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Questions on the analytic and numerical equation solving functions of Mathematica (Solve, Reduce, NSolve, FindRoot, DSolve, RSolve, etc.).

0
votes
Solving first for $t$ solt = Solve[y == a (Sin[t/b - h] + 1) + k, t] /. {C[1] -> 0} and then after substitution x - t - a (Sin[t/b - h] + 1) Tan[theta] /. solt // FullSimplify we get at the imp …
answered Jan 6 by Cesareo
2
votes
Try Y = A k^a L^b; PROF = P Y - r k - w L; z1 = D[PROF, k]; z2 = D[PROF, L]; sol=Together[Simplify[Solve[{z1 == 0, z2 == 0}, {k, L}], {0 <= a <= 1 && 0 <= b <= 1 && a + b + c == 1 && A > 0 && k > 0 & …
answered Aug 12 by Cesareo
3
votes
Think that $$ (d_1-i d_2)b(y) = \lambda a(y)\\ (d_1+i d_2)a(y) = \lambda b(y) $$ can be handled as $$ (d_1+i d_2)(d_1-i d_2)b(y) = \lambda (d_1+i d_2) a(y) = \lambda^2 b(y) $$ then $$ (d_1^2+i(d_ …
answered Jan 21 by Cesareo
2
votes
$\left(a^2-b^2\right) y(x)-2 b^2 x \left(x y''(x)+2 y'(x)\right)=0$ is an Euler-type DE so substituting $y = x^r$ into the DE we get $x^r \left(a^2-b^2 (2 r (r+1)+1)\right)=0$ and solving for $r$ we g …
answered Jan 2 by Cesareo
2
votes
With V = m/Sqrt[(x - m)^2 + (y - m)^2 + (z - m)^2] + m/Sqrt[(x + m)^2 + (y + m)^2 + (z + m)^2] + 1/2*(x^2 + y^2); The system $$ V_x = 0\\ V_y = 0\\ V_z = 0 $$ has notoriously three solution …
answered Jan 16 by Cesareo