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7

Avoid subscripts and user-defined names starting with capitals. Format[a[n_]] := Subscript[a, n] Format[b[n_]] := Subscript[b, n] Format[c[n_]] := Subscript[c, n] equ1 = 0.0004666666666666666 (Cosh[0.24988836793370642 d] a[2] + c[2]) == c[3]; equ2 = 0.30000000000000004 == 1/4 (12.7` - d)^2 a[3] + 1/2 (12.7 - d) b[3] + c[3]; equ3 = ...


4

You have the irreducible polynomial of the 7th degree: -x + a^3 x - a^3 x^2 - a^4 x^2 - a^5 x^2 + 2 a^4 x^3 + 2 a^5 x^3 + 2 a^6 x^3 - a^4 x^4 - a^5 x^4 - 6 a^6 x^4 - a^7 x^4 + 6 a^6 x^5 + 4 a^7 x^5 - 2 a^6 x^6 - 6 a^7 x^6 + 4 a^7 x^7 - a^7 x^8 The general algebraic solution doesn't exist for polynomials of the 5th degree and higher with arbitrary ...


3

The reference provided in the question recommends m = 3, so that is what I consider. (Edit: m = 2 has no real solution, as can be seen from NSolve[sys, var].) The equations with minor simplification then become n = 2; m = 3; p1 = Tuples[{1, Subscript[x, i], Subscript[y, i]}, {n}]; p3 = DeleteDuplicates[Times @@@ p1]; p4 = Integrate[#, {Subscript[x, i], ...


2

Given eqn = {vo == r1 + v2/(r1 + 1/(I w + c1)), v1 == ((rs + 1/(I w + cs)) v2 + rf + vin)/(rs + rf + 1/(I w + cs)), (vin - v2)/(rs + rf + 1/(I w + cs)) == 2 gm v1 + v2/(r1 + 1/(I w + c1))}; then Solve[eqn, {vo, v1, v2}]// FullSimplify gives the following solution


2

NMinimize requires a function rather than a list of equations. Let's take your equations and try three things. Use eqn11 to solve for c3. Replace c3 with 21/5000 in the other equations. Replace the equality in the remaining equations with a subtraction. eqn1 = c2 + a2 Cosh[(20048005 d)/80227844] - 9; eqn2 = 1/400 (127 - 10 d) 2 a3 + (127/20 - d/2) b3 + ...


2

Now that we have your initial conditions, the problem turns out to be simple and not CPU intensive. eqn = Derivative[0, 0, 2, 0][u][x, y, z, t] + Derivative[0, 2, 0, 0][u][x, y, z, t] + Derivative[2, 0, 0, 0][u][x, y, z, t] == D[u[x, y, z, t], t]; inti = u[x, y, z, 0] == 314; bon1 = DirichletCondition[u[x, y, z, t] == 304, x == 0]; bon2 = ...


2

n = 0; a = 1; L = 0; b = 1/100; m = 1/2; g = (2 m a r)/(n + L + 1); f[r_] = AiryAi[(2 m b)^(1/3) r]; r0 = r /. FindMaximum[r^(L + 1) f[r], r, WorkingPrecision -> 20][[2]] // Rationalize[#, 0] &; r0 // N // InputForm (* 4.103398736759 *) r1inv = Series[1/r, {r, r0, 4}] // Normal // Simplify; r0 == r /. Solve[r1inv - 1/r == 0, r, ...


2

NDSolve as used in the question has too many boundary conditions in x and none in t. (One periodic boundary condition in x for each dependent variable fully meets the need for boundary conditions in that dimension.) Not knowing what boundary conditions are desired in t, I made some up. e = 0.1; k = -4.47675; eqns = {D[y1[t, x], {t, 1}] + k/Pi * D[y1[t, ...


2

TD = 200; Debye[x_] := 3 (x/TD)^3 Integrate[y^3/(Exp[y] - 1), {y, 0, TD/x}, Assumptions -> x > 0] Fa[x_] := 8.6173324*10^(-5) x (9*TD/(8 x) + 3 Log[1 - Exp[-TD/x]] - Debye[x]) ser = Series[Fa[x], {x, 0, 1}] // Normal (* 0.00025852 x Log[1 - E^(-200/x)]*) Plot[ser, {x, 1, 100}]


1

Try fulleq=Append[eqnsMain, eqnsAdd]; And then Eliminate[fulleq, c] Using the {} parenthesis to fuse them would require Flatten on the result, as Mathematica will combine the vectors, not their components.


1

eqns = {AA == 2 a + 4 b + 45 c + 5, BB == 4 a + 45 b + 31 c + 78, CC == 0.23 a + 0.4 b + 4.35 c + 0.12, DD == 0.73 a + 0.2 b + 43.455 c + 3.12}; Substituting for c eqns2 = eqns /. c -> 43.5 AA + 34 b + 32 (* {AA == 5 + 2 a + 4 b + 45 (32 + 43.5 AA + 34 b), BB == 78 + 4 a + 45 b + 31 (32 + 43.5 AA + 34 b), CC == 0.12 + 0.23 a + 0.4 b + ...


1

In response to your comment, assign an intermediate variable With[{a1 = 2, a2 = 5, b1 = -4, b2 = 2}, solxy = Solve[ {a1*x + b1*y == 4, a2*x + b2*y == 6}, {x, y}][[1]]; p = a1*x + b2*y /. solxy] (* 2 *) The values of x and y are solxy (* {x -> 4/3, y -> -(1/3)} *)



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