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Let Aly) be a smooth function and letJ(y) = J" A6)y' &xandIy) =Vi+y2 dr_Formulate the Euler-Lagrange equations for the isoperimetric problem with y(0...

Question

Let Aly) be a smooth function and letJ(y) = J" A6)y' &xandIy) =Vi+y2 dr_Formulate the Euler-Lagrange equations for the isoperimetric problem with y(0) 0, 9(1) = 1, and I(y) = L > V2. Show that A = 0, and that there are an infinite number of solutions to the problem. Explain without using the Euler-Lagrange equations (or any conservation laws) why there must be an infinite number of solutions to this problem:

Let Aly) be a smooth function and let J(y) = J" A6)y' &x and Iy) = Vi+y2 dr_ Formulate the Euler-Lagrange equations for the isoperimetric problem with y(0) 0, 9(1) = 1, and I(y) = L > V2. Show that A = 0, and that there are an infinite number of solutions to the problem. Explain without using the Euler-Lagrange equations (or any conservation laws) why there must be an infinite number of solutions to this problem:



Answers

Find the solution to the Cauchy-Euler equation on the interval $(0, \infty) .$ In each case, $m$ and $k$ are positive constants. $$x^{2} y^{\prime \prime}+x y^{\prime}-m^{2} y=0$$

So here we need to first solve the initialization. That's gonna be our square. And then we have a plus. A one minus one are term. So anyone here is just one. So we have one minus one are zero r. Then we're gonna haven't plus a to A to here is negative m squared. So we're gonna have minus M squared. Is he? Will this euro this condemned factor into r equals R minus M and R plus M is it with a zero? So we got that are is gonna be plus or minus m. So our general solution it's gonna be y of x equal. C one exit e m plus C two x to the negative m

So you solve this problem here? Where first going to solve the additional question. That's gonna be r squared. And then we have a plus a one minus one. Our time. A one here is the negative two M minus one like that. So we're gonna have minus to em. Um, some minus two M plus one and then minus one. So I'm just gonna be minus two m r and then we have Oh, sorry. This should be a plus. I'm squared. So then plus a two, which is M squared, is equal to zero. This can factor into n minus R. Well, sorry. Ar minus M squared is equal to zero. So we get that are is able to em with a multiplicity of to. Okay, so then our general solution why of acts is gonna be go to X to the foreign started. See, one adds to the M plus C two and then exit e m times Ellen of X for that second multiplicity

To start solving this problem here. First, we need to find the indicia location that starts with R squared and then plus a one minus one are so here. A one is able to this here which is equal to negative to m plus one. And then we subtract one. So that becomes just minus two m. We have minus two m r here and then plus a two which is M squared plus K squared is equal to zero. We don't need to use quadratic equation to find are so it's gonna be people to negative be which is to m plus or minus square root of. And then we have B squared, which is forever squared minus four and then a C is just one and then we have see Okay, that's a square root of all of that. There were then divided by to a so outfront we have ah on em and then plus or minus, the M squared is here will cancel out and then we're just left before negative four k squared square to that is going to be equal to, uh, that's gonna be able to to k i. And then we have a divided by two on the bottom. So we're just gonna have plus or minus K I So our general solution Why of X is gonna be able to see one and then exiting A, which is M and then co sign of B, which is K. You're OK, Ellen of X and then plus C two x to the M and then sign clips. Sign of K Ln X

In this video, we're gonna go through the answer to question number 32 from Jack 10.2. Eso were asked to and show that for the path of differential equation given at the top here. Andi, assuming that the boundary conditions one too satisfied, the non trivial solutions must satisfy a different. So have boundary conditions best in show. Those That's our traditions equivalent to these. If we assume a separable for Okay, so separate form is I'm gonna let you, Which the function of X Why t equal Thio a product of functions of each variable us. Okay, so now that substitute this form into the first round of condition, I was gonna say the ex evaluated at zero times why, But why? And t t he's got to be equal to X. So I said, eh, times why Time t are equal to zero for the full domain of variety. As this is to be true for all values of wine over UT given range. So therefore, it must be true that x zero on Becks, eh? Equal to zero. Okay, so first rounder condition straight into the second round of position. We have X Molly rated x the first year. Save it. Why? Evaluated Xaver Times t vote fatigue is equal to x X. Why everybody has to be where? White power bottom It'd be times t that tea he was zero. And that reasoning, um, given the this has to be true for all the whole domain of X and the whole debate t it must be the case that why crime that zero is equal to wide crime could be which sequences era, and that completes the question.


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