bloodhound Posted November 19, 2004 Share Posted November 19, 2004 [math]\frac{dy}{dx}=\cos(y+f(x))[/math] here y and f are functions of x . highly non linear. tried maple. didnt give jack Link to comment Share on other sites More sharing options...
bloodhound Posted November 19, 2004 Author Share Posted November 19, 2004 [math]\frac{dy}{dx}=\cos(y+f(x))[/math] here y and f are functions of x . highly non linear. tried maple. didnt give jack Link to comment Share on other sites More sharing options...
Dave Posted November 20, 2004 Share Posted November 20, 2004 I know it's a long shot, but the first thing that springs to mind is a Taylor expansion of cos? Might be worth a shot, although looking at it, it might not do any good. Link to comment Share on other sites More sharing options...
Dave Posted November 20, 2004 Share Posted November 20, 2004 I know it's a long shot, but the first thing that springs to mind is a Taylor expansion of cos? Might be worth a shot, although looking at it, it might not do any good. Link to comment Share on other sites More sharing options...
bloodhound Posted November 20, 2004 Author Share Posted November 20, 2004 good idea . will give it a go. then i assume the the solution will be a series solution. it is rigorous to change the order of integration and sum of a series. i.e is [math]\int (\sum a_n)=\sum(\int a_n)[/math]? Link to comment Share on other sites More sharing options...
bloodhound Posted November 20, 2004 Author Share Posted November 20, 2004 good idea . will give it a go. then i assume the the solution will be a series solution. it is rigorous to change the order of integration and sum of a series. i.e is [math]\int (\sum a_n)=\sum(\int a_n)[/math]? Link to comment Share on other sites More sharing options...
premjan Posted November 20, 2004 Share Posted November 20, 2004 you have two different unknowns: y and f, so you don't appear to have enough information to produce a solution with only one equation. Link to comment Share on other sites More sharing options...
premjan Posted November 20, 2004 Share Posted November 20, 2004 you have two different unknowns: y and f, so you don't appear to have enough information to produce a solution with only one equation. Link to comment Share on other sites More sharing options...
bloodhound Posted November 20, 2004 Author Share Posted November 20, 2004 f will be given.. so just want to find y in terms of that f along with composition of other fucntions Link to comment Share on other sites More sharing options...
bloodhound Posted November 20, 2004 Author Share Posted November 20, 2004 f will be given.. so just want to find y in terms of that f along with composition of other fucntions Link to comment Share on other sites More sharing options...
Dave Posted November 21, 2004 Share Posted November 21, 2004 Another idea that occurred might be to use Fourier transforms, but similarly to the other problem, I don't know whether that would help any (because f is just an arbitrary function). Link to comment Share on other sites More sharing options...
CPL.Luke Posted December 24, 2004 Share Posted December 24, 2004 if you want to have it in terms of y replace f(x) with u then you have dy/dx=cos(y+u) which is equal to dy/dx=cos(y)cos(u)-sin(y)sin(u) thats as far as I can go hope it helps Link to comment Share on other sites More sharing options...
Dave Posted December 25, 2004 Share Posted December 25, 2004 Maybe you can find something that'll differentiate to give that angle addition? Link to comment Share on other sites More sharing options...
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