Can this ODE system be solved?x′(t)=sin⁡(x(t)(y(t)2+1);y′(t)=−cos⁡(x(t))cos⁡(y(t))2&nbsp;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&nbsp;x(0)=x0&nbsp;and&nbsp;y(0)=y0.

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Can this ODE system be solved?x′(t)=sin⁡(x(t)(y(t)2+1);y′(t)=−cos⁡(x(t))cos⁡(y(t))2&amp;nbsp;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&amp;nbsp;x(0)=x0&amp;nbsp;and&amp;nbsp;y(0)=y0.

Saul Murray · Accepted Answer

Step 1I 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 2This is a separable ODE.cos⁡(x)sin⁡(x)&amp;nbsp;dx&amp;nbsp;=−y+2cos⁡(y)&amp;nbsp;dy&amp;nbsp;ln⁡(sin⁡(x))=−∫y+2cos⁡(y)&amp;nbsp;dy&amp;nbsp;+constantThis is the equation of the trajectory on implicite form.x(y)=arcsin⁡(C:exp⁡(−∫y+2cos⁡(y)&amp;nbsp;dy&amp;nbsp;))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&amp;nbsp;&amp;nbsp;dx&amp;nbsp;&amp;nbsp;dt&amp;nbsp;=sin⁡(x)(y(x)2+1)&amp;nbsp;would probably be very difficult to integrate analytically for t(x).

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