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(20 points) Letu(r,t) =e n"#(C1" cos(Ax) Cz sin(Az)) Find all values of C1; Cz and for which satisfies the boundary conditionsUz (0,t) = 0andu(L,t) = 0for...

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(20 points) Letu(r,t) =e n"#(C1" cos(Ax) Cz sin(Az)) Find all values of C1; Cz and for which satisfies the boundary conditionsUz (0,t) = 0andu(L,t) = 0for every

(20 points) Let u(r,t) =e n"#(C1" cos(Ax) Cz sin(Az)) Find all values of C1; Cz and for which satisfies the boundary conditions Uz (0,t) = 0 and u(L,t) = 0 for every



Answers

Solve the boundary value problem. $$p^{\prime \prime}+2 p^{\prime}+2 p=0, \quad p(0)=0, \quad p(\pi / 2)=20$$

Alright for this problem is to find the solution to this differential equation. And so let's go ahead and start off by replacing these wide terms with our terms instead. So we'll have r squared minus 10 R plus 25 equals to zero. And we can do a little bit of factoring So we'll have our months five whole squared equals zero From this. We can extract that our values are five and 5 Or that five is a repeating group. And so from this we can actually build our solution so our solutions reform C one E. To the five X plus C. Two X. E. +25 X. And we add this X term because it's repeating and since we're giving initial value conditions with wives zero equaling one and why of one equaling zero, We can actually sell for c. one and c. two. So let's go ahead and start off by doing that. So we'll have c. One so we're gonna plug in one for X. Will have c. one Eat to the zero which is one plus C. Two times zero Which is zero. So I'll just have c. one equals to one. And for the for the next thing we have one so where we see one, we're going to actually see we're gonna plug in one. So F. C. One to the fifth power plus C. Two. U. to the 5th power equals to zero. And from this we can go ahead and extract if C one equals to one and C two has to equal negative one. And so we can build our solution So it's going to be y equals five x minus X. E. to the five x. And so that's going to be your answer.

For this one we got our Squire plus 100 equals to zero. So our square equals to 9100. If you take a square rules on both sides Our address vehicles to post a possible Matthew 10 I. So the general function for this 1 C1 call sign 10 x. See to sign 10 x. Mhm. So we got the initial condition when X equals to zero, y equals to two. So Coulson zero just equals to one son, zero equals to zero. So C one equals 22. And when X equals two pi Y equals 25. So we got Coulson 10 temp. I wish just equals to Coulson to pi equals to Coulson zero equals to one. Saw the same for Santa and Pie, which equals to zero. So we can see in those situations equals to two C equals 25. C has two different values. So we can. So those questions.

For this one we got our Squire minus six hour plus 25 week holds +20. So AR -3 Squire Plus 16,000. So we got AR -3 Squire equals 2 1916. AR -3 equals a positive. Thank you for I Are just echoes to three plus or not you for I. So we got a general solution. Mhm. So we got to initial condition that is when X equals to zero, Y equals to one. Coulson zero equals to one. Send zero equals zero. So see one dress equals to one and when actually called soup I. Y. Equals to two, closing for pi equals to one and sign for pi equals to zero. Here we go. See one nickels to to our YouTube Taiwan three pies. So we can see see um got two different values so we can solve this crime.

So for those, well we got Night Our Squire -18. Our last 10 equals to zero. So three AR -3 Squire. Last one equals to 03. Ar minus three spire. He calls you next to one. So three AR -3. We take us where rules on both sides just equals to post your nephew. I and then we can move three on the side and then you are the both sides of the street. Yeah. So by the general rule we have Y equals to buy the for the general solution. Seven actually calls to zero Y equals to zero. Got it a powerful Acsi cost one. Close and 00 equals one sign zero just equals to zero. And also we got an X equals two pi Y equals to what? Oh yeah. Mhm. So Cosan pi over three just equals to 1/2 son. Pi over three equals to square roots out 3/2. Because we know see one just equals to zero. We all got one just equals two Spare results three over to eat too high c. two. So if we divide both sides by all those terms we're gonna see too just equals to two hours square result three E. To the power of 95


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