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I found some problems with your code. In `Solve[D[eqn2, k] == 0, k]` the equation is quite difficult to solve exactly. I would suggest `NSolve` instead of `Solve`. When you write `filtered_k2`, it is interpreted as a pattern. I would suggest a...
The numerical solution with `NDSolve` misses a singular point with infinite derivative. Here is a derivation of the solution: diffEq = y'[x] == (7 x + 4 y[x])/(7 x + 4 y[x] + 4); solY = DSolveValue[diffEq, y, x] eq = y == solY[x]...
Sorry, I see your text only now. Here is a solution: {reMp, imMp} = Simplify[ComplexExpand[ReIm[Mp]]] absMp = Sqrt[reMp^2 + imMp^2] `ReIm` separates the two parts, so that `Simplify` does cannot remix them again.
I am no expert in numerical methods, but when `y == 0` the differential equation give no information on what direction to go as `y` increases. Something like `DSolve[{y f'[y] == 3, f[0] == 1}, f, y]`. By the way, I suspect that `ClearAll;`...
I had copied and pasted from your second line. Now it's better: f[t_] = InverseLaplaceTransform[(12 E^(24 s) (1 - E^(168 s) + 168 E^(180 s) s))/ (-1 + E^(12 s) + E^(168 s) - E^(180 s) + 2016 ...
Your syntax `{t, 0, 12}` is typical of `NDSolve`, not of `DSolve`. Unfortunately, `NDSolve` does not support `CaputoD` at the moment, apparently. `DSolve` does support it, but in a very limited way: try these two independent equations, that `DSolve`...
I am confused. I tried the new `SurfaceIntegrate`, and found that these inputs SurfaceIntegrate[Curl[vf, {x, y}], Element[{x, y}, reg]] Integrate[Curl[vf, {x, y}], Element[{x, y}, reg]] give different outputs. So what does...
I had another look at your code. You make the following replacement: Cos[\[Theta]] Dt[\[Theta]] /. {\[Theta] -> \[Pi]/3} Perhaps you expected `theta` to be replaced inside `Cos[theta]` alone. Unfortunately, it got replaced into `Dt[theta]`...
On my Mac I get an `ImageDimensions` of `{400, 400}`. It does not depend on the window magnification. It is funny that ImageQ[Graphics[Circle[], ImageSize -> 200]] gives `False`, but still `ImageDimensions` does not complain.
Check your syntax. You have used square brackets `[]` where you probably should have written `()`. Also, to get the square of `(r - 0.0001)` do not type `(r - 0.0001) ²`, which is not interpreted as square.