Path: news.mathworks.com!not-for-mail From: <HIDDEN> Newsgroups: comp.soft-sys.matlab Subject: Re: Solving second order differential equations Date: Thu, 25 Nov 2010 18:19:05 +0000 (UTC) Organization: The MathWorks, Inc. Lines: 31 Message-ID: <icm9ap$5nr$1@fred.mathworks.com> References: <icm404$2pn$1@fred.mathworks.com> Reply-To: <HIDDEN> NNTP-Posting-Host: webapp-02-blr.mathworks.com Content-Type: text/plain; charset=UTF-8; format=flowed Content-Transfer-Encoding: 8bit X-Trace: fred.mathworks.com 1290709145 5883 172.30.248.37 (25 Nov 2010 18:19:05 GMT) X-Complaints-To: news@mathworks.com NNTP-Posting-Date: Thu, 25 Nov 2010 18:19:05 +0000 (UTC) X-Newsreader: MATLAB Central Newsreader 1187260 Xref: news.mathworks.com comp.soft-sys.matlab:690029 "Sridatta Pasumarthy" <sridatta1988@gmail.com> wrote in message <icm404$2pn$1@fred.mathworks.com>... > Hi, > How do i solve four differential equations of second order involving many variables (numerically)? > I have referred van der pol (non-stiff) example in Matlab, but it didn't help much because i couldn't > figure a way to convert these to first order right away. Please help me find a solution. > > Problem Structure > ---------------------------------- > w' = [k*f(z)] * y' > w * x'' = c*g(z) * y' > w * y'' = f(z) + g(z) * x' + j(z) * z' > w * z'' = c*j(z) * y' > 1-(1/sqrt(w))=x'^2 + y'^2 + z'^2 > > x(0)=0, y(0)=0, z(0)=0, dx/dt(0)=0, dy/dt(0), dz/dt(0)=v > c,k,v are constants. > It also doesn't help that f,g,j are complicated exponential functions based upon z. > > I am supposed to plot 'dw/dz' vs 'z' > ---------------------------------- > > Thanks, > Sridatta - - - - - - - - Hello again Sridatta. You said "four differential equations" but I count five! With only the four variables x, y, z, and w, that seems to be one too many equations to satisfy. Think of it this way. If w were being held fixed, the three middle equations involving x'', y'', and z'' would suffice for solving for x, y, and z. Allowing w to vary according to the first equation would again constitute a solvable problem of four equations and four unknowns. I see no reason why that fifth equation should then hold true. How do you explain this? I am assuming that the functions f, g, and j are all already determined and not unknown functions. Roger Stafford