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 September 22nd, 2016, 02:27 PM #1 Member   Joined: Oct 2014 From: Norway Posts: 42 Thanks: 1 Separation of variables, ODE. Is it possible to get a factor t in a solution? Here's my problem: http://i.imgur.com/XT7Mjw1.png I tried a). But when I look at c), it says $\displaystyle u(x,t) = te^{-t}\sin(x)$ should be a solution and I cannot make that solution; I can only get $\displaystyle e^{-t}\sin(x)$ with the solution I found. So I just need to know: where would a 't' factor come from? By separation of variables, I solved for F first, then for G, and for G I got the ODE $\displaystyle G'' + 2G' + n^2 G = 0$ for n=1,2,3... So solving this I assume that $\displaystyle G = e^{rt}$ and solve the characteristic equation and all that, right. But is it possible that this yields a $\displaystyle t$ factor, or t*e^something at all? --- A little more about my solution: I separate the variables so that $\displaystyle (G'' + 2G')/G = c^2 F''/F = \text{some constant k}$. Found out that the constant k has to be negative: $\displaystyle k = - \lambda^2$. Found that $\displaystyle F_n(x) = B \sin(nx)$, for n = 1,2,3,... The ODE in G that I mentioned follows. And the solution to that is $\displaystyle G_n(t) = c_1 e^{(\sqrt{1-n^2}-1)*t} + c_2 e^{(-\sqrt{1-n^2}-1)t}$ According to my calculations. So the general solution I get is $\displaystyle u(x,t) = \sum_{n=1}^{\inf}[(\cos(\sqrt{n^2-1}t)(B+A) + i*\sin(\sqrt{n^2-1}t)(B-A))e^{-t}\sin(nx)]\\$ And of course if we want $\displaystyle u(x,t) = te^{-t}\sin(x)$, then we only want n=1. But choosing B and A, I can't get that solution. Last edited by skipjack; September 23rd, 2016 at 12:16 PM. September 22nd, 2016, 03:18 PM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,675 Thanks: 2655 Math Focus: Mainly analysis and algebra I need to revise the method for this type of equation, but: On separating variables we get $F''(x)=kF(x)$ and $G''(t)+2G'(t)-kG(t)=0$. For negative $k=-m^2$ we have $F(x)=a_1\sin mx + a_2\cos mx$ and for positive integer $m$ the boundary conditions are satisfied by $F(x)=a_1\sin mx$. Now for the particular case $m=1$ we then have $G''(t)+2G'(t)+G(t)=0$. The characteristic polynomial for this is $(r+1)^2$ so the general solution is $G(x)=(b_1+b_2t)e^{-t}$ which has your factor of $t$. Edit: I apologise for failing to read the second half of your post, but essentially I agree with most of what you have. You've just missed the particular case of the equation for $G(t)$. Thanks from topsquark and uint Last edited by skipjack; September 23rd, 2016 at 12:12 PM. September 22nd, 2016, 07:52 PM #3 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 For n=1, G(t) = ce$\displaystyle ^{-t}$ and you lose an independent solution. The characteristic equation, as V8archie points out, is (r+1)$\displaystyle ^{2}$=0, which gives only the one independent solution above for r=-1. To get a second independent solution, someone discovered that t$\displaystyle e^{-t}$ was also a solution (try it), so G(t) = c$\displaystyle _{1}$e$\displaystyle ^{-t}$ + c$\displaystyle _{2}$te$\displaystyle ^{-t}$ So the point is, what you should look out for in part a) , and correct for in the chapt, is repeated roots. Something similar applies for a more general case of repeated roots but I simply can't remember where I saw it. Thanks from uint September 23rd, 2016, 01:26 AM #4 Member   Joined: Oct 2014 From: Norway Posts: 42 Thanks: 1 Thank you very much both! Tags factor, ode, separation, solution, variables Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Njprince94 Differential Equations 7 December 9th, 2015 06:20 AM philm Differential Equations 5 May 6th, 2015 10:41 AM engininja Calculus 2 February 22nd, 2011 01:06 PM engininja Calculus 4 September 22nd, 2010 11:58 PM zaserov Calculus 1 October 25th, 2007 01:11 PM

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