lists of negative seven I, which is two plus seven or nine. I'm not going to bring my dealer with finally the two distance from UtV. By definition, this is the norm to norm of U minus V. Which is the square root of the sum of the squares of the modular. So the square root of the modulates of two I squared plus the modulates of negative seven I squared, which is the square root of four plus 49, which is the square root of 53. Mhm. Okay, mm.

We want to solve this system of the questions using Go Jordan in ammunition method. First, we exclude X one from the second equation here. We can do this by multiplying the first equation. Oh, you minus one provided by one minus I and we had the first equation toe the second vanquish. First we have the first equation. As it is, the change is only for the second equation. We have eliminated X one from the first from the second equation. And we have on the extra room we have minus two, divided by one minus I my enough When less I these old is multiplied by X two equals you. We can simplify the second equation, Toby. 24 that the minuto You have one minus all you and you. The first thing we have minus two. And for the 70 we have minus when bliss I'm a tabloid by one minus soy, which is one minus I square, which is minus one bye bye x two. And for further simplification, we can have we can about the boy, you know, one minus soy here on here the lift, insight and right hand side. And we have no minus two minus one minus one. Which is you. We have minus two equals minus one and minus minus one, which is 21 plus one. This old is multiplied by X two. Finally, we have that extra room equals zero after we have eliminated X one from the second equation. We can't go backwards to eliminate exit from the first equation here by substituting back by its very, which is you. Now we have from the fittest equation you have eliminated X do may have one minus. I have deployed by excellent equals zero. And to simplify this equation, we can write. Exxon equals zero, and we have the second equation as it is, which is X two equals zero. And this is the final answer off our problem.

Okay, This question wants us to solve this equation for you. So, two star, I'll clear the fraction by multiplying by u plus one on both sides. So on the left, the fraction just goes away and then the right side gets multiplied entirely by U plus one. So now we get to mine. Issue is equal to then distributing the five. We get five you plus five. And now I prefer the variables on the left side. So subtracting five you, we get tu minus six. You when we combine like terms and that's equal to five and then getting all the constants on the right side. So subtracting two we're left with negative six u is equal to really or U is equal to negative three divided by six or negative 1/2.