Author 
Message 
Theo P New poster
Post Number: 1
 Posted on Saturday, 10 November, 2012  02:44 pm:  
Hello! I'm doing some coursework on differential equations and have come across a pair of simultaneous equations that I can't work out how to solve. They're of the form [display]a = b\sqrt{x} \tan{\left(c \sqrt{x} + yb\sqrt{x}\right)}[/display] and [display]0 = \tan{\left(d \sqrt{x} + yb\sqrt{x}\right)}[/display] where a, b, c, and d are known (positive) constants which I have the values of, and x and y are to be found. I'm not really sure how to attack this problem. I've tried getting WolframAlpha to solve them, but the computation timed out. I also tried plotting on Autograph to get a numerical solution, but the graph was some horrendous discontinuous mess! I beginning to wonder whether it is indeed possible to solve these, or whether I should go about the solution in a different way ... Thank you in advance for your help. 
Simon Taylor Veteran poster
Post Number: 707
 Posted on Saturday, 10 November, 2012  04:06 pm:  
I don't think that there are any exact solutions to this. But you can do something to the second equation to get rid of the tan, and that should allow you to sub into the first equation to get something you can solved numerically or graphically. 
Theo P New poster
Post Number: 2
 Posted on Saturday, 10 November, 2012  04:51 pm:  
Thanks for your reply. So taking arctan of both sides of the second equation, I get: [display]d\sqrt{x} + yb\sqrt{x} = n\pi[/display] for integer values of n. If I take n to be 0, then: [display]yb\sqrt{x} = d\sqrt{x}[/display] Dividing by [inline]\sqrt{x}[/inline] assuming it's not 0: [inline]yb = d[/inline] so, [inline]y = \frac{d}{b}[/inline] Then, if I substitute this into the first equation: [display] \frac{a}{b} = \sqrt{x}\tan{\left(\sqrt{x}\left(d  c\right)\right)}[/display] So, basically, I need to solve [display]\mu = xtan(\lambda x)[/display] Is there an analytical solution to this? 
Simon Taylor Veteran poster
Post Number: 708
 Posted on Sunday, 11 November, 2012  05:04 pm:  
There is no analytic solution to that equation  you can investigate solutions graphically or numerically. Equations with both trig and nontrig terms crop up quite frequently in theoretical physics, and in almost all cases don't have any analytic solutions. Once you have done that you may want to look at what happens for other integer values of n, but it isn't pretty. 
Theo P New poster
Post Number: 3
 Posted on Sunday, 11 November, 2012  07:46 pm:  
I've solved it graphically, which gave an almost perfect solution to the original question. It's a shame that there aren't any analytical solutions, since they're much nicer to work with. Never mind  I guess that's the price you pay for trying to apply Maths to the real world ... I may well have a look at other values of n once I've met the deadline for the coursework! Thank you so much for your help. 
azerbajdzan Veteran poster
Post Number: 1278
 Posted on Sunday, 11 November, 2012  09:10 pm:  
Tangent function is periodic function with period "pi". So tan(x)=tan(x+n*pi) for all "x" and integer "n". So any integer "n" will give the same result. There seems to be too many (redundant) variables. If b=0 then system of equations has solution only in case "a" is also a=0, in which case the solutions are given by [inline]x=\left(\frac{\pi k}{d}\right)^2[/inline], ("k" is any integer) totally independent on the value of "c". If "b" is not 0 you can substitute A=a/b and B=dc and the equation you have to solve would look like: [inline]A=\sqrt{x}\tan{\left(B\sqrt{x}\right)}[/inline] Which is much simpler then the equations you have started with but anyway it can not be solved analytically. But for some luckily chosen A, B (a, b, c, d) there still can exist a nice solution. 
