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6

In short: There is no closed solution, but you can get solutions for specific values of ϵ. Detailed explanation: First, we can transform your equation into a more compact form by using FullSimplify (and replacing ϵ with e for less typing within Mathematica): eq = (x ϵ + (-ϵ) x^-ϵ - c (-ϵ) x^(-ϵ - 1) /. ϵ -> e //FullSimplify) == 0 (* e x^(-1 - e) (c - x ...


4

In this case, I recommend Manipulate to investigate the function. Note as x, c and ϵ are used. With the information found You can start better studies Manipulate[ Plot[x ϵ + c x^(-1 - ϵ) ϵ - x^-ϵ ϵ , {x, -6., 6.}] , {c, -6., 6.} , {ϵ, -6, 6}]


3

According to your statement, I think what you need is just 4th-order Runge-Kutta method, and a completely self-made implementation of 4th-order Runge-Kutta method isn't necessary, then the answer from J.M. has shown you the optimal direction: (* Unchanged part omitted. *) ClassicalRungeKuttaCoefficients[4, prec_] :=With[{amat = {{1/2}, {0, 1/2}, {0, 0, ...


2

Your equation is easier to handle if it put into an equivalent form. eq1 = Sqrt[l1] + Sqrt[t1] + Sqrt[w1] == d; It is also to convenient to write the rules as r1 = {l1 -> (pw^2 w1)/((1 + a e1)^2 w^2)}; r2 = {t1 -> (pw^2 w1)/((1 + a e1)^2 r^2)}; and substitute u^2 for w1 getting a new equation eq2 = eq1 /. Join[r1, r2] /. w1 -> u^2 ...


2

This expression can be simplified substantially as follows. Beginning with ((a - 1) (1 + d x^2))/(a (Exp[b x + c] - 1)) - e multiply the expression by its denominator and Simplify Numerator[Together[%]] (* -1 + a + a*e - a*e*E^(c + b*x) - d*x^2 + a*d*x^2 *) Collect[b^2/(d (a - 1)) %, x^2, Simplify] (* (b^2*(-1 + a*(1 + e - e*E^(c + b*x))))/((-1 + a)*d) ...


2

gradient[g_, vars_] := Table[D[g@@vars, vars[[j]]], {j, 1, Length[vars]}] system1[lstConst_, vars_] := Join[ Join@@ Table[gradient[lstConst[[j]], vars], {j, 1, Length[lstConst]}], Table[lstConst[[j]]@@vars,{j,1,Length[lstConst]}]]; system2[f_, lstConst_, vars_, lambda_] := Join[ gradient[f, vars] - Sum[ lambda[[j]]*gradient[lstConst[[j]], vars], ...


2

There's a misunderstanding here. The third "dom" argument is not simply a set over which we solve the equation. There are only a few choices that can be used for the domain argument, and they have very specific effects on how Solve works. An example from the documentation: If dom is Reals, or a subset such as Integers or Rationals, then all constants ...


1

For given numeric values of ϵ and c you can use FindRoot: eq = x ϵ + (-ϵ) x^-ϵ - c (-ϵ) x^(-ϵ - 1) == 0; FindRoot[eq /. {c -> 2, ϵ -> 3/4}, {x, 1}] (* {x -> 1.38718} *) For arbitrary analitycal values of ϵ and c the equation seems to be unsolvable. However, if you can threat ϵ as a small parameter (like in perturbation theory), I would suggest to ...


1

First (for a given Epsilon): eq = x ϵ + (-ϵ) x^-ϵ - c (-ϵ) x^(-ϵ - 1) == 0 then: FullSimplify @ eq Thus it suffices to solve: c - x + x^(2 + ϵ) == 0 Then sol = Solve[(c - x + x^(2 + ϵ)) == 0, x] delivers the solutions. If you want the solutions for e.g.,c = 5, then c = 5; x /. sol will show them. When Epsilon is is not given, Solve is not ...


1

{Solve [ x^2 == 1, {x}], Solve [ x^2 == 1 && x > 0, {x}]} (* {{{x -> -1}, {x -> 1}}, {{x -> 1}}}*)


1

So I think I found a way following this post: Find Roots in Do loop I added these commands: t = List[0, 0.5, 1, 1.5, 2] Table[FindRoot[{e1, e2, e3, e4} /. dat, {{w, 0.5}, {r, 2}, {p2, 0.8}, {pw, 1.8}}], {a, t}] And it does calculate FindRoot over the different values of a.



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