My Math Forum Separation of variables, ODE. Is it possible to get a factor t in a solution?

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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

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