Recent content by wahoo2000

Hmm, I don't really understand. I think that you rewrite the RHS in a smart way so that both cases 0<x<=pi and x>pi are "built in" in one expression, right? Was my approach wrong (I guess so), but why? What is U(x)? I have only solved more simple and straightforward IVPs with Laplace...

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.

Aha.. but then 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}

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 :rolleyes: u(1,0)= -\frac{32}{\pi^{4}} \sum_{n=1}^{\infty} \frac{(-1)^{n}}{(2n-1)^{4}}*-(-1)^{n}*1-\frac{1}{6}+\frac{1}{2}+1 but I don't have...

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. A_{n}D_{n}=+\frac{4}{(2n+1)\pi}\int^{1}_{0}(\frac{3x^{2}}{6}-\frac{1}{2})(cos(\frac{(2n-1)\pi}{2}x)dx =...

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

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

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

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...

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

Hmm, does that mean we are looking for Fourier sinus series of f(x)=x^3/6-x/2? Is it A_{n}D_{n}=2\int^{1}_{0}(\frac{x^{3}}{6}-\frac{x}{2})(\frac{(2n-1)\pi}{2}x)dx

Homework Statement Solve the IVP for t>0 y'(t)+\int^{t}_{0}y(\tau)d\tau = =sin(t) for 0<t<=pi =0 for t>pi y(0)=-1 The Attempt at a Solution The solution depends on how large t is and therefore the solution consist of two parts depending on the size of t.. Let's call them y1(t)...

Well... that means that \Rightarrow w(x,0)= \sum_{n=1}^{\infty} A_n D_n sin \left( \frac{(2n-1) \pi}{2} x \right) cos \left( 0\right) \Rightarrow \sum_{n=1}^{\infty} A_n D_n sin \left( \frac{(2n-1) \pi}{2} x \right)=\frac{x^{3}}{6}-\frac{x}{2}

that k=(n*Pi)/2... so \lambda=-\frac{n^{2}\pi^{2}}{4}

Anyway.. if I keep the A (ii) \Rightarrow w_{x}(1,t)=Akcos(kx)(Csin(kt)+Dcos(kt))=0 then cos(kx) should be zero because k and (Csin(kt)+Dcos(kt) are not zero, (and neither is A..), because if A=0 we have only trivial solution X(x)T(t)=0, and same if (Csin(kt)+Dcos(kt)=0, then per definition k...