From: nma@12000.org   
      
   I am solving each Kamke ODE by hand, and I am up to problem #22   
   (it is slow process) which I have question on.   
      
   Maple gives the answer to Kamke #22. The problem is   
      
   eq:=diff(y(x),x)-y(x)^2-y(x)*sin(2*x)-cos(2*x) = 0;   
      
   Maple answer is in terms of Heun Confluent function   
   and its derivative. (see (2) below). This ODE is   
   Riccati first order non-linear.   
      
   But the solution done by hand, and it matches the one given   
   in Kamke, only contains tan(x) and one unresolved integral   
      
        u:=cos(x)   
        I:=int( exp(-u^2)/u^2, x)   
      
   This integral has no closed form. (or does it?) So   
   the general solution to the ode is in terms of I. Here it is   
      
       y:= tan(x) + exp(-u^2)/(u^2 * (I + C))    (1)   
      
   The above is also what Kamke shows in the book.   
      
   C is integration constant. Maple answer on the other hand,   
   is very complicated using HeunC and its derivative. Here it is   
      
   restart;   
   eq:=diff(y(x),x)-y(x)^2-y(x)*sin(2*x)-cos(2*x) = 0;   
   sol:=dsolve(eq,y(x));   
      
   sol := y(x) = (2*HeunCPrime(1, 1/2, -1/2, -1, 7/8, (1/2)*   
   cos(2*x)+1/2)*_C1*cos(2*x)/((2*cos(2*x)+2)^(1/2)*   
   (_C1*HeunC(1, 1/2, -1/2, -1, 7/8, (1/2)*cos(2*x)+1/2)*   
   (2*cos(2*x)+2)^(1/2)+HeunC(1, -1/2, -1/2, -1, 7/8, (1/2)   
   *cos(2*x)+1/2)))+(HeunCPrime(1, -1/2, -1/2, -1, 7/8, (1/2)   
   *cos(2*x)+1/2)*(2*cos(2*x)+2)^(1/2)+2*   
   HeunCPrime(1, 1/2, -1/2, -1, 7/8, (1/2)*cos(2*x)+1/2)*_C1+   
   2*HeunC(1, 1/2, -1/2, -1, 7/8, (1/2)*   
   cos(2*x)+1/2)*_C1)/((2*cos(2*x)+2)^(1/2)*   
   (_C1*HeunC(1, 1/2, -1/2, -1, 7/8, (1/2)*cos(2*x)+1/2)*   
   (2*cos(2*x)+2)^(1/2)+HeunC(1, -1/2, -1/2, -1, 7/8, (1/2)*cos(2*x   
   +1/2))))*sin(2*x)   
      
                                              (2)   
      
   So it must be then that (1) == (2) ?   
      
   Does this mean there is analytical solution to "I" in terms   
   of special function HeunC? Since I do not see the integral "I" in   
   Maple solution.   
      
   If the solution to the ODE is unique (up to arbitrary constant),   
   then one must be able to show that (1)<=>(2)?   
      
   Is this correct?   
      
   Can someone shed some light on this? And does this mean   
   integral "I" can be expressed in closed form using these HeunC functions?   
      
   So how did Maple come up with answer it did if this   
   integral has no closed form solution?   
      
   The answer by Maple must be correct, since Maple odetest()   
   returns 0.   
      
   Here is a link to the problem with detailed solution and Maple answer   
      
   http://12000.org/my_notes/kamek/version_3/KEsu22.htm#subsection_22   
      
   ps. Mathematica unable to solve this one for some reason.   
      
   Can your CAS solve it? What solution does it give?   
   The book solution or something else?   
      
   --Nasser   
      
   --- SoupGate-Win32 v1.05   
    * Origin: you cannot sedate... all the things you hate (1:229/2)