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6

There is unlikely to be a "nice" closed-form solution for this equation, since it involves both transcendental and algebraic expressions for $v$. However, it can be solved in terms of special functions with a bit of work. We can make the substitution $v = \alpha x + b$ to rewrite the equation (after some algebra) as $$ x e^x < - e^{-b/\alpha }. ...


1

You can do either of the following result1 = 16 (y - x); result2 = x (45 - z) - y; result3 = x*y - 4 z; F[{x_, y_, z_}] := Evaluate[{result1, result2, result3}]; x = Array[a, 3]; F[x] Or result1 = 16 (y - x); result2 = x (45 - z) - y; result3 = x*y - 4 z; F[{x_, y_, z_}] = {result1, result2, result3}; x = Array[a, 3]; F[x] Both give same result you showed. ...


3

We can include derivative in solution as follows sol = ParametricNDSolveValue[{e1'[t] == s e2[t], e2'[t] == -s e1[t], e1[0] == 1, e2[0] == 0}, {e1, e2, e1', e2'}, {t, 0, 1}, s] fun[t_, s_] := Through[sol[s][t]][[1]]; Plot[fun[t, 1], {t, 0, 10}] dfun[t_, s_] := Through[sol[s][t]][[3]]; Plot[dfun[t, 10], {t, 0, 1}]


4

Considering the lagrangian L which follows, Qk[p_] := 1 - (p - f - theta (alpha + beta))/theta/(1 - beta) Qr[p_] := (p - f - theta (alpha + beta))/ theta/(1 - beta) - (p + (1 - m)/m f - (alpha + beta))/(1 - beta) R[p_] := (1 - m) Qr[p] Q[p_] := Qr[p] + Qk[p] obj = Q[p] (p - k) + R[p] (f + r - p) L = obj + l1 (p - f mu/m/(1 - theta) - s1^2) + l2 (p ...


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