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July 4

What no questions today? Is everyone taking a holiday? —Preceding unsigned comment added by 122.107.192.187 (talk) 13:02, 4 July 2010 (UTC)[reply]

I've done an IP lookup on the last 4 anons who posted questions here. Two are from Australia, one from UK and one from Germany. This suggests that a lot less people ascribe special significance to the 4th of July than you'd think. Also, the day is still young, of course. -- Meni Rosenfeld (talk) 14:23, 4 July 2010 (UTC)[reply]

Second Order Differential Equations

Hi. I'm currently working through a course on DEs and now have some SODEs to solve, two examples being:

$y''+y=H(t-\pi )-H(t-2\pi )$

subject to y(0)=y'(0)=0 and y(t), y'(t) continuous at $\pi ,2\pi$

and

$y''-4y=\delta (x-a)$

subject to y being bounded as $|x|\to \infty$ , with $\delta (x)$ being the Dirac delta function.

My problems are as follows. For each SODE I can find the complementary function and with the first one, I can find the particular integral in the cases $t<\pi$ , $\pi <t<2\pi$ and $t>2\pi$ but I don't know how to ensure that the solution is continuous at pi and 2pi. Then, for the second one, I haven't got a clue what to try for the particular integral. Am I supposed to do it in the same fashion as I did for the Heaviside step function and consider what happens for x=a and x≠a separately? Thanks 92.11.130.6 (talk) 15:09, 4 July 2010 (UTC)[reply]

Partial Derivatives

A quick one on PDs. If we set $x=e^{-s}\sin t$ and $y=e^{-s}\cos t$ subject to u(x,y)=v(s,t), I have to find ${\frac {\partial v}{\partial s}}$ and ${\frac {\partial v}{\partial t}}$ in terms of x, y, ${\frac {\partial u}{\partial x}}$ and ${\frac {\partial u}{\partial y}}$ .

Via the chain rule I get that ${\frac {\partial v}{\partial s}}$ = $-x{\frac {\partial u}{\partial x}}-y{\frac {\partial u}{\partial y}}$ and ${\frac {\partial v}{\partial t}}$ = $y{\frac {\partial v}{\partial x}}-x{\frac {\partial v}{\partial y}}$ . Is this correct? I then have to find ${\frac {\partial ^{2}v}{\partial t^{2}}}$ . Do I do this by finding ${\frac {\partial }{\partial t}}({\frac {\partial v}{\partial t}})={\frac {\partial }{\partial t}}(y{\frac {\partial v}{\partial x}}-x{\frac {\partial v}{\partial y}})$ ? It's quite messy when I try it and so I doubt I'm going about this the right way. Thanks 92.11.130.6 (talk) 19:21, 4 July 2010 (UTC)[reply]

Hint: y+ix=e^−s+it. Bo Jacoby (talk) 07:04, 5 July 2010 (UTC).[reply]

Thanks, I hadn't actually spotted that and it's quite clever, but the final part of the question is about finding a partial differential equation for v, so I'm not allowed to know what the function is yet. 92.11.130.6 (talk) 10:04, 5 July 2010 (UTC)[reply]

Never mind, think I've got this one now. 92.11.130.6 (talk) 14:37, 5 July 2010 (UTC)[reply]