Thank you so much for this response. I just think that it's unreliable because some examples work on the classical application and don't on the non-classical one. When I changed the range a little bit, I still got a plot but this time it's a cusp!! Plus, I don't understand the numbers on the axes as they don't agree with the 'range' maple was able to plot. It's all just confusing!!

Thank you so much though I was expecting to get a sort of a closed curve like an ellipse. Does specifing the range has an effect on the general shape. The curve looks like half an ellipse, but I guess that's just a coincidence!! Is there another way to figure out whether the differential equation is of an ellipse or not other than polarplotting it? I have started to get convinced that maple is unreliable and it makes things difficult rather than easying them.

Here is the code: > dEq := diff(r(theta), theta, theta) = r(theta)^4*(r(theta)^2-r(theta))+1/r(theta); > initC := r(0) = 2*10^9, (D(r))(0) = 0; > soln := dsolve({dEq, initC}, numeric); > R := u -> eval(r(theta), soln (u)) > with(plots); > polarplot([R, 0 .. 2*Pi], color = gold); > with(plots); > polarplot([soln[1], 0 .. 2*Pi], color = gold); thanking you in advance

Thank you. I tried: polarplot([soln[1], 0 .. 2*Pi],color = gold) and I got the error : empty plot I tried your previous code: R := u -> eval(r(theta), soln (u)) polarplot(R,0 .. 2*Pi],color = gold) and got the same error: Empty plot

Thanks

But that would be a static solution. How would I know where the sol stops when I change some parameters in the equation or even change some of the inital conditions?

That's right can you think of a solution to this problem? I changed the requirement to fsolve('f'(u)=0.0001); which is close to zero to avoid having a singularity but it still doesn't give an answer

I've just tried it in the classical worksheet and although it didn't error out, it gave no solution to the eq. dEq := diff(r(t), t, t) = -1/r(t)^2; 2 d 1 dEq := --- r(t) = - ----- 2 2 dt r(t) > i := r(0) = .87, (D(r))(0) = 0; > soln := dsolve({dEq,i}, numeric); > f:=u->eval(r(t),soln(u)): > fsolve('f'(u)=0); > i := r(0) = 0.87, D(r)(0) = 0 soln := proc(x_rkf45) ... end proc fsolve(f(u) = 0, u)

This is the new equation and the relevant steps to find the sol: > dEq := diff(r(t), t, t) = -1/r(t)^2; > i := r(0) = .87, (D(r))(0) = 0; > soln := dsolve({i, dEq}, numeric); > f := proc (u) options operator, arrow; eval(r(t), soln(u)) end proc; > fsolve(('f')(u) = 0); print(fsolve(f(u) = 0, u)); Error, (in f) invalid input: eval received soln(u), which is not valid for its 2nd argument, eqns

It gives the same error!! > fsolve(f); print(fsolve(f(r), r)); Error, (in f) invalid input: eval received soln(r), which is not valid for its 2nd argument, eqns That's weird!

I spelt it correctly > u -> eval(r(t), soln(u)) > f(.1); print(0.864387265289590264e-2); 0.00864387265289590264 > soln(.1); print([t = .1, r(t) = 0.864387265289590264e-2, diff(r(t), t) = -0.327837117738825510e-3]); [ d ] [t = 0.1, r(t) = 0.00864387265289590264, --- r(t) = -0.000327837117738825510] [ dt ] > fsolve(('f')(u) = 0); print(fsolve(f(u) = 0, u)); Error, (in f) invalid input: eval received soln(u), which is not valid for its 2nd argument, eqns

I used the same two statements to find a sol when the function equals to zero but changed the equation this time and the statements ceased to function. I wonder why!!! f:=u->eval(r(t),soln(u)): fsolve('f'(u)=0); Error, (in f) invalid input: eval received soln(u), which is not valid for its 2nd argument, eqns All the other steps of finding and plotting the solution worked out find excpet for this step!! Thanks again in advance

Wow!! Thanks!!

Thanks so much that was very helpful the eq had many solutions so I set an interval from which fsolve finds a sol. and it worked just as I wanted thanks again

That makes sense. Is there a way that after finding the solution theta(t)=blah blah to find t, by setting theta = some value, ie to find the time at which theta will equal to that value? Thank you all