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Hint. Try to solve first eqn2 = -dp/L + \[Mu]*D[v[x, y], x, x] - Subscript[\[Sigma], e]*Subscript[B, x0]^2*v[x, y] + Subscript[\[Sigma], e]*Subscript[E, z]* Subscript[B, x0] + \[Zeta] Co …

We can have an insight about the solutions behavior for each m by doing tmax = 1000; solution = ParametricNDSolve[{-((m (1 + m) + 4/(9 (-2/3 + t) t)) y[t]) + 2 (-1/3 + t) y'[t] + (-2/3 + t) t y''[t] = …

Hint. Assuming you have an approximation $(u_k,\psi_k)$ then you can proceed linearly as $$ \cases{ \mathcal{D}_1[u_{k+1}]+6(u_k \psi_k +u_k(\psi_{k+1}-\psi_k)+\psi_k(u_{k+1}-u_k))+(x^2+y^2)u_{k+1}=i …

An iterative approach based on the fixed point existence. $$ \epsilon_{k+1}(t)=\Phi\left(\epsilon_{k}(t),t\right) $$ f = c0; sols = {f} n = 9; For[k = 1, k <= n, k++, Clear[ϵ]; ϵ[t_] := f; f = (β*I …

As a product of visual inspection, taking data from $\approx 80$ to $120$ and using the model $$ f(a,b,\sigma_1,\sigma_2,x_1,x_2,x)=a e^{-\left(\frac{x-x_1}{\sigma_1}\right)^2}+b e^{-\left(\frac{x-x_2 …

Solve it as DSolve[{dcA[r] == cA'[r], 2/r dcA[r] + dcA'[r] == \[Phi]^2/R^2*cA[r], cA[R] == cAR, dcA[0] == 0}, {cA, dcA}, r]

Try this R1 = 1; R1 = 100; R5 = 45; L1 = 346*10^(-3); L2 = 7169*10^(-9); c = 360*10^(-5); Vi = LaplaceTransform[230*Sqrt[2]*Sin[100*Pi*t], t, s]; R2 = s*L1; R3 = 1/(s*c); R4 = s*L2; x = (R2 R3 R5 Vi)/ …

Hint. $$ \frac{(1+a\,x)^{1/4} - (1+b\,x)^{1/4}}{x}=(a-b)\frac{(1+a\,x)^{1/4} - (1+b\,x)^{1/4}}{(1+a\ x)-(1+b\ x)} $$ and now $$ \frac{u^4-v^4}{u-v} = (u^2+v^2)(u+v) $$

Try this: Gxyz = z - x^2 - y^2; p = {x, y, z}; p1 = {1, 0, 0}; p2 = {0, 1, 0}; n = Grad[Gxyz, p] equ1 = n.(p - p1) == 0 equ2 = n.(p - p2) == 0 equ3 = Gxyz == 0 sol = Solve[{equ1, equ2, equ3}, p] n0 = …

G = n[t] - n0 Exp[b[n[t]] t]; dG = D[G, t] /. {E^(t b[n[t]]) n0 -> n[t]}; solnt = n'[t] /. Solve[dG == 0, n'[t]][[1]] // Simplify $$ \frac{n(t) b(n(t))}{1-t n(t) b'(n(t))} $$

Take real numbers k = 9.; l = 12.; m = 2.; M = 4.; mat = {{m*w^2 - 2*k, k, k*zeta}, {k, M*w^2 - (l + k), l}, {-k*zeta, l, M*w^2 - (k - l)}}; mydet = ExpToTrig[Det[mat]]; sol = Quiet@Solve[mydet == 0 …

As long as this system is the Lorenz attractor, you have a changed sign in the first equation, so it blows up. Now it is fixed. s = Quiet @ NDSolve[{X'[t] == -10 (X[t] - Y[t]), Y'[t] == X[t] (28 - Z[ …

Once we have the equilibrium points as equs = {sigma (X - Y), X (rho - Z) - Y, X Y - beta Z} sols = Solve[equs == 0, {X, Y, Z}] you can calculate the associated jacobian to each equilibrium point a …