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System of two differential equations with trigonometric functions 
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#1
Feb2312, 09:45 AM

P: 625

Hello,
do you have any strategy to suggest in order to solve the following system of partial differential equations in x(s,t) and y(s,t)? [tex]\frac{\partial x}{\partial t} = x  \frac{1}{2}\sin(2x)[/tex] [tex]\frac{\partial y}{\partial t} = y \; \sin^2(x)[/tex] (note that the partial differentiation is always with respect to t). In case it might be useful, I can safely assume that the codomain of x(s,t) and y(s,t) is [1,1]. I already tried with Maple and Mathematica but they only give me numerical solutions. An approximation would be ok for me, as long as I get a closed form for x and y. I was also wondering if you think there might exist another system of coordinates in which this system is easier to solve Thanks. 


#2
Feb2412, 02:32 AM

P: 285

You might try dividing one equation by the other (left hand side by left hand side and right hand side by right hand side) and get a single ode for dy/dx. The result is of the form y'=f(x)*y.
I am surprised that mathematica is not able to solve this. 


#3
Feb2412, 03:34 AM

P: 625

...it is also possible that I don't know how to use properly those packages, as I am a total beginner. For Mathematica I used the following syntax:
 DSolve[{D[x[t], t] == x[t]  1/2*Sin[2*x[t]], D[y[t], t] == y[t]*Sin[x[t]]^2}, {x[t], y[t]}, t] while for Maple:  dsolve({diff(x(t), t) = x(t)(1/2)*sin(2*x(t)), diff(y(t), t) = y(t)*sin(x(t))^2}, {x(t), y(t)}) With both packages I obtain a numerical solution involving integrals, which I guess indicates a failure in finding a closed form solution. *********** *** EDIT: *** *********** I tried to divide the first equation by the second, as you suggested and obtained: [tex]y(x)=\pm \sqrt{\frac{2x}{\tan(x)}+c}[/tex] but now we only have y(x), and I guess we still need to solve the first differential equation in order to obtain x(t), am I right? Am I also free to set c=0 ? 


#4
Feb2412, 02:36 PM

P: 57

System of two differential equations with trigonometric functions
x(s, t) = xt  t/2 sin(2x) + f(s) y(s,t) = y t sin^2 (2x) + f(s) where f(s) is a function of s. 


#5
Feb2512, 11:52 AM

P: 625

Hi hawaiifiver...I am afraid something went wrong with your solution, because you treated x(s,t) as a constant, and did not take into account that x is a function of t.
For instance the solution of the simper differential equation [tex]\frac{\partial x}{\partial t} = x(s,t)[/tex] is not tx(t,s). It is instead [itex]x(s,t)=e^t c(s)[/itex] 


#6
Feb2512, 03:59 PM

P: 57




#7
Feb2612, 11:19 AM

P: 199

The first equation, in x, is separable. Once you have x, including an arbitrary function of s, substitute that into the equation for y which become separable. I don't see how to do the first integration of 1/{xsin(2x)/2}.



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