Can this ODE system be solved? \(\displaystyle{x}'{\left({t}\right)}={\sin{{\left({x}{\left({t}\right)}{\left({\frac{{{y}{\left({t}\right)}}}{{{2}}}}+{1}\right)};{y}'{\left({t}\right)}={\frac{{-{\cos{{\left({x}{\left({t}\right)}\right)}}}{\cos{{\left({y}{\left({t}\right)}\right)}}}}}{{{2}}}}\right.}}}\) Is there

WesDiectstemiwxg

WesDiectstemiwxg

Answered question

2022-03-19

Can this ODE system be solved?
x(t)=sin(x(t)(y(t)2+1);y(t)=cos(x(t))cos(y(t))2 
Is there a method to solve the following ODE system?

{ x ( t ) = sin ( x ( t ) ) ( y ( t ) 2 + 1 ) y ( t ) = cos ( x ( t ) ) cos ( y ( t ) ) 2 }

with initial conditions x(0)=x0 and y(0)=y0.

Answer & Explanation

Saul Murray

Saul Murray

Beginner2022-03-20Added 4 answers

Step 1
I am afraid that analytical solving involves too complicated special functions and that numerical methods are recomended in practice.
Nevertheless one can start the analytical solving on this way:

{ d x d t = sin ( x ( t ) ) ( y ( t ) 2 + 1 ) d y d t = cos ( x ( t ) ) cos ( y ( t ) ) 2 d x d y = sin ( x ) ( y 2 + 1 ) 2 cos ( x ) cos ( y )

Step 2
This is a separable ODE.
cos(x)sin(x) dx =y+2cos(y) dy 
ln(sin(x))=y+2cos(y) dy +constant
This is the equation of the trajectory on implicite form.
x(y)=arcsin(C:exp(y+2cos(y) dy ))
The closed form of the integral involves the Polylogarithm function. Then, it is doubtfull that one could invert the function x(y) in order to find y(x). So we are stuck here.
Even if this was possible, the full solving would be still far away. Putting y(x) into  dx  dt =sin(x)(y(x)2+1) would probably be very difficult to integrate analytically for t(x).

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