## 152 Reputation

16 years, 156 days

## Thank you so much for this...

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...

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 :=...

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, 0 .. 2*Pi], color = gold); ``` thanking you in advance

## Thank you. I...

Thank you. I tried: polarplot([soln, 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...

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...

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...

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...

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!! >...

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 ->...

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...

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...

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...

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
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