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Solving inhomogenous wave equation 
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#37
Sep2108, 01:01 PM

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#38
Sep2108, 01:21 PM

P: 37

I think I made a typing error there, of course it should be
[tex] A_{n}D_{n}=2\int^{1}_{0}(\frac{x^{3}}{6}\frac{x}{2})sin(\frac{(2n1)\pi}{2}x)dx [/tex] , right? I'll attack it right now :) 


#40
Sep2108, 02:08 PM

P: 37

Ok..puh, now I have tried.. I am not too confident it is OK though :( so much to write if I write everything, but I can show how I did and thought if this is completely wrong..
This is what I got: [tex]A_{n}D_{n}=\frac{2}{(2n+1)^{2}\pi^{2}}((1)^{n})\frac{16}{(2n+1)^{2}\pi^{2}}\frac{64}{(2n+1)^{4}\pi^{4}}[/tex] 


#41
Sep2108, 02:40 PM

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#42
Sep2108, 03:43 PM

P: 37

first time was [tex]2[(\frac{x^{3}}{6}\frac{x^{2}}{2})(cos(\frac{(2n1)\pi}{2}x)]^{1}_{0}
+\frac{4}{(2n+1)\pi}\int^{1}_{0}(\frac{3x^{2}}{6}\frac{2x}{2})(cos(\frac{(2n1)\pi}{2}x)dx[/tex] But I saw now when writing this that I used x^2/2 instead of x/2 in my calculations.. :( :( 


#44
Sep2108, 04:20 PM

P: 37

Here is the first step.. if you are online, just tell me if that step i correct before I do it all again...
[tex]A_{n}D_{n}=2[(\frac{x^{3}}{6}\frac{x}{2})(cos(\frac{(2n1)\pi}{2}x)]^{1}_{0} +\frac{4}{(2n+1)\pi}\int^{1}_{0}(\frac{3x^{2}}{6}\frac{1}{2})(cos(\frac{(2n1)\pi}{2}x)dx [/tex] where the first term, [tex]2[(\frac{x^{3}}{6}\frac{x}{2})(cos(\frac{(2n1)\pi}{2}x)]^{1}_{0}=0[/tex] 


#45
Sep2108, 04:49 PM

P: 37

I worked it through anyway... all terms except the last became zero.
I now got [tex]A_{n}D_{n}=\frac{32}{(2n1)^{3}\pi^{3}}*(1)^{n}[/tex] 


#46
Sep2108, 05:12 PM

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[tex] A_{n}D_{n}=\frac{32}{(2n1)^{4}\pi^{4}}*(1)^{n} [/tex] ? If so, then your right. What are w(x,t) and u(x,t) then? 


#47
Sep2108, 05:40 PM

P: 37

Hmm.. no, actually i got ^3... I will show you how I did and then you can see where it went wrong. I don't write out the terms that went zero.
[tex] A_{n}D_{n}=+\frac{4}{(2n+1)\pi}\int^{1}_{0}(\frac{3x^{2}}{6}\frac{1}{2})(cos(\frac{(2n1)\pi}{2}x)dx = [/tex] [tex]=\frac{16}{(2n+1)^{2}\pi{2}}\int^{1}_{0}(x*sin(\frac{(2n1)\pi}{2}x)dx = \frac{32}{(2n+1)^{3}\pi{3}}\int^{1}_{0}(cos(\frac{(2n1)\pi}{2}x)dx=[/tex] [tex]=\frac{32}{(2n+1)^{3}\pi{3}}[(sin(\frac{(2n1)\pi}{2}x)]=\frac{32}{(2n1)^{3}\pi^{3}}*(1)^{n}[/tex] edit: I think I lost one inner derivative there at the end...but it doesn't make sense to me anyway..in that case it would be 64/((2n+1)^4*pi^4) :/ Hope we can find the mistake. I go on with "your" AnDn meanwhile. Then it means that [tex] w(x,t)=\frac{32}{\pi^{4}} \sum_{n=1}^{\infty} \frac{(1)^{n}}{(2n+1)^{4}}sin(\frac{(2n1)\pi}{2}x)cos(\frac{(2n1)\pi}{2}t) [/tex] and [tex]u(x,t)=w(x,t)+S(x)= w(x,t)=\frac{32}{\pi^{4}} \sum_{n=1}^{\infty} \frac{(1)^{n}}{(2n+1)^{4}}sin(\frac{(2n1)\pi}{2}x)cos(\frac{(2n1)\pi}{2}t)\frac{x^{3}}{6}+\frac{x}{2}+1[/tex] 


#48
Sep2108, 06:00 PM

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[tex]u(x,y)= \frac{32}{\pi^{4}} \sum_{n=1}^{\infty} \frac{(1)^{n}}{(2n1)^{4}}sin(\frac{(2n1)\pi}{2}x)cos(\frac{(2n1)\pi}{2}t)\frac{x^{3}}{6}+\frac{x}{2}+1[/tex] or if you make the substitution n=m+1 you get: [tex]u(x,y)= \frac{32}{\pi^{4}} \sum_{m=0}^{\infty} \frac{(1)^{m}}{(2m+1)^{4}}sin(\frac{(2m+1)\pi}{2}x)cos(\frac{(2m+1)\pi}{2}t)\frac{x^{3}}{6}+\frac{x}{2}+1[/tex] Now, what is u(1,0) according to the boundary conditions? What is it according to the above formula? 


#49
Sep2108, 06:37 PM

P: 37

Ah, I see.. The missed inner derivative I found myself, but I squared once instead of multiplying by 2.. Time to go to bed soon
[tex] u(1,0)= \frac{32}{\pi^{4}} \sum_{n=1}^{\infty} \frac{(1)^{n}}{(2n1)^{4}}*(1)^{n}*1\frac{1}{6}+\frac{1}{2}+1 [/tex] but I don't have u(1,0) in my BC? EDIT: and (1)^n*(1)^n = 1 ..... 


#51
Sep2108, 07:07 PM

P: 37

Aha.. but then
[tex] u(1,0)= \frac{32}{\pi^{4}} \sum_{m=0}^{\infty} \frac{(1)^{m}}{(2m+1)^{4}}*(1)^{m}*1\frac{1}{6}+\frac{1}{2}+1=1 \Rightarrow \sum_{m=0}^{\infty} \frac{1}{(2m+1)^{4}}=(1\frac{4}{3})*(\frac{\pi^{2}}{32})=\frac{\pi^{4}}{96} [/tex] 


#52
Sep2108, 07:11 PM

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Yep, and
[tex]\sum_{n=1}^{\infty} \frac{1}{(2n1)^{4}} =\frac{\pi^{4}}{96}[/tex] can be used to answer your question by substituting n=m+1 : [tex]\Rightarrow \sum_{n=1}^{\infty} \frac{1}{(2n1)^{4}} = \sum_{m+1=1}^{\infty} \frac{1}{(2(m+1)1)^{4}} = \sum_{m=0}^{\infty} \frac{1}{(2m+1)^{4}}=\frac{\pi^{4}}{96}[/tex] 


#53
Sep2108, 07:14 PM

P: 37

Thank you SO much for your patience. Tomorrow I am going through this problem from the start again to make sure I get each step.



#54
Sep2108, 07:25 PM

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