🧠:<think>Alright, so I've got this differential equation to solve. It's about job satisfaction among TSA officers, which is interesting. The equation is:[ frac{dS}{dt} = kS(1 - S) - m(1 - e^{-nt}) ]And the initial condition is ( S(0) = S_0 ). I need to solve this differential equation and then figure out what happens to ( S(t) ) as time goes to infinity. Let me break this down step by step.First, let's understand the equation. The term ( kS(1 - S) ) looks like a logistic growth model, which typically models population growth where there's a carrying capacity. In this case, it might represent internal factors affecting job satisfaction, where satisfaction grows until it reaches a certain point. Then, the term ( -m(1 - e^{-nt}) ) seems to represent external stressors. The ( e^{-nt} ) part suggests that the effect of these stressors diminishes over time, but initially, they have a full impact.So, the equation is a combination of a logistic growth term and a decaying external stress term. It's a non-linear differential equation because of the ( S(1 - S) ) term. Non-linear equations can be tricky, but maybe I can find an integrating factor or use substitution to simplify it.Let me write the equation again:[ frac{dS}{dt} = kS(1 - S) - m(1 - e^{-nt}) ]This is a first-order ordinary differential equation (ODE). It's non-linear because of the ( S^2 ) term. Hmm, solving non-linear ODEs can be challenging. Maybe I can rewrite it in a way that allows me to separate variables or use an integrating factor.Alternatively, perhaps I can use substitution. Let me see.Let me consider the homogeneous part first, ignoring the external stress term:[ frac{dS}{dt} = kS(1 - S) ]This is the logistic equation, and its solution is well-known. The solution is:[ S(t) = frac{1}{1 + left(frac{1 - S_0}{S_0}right) e^{-kt}} ]But in our case, we have an additional term ( -m(1 - e^{-nt}) ). So, the equation is non-homogeneous. I might need to use the method of integrating factors or variation of parameters.Wait, the equation is:[ frac{dS}{dt} + kS^2 - kS = -m(1 - e^{-nt}) ]Hmm, that's a Riccati equation because of the ( S^2 ) term. Riccati equations are generally difficult to solve unless we have a particular solution. Maybe I can find a particular solution for the non-homogeneous part.Alternatively, perhaps I can rewrite the equation in a linear form. Let me see.Wait, if I rearrange the equation:[ frac{dS}{dt} + (-k)S + kS^2 = -m(1 - e^{-nt}) ]This is a Bernoulli equation because of the ( S^2 ) term. Bernoulli equations can be linearized by substituting ( y = S^{1 - 2} = frac{1}{S} ). Let me try that.Let ( y = frac{1}{S} ). Then, ( frac{dy}{dt} = -frac{1}{S^2} frac{dS}{dt} ).Substituting into the equation:[ -frac{1}{S^2} frac{dS}{dt} + (-k)frac{1}{S} + k = -m(1 - e^{-nt}) ]Multiply both sides by ( -S^2 ):[ frac{dS}{dt} + kS + (-k) = m S^2 (1 - e^{-nt}) ]Wait, that doesn't seem to help much. Maybe I made a mistake in substitution.Wait, let's do it step by step.Given:[ frac{dS}{dt} = kS(1 - S) - m(1 - e^{-nt}) ]Let me rewrite it as:[ frac{dS}{dt} + kS^2 - kS = -m(1 - e^{-nt}) ]This is a Riccati equation of the form:[ frac{dS}{dt} = a(t) + b(t)S + c(t)S^2 ]Where ( a(t) = -m(1 - e^{-nt}) ), ( b(t) = -k ), and ( c(t) = k ).Riccati equations are tough because they don't have a general solution method unless we know a particular solution. Maybe I can guess a particular solution.Alternatively, perhaps I can look for an integrating factor. Let me consider the equation as:[ frac{dS}{dt} + (-k)S + kS^2 = -m(1 - e^{-nt}) ]This is a Bernoulli equation with ( n = 2 ). The standard substitution for Bernoulli is ( y = S^{1 - n} = S^{-1} ). So, let me set ( y = 1/S ).Then, ( frac{dy}{dt} = -frac{1}{S^2} frac{dS}{dt} ).Substitute into the equation:[ -frac{1}{S^2} frac{dS}{dt} + (-k)frac{1}{S} + k = -m(1 - e^{-nt}) ]Multiply both sides by ( -S^2 ):[ frac{dS}{dt} + kS + (-k) = m S^2 (1 - e^{-nt}) ]Wait, that doesn't seem to help. Maybe I need to rearrange differently.Wait, let's substitute ( y = 1/S ). Then:[ frac{dy}{dt} = -frac{1}{S^2} frac{dS}{dt} ]So,[ frac{dS}{dt} = -S^2 frac{dy}{dt} ]Substitute into the original equation:[ -S^2 frac{dy}{dt} = kS(1 - S) - m(1 - e^{-nt}) ]Divide both sides by ( -S^2 ):[ frac{dy}{dt} = -frac{k}{S}(1 - S) + frac{m}{S^2}(1 - e^{-nt}) ]But since ( y = 1/S ), ( frac{1}{S} = y ) and ( frac{1}{S^2} = y^2 ). So,[ frac{dy}{dt} = -k y (1 - frac{1}{y}) + m y^2 (1 - e^{-nt}) ]Simplify:[ frac{dy}{dt} = -k y + k + m y^2 (1 - e^{-nt}) ]Hmm, that seems more complicated. Maybe this substitution isn't helpful.Alternatively, perhaps I can use an integrating factor for the linear part. Let me try to write the equation in linear form.Wait, the equation is:[ frac{dS}{dt} + (-k)S + kS^2 = -m(1 - e^{-nt}) ]It's non-linear because of the ( S^2 ) term. Maybe I can consider this as a linear equation in ( S ) if I treat ( S^2 ) as a source term. But that might not be straightforward.Alternatively, perhaps I can use the method of variation of parameters. Let me recall that for a Riccati equation, if we have one particular solution, we can find the general solution.But I don't have a particular solution here. Maybe I can assume a particular solution of a certain form. Let me think.Suppose the particular solution ( S_p(t) ) is a constant. Let's see if that's possible.If ( S_p ) is constant, then ( frac{dS_p}{dt} = 0 ). So,[ 0 = k S_p (1 - S_p) - m(1 - e^{-nt}) ]But the right-hand side depends on ( t ), so unless ( m(1 - e^{-nt}) ) is zero, which it isn't except at ( t = 0 ), a constant particular solution isn't possible.Alternatively, maybe the particular solution has an exponential form. Let me assume ( S_p(t) = A e^{-nt} + B ). Let's try that.Compute ( frac{dS_p}{dt} = -n A e^{-nt} ).Substitute into the equation:[ -n A e^{-nt} = k (A e^{-nt} + B)(1 - A e^{-nt} - B) - m(1 - e^{-nt}) ]This seems messy, but let's expand the right-hand side.First, compute ( (A e^{-nt} + B)(1 - A e^{-nt} - B) ):Let me denote ( x = A e^{-nt} ), then it becomes ( (x + B)(1 - x - B) = (x + B)(1 - B - x) = (x + B)(1 - B - x) ).Multiply out:= ( x(1 - B - x) + B(1 - B - x) )= ( x - Bx - x^2 + B - B^2 - Bx )= ( x - 2Bx - x^2 + B - B^2 )= ( (1 - 2B)x - x^2 + B(1 - B) )Substitute back ( x = A e^{-nt} ):= ( (1 - 2B) A e^{-nt} - A^2 e^{-2nt} + B(1 - B) )Now, multiply by ( k ):= ( k(1 - 2B) A e^{-nt} - k A^2 e^{-2nt} + k B(1 - B) )So, the equation becomes:[ -n A e^{-nt} = k(1 - 2B) A e^{-nt} - k A^2 e^{-2nt} + k B(1 - B) - m + m e^{-nt} ]Now, let's collect like terms.Left-hand side: ( -n A e^{-nt} )Right-hand side:- Terms with ( e^{-2nt} ): ( -k A^2 e^{-2nt} )- Terms with ( e^{-nt} ): ( k(1 - 2B) A e^{-nt} + m e^{-nt} )- Constant terms: ( k B(1 - B) - m )So, equating coefficients for each term:1. For ( e^{-2nt} ):Left-hand side: 0Right-hand side: ( -k A^2 )So, ( -k A^2 = 0 ) => ( A = 0 )But if ( A = 0 ), then ( S_p(t) = B ), which is a constant. But earlier, we saw that a constant particular solution isn't possible unless ( m(1 - e^{-nt}) ) is zero, which it isn't. So, this approach might not work.Alternatively, maybe I need to consider a particular solution that includes terms up to ( e^{-nt} ). Let me try ( S_p(t) = A + B e^{-nt} ).Compute ( frac{dS_p}{dt} = -n B e^{-nt} )Substitute into the equation:[ -n B e^{-nt} = k (A + B e^{-nt})(1 - A - B e^{-nt}) - m(1 - e^{-nt}) ]Again, let's expand the right-hand side.First, compute ( (A + B e^{-nt})(1 - A - B e^{-nt}) ):Let me denote ( x = B e^{-nt} ), then it becomes ( (A + x)(1 - A - x) ).Multiply out:= ( A(1 - A - x) + x(1 - A - x) )= ( A - A^2 - A x + x - A x - x^2 )= ( A - A^2 - 2A x + x - x^2 )Substitute back ( x = B e^{-nt} ):= ( A - A^2 - 2A B e^{-nt} + B e^{-nt} - B^2 e^{-2nt} )Multiply by ( k ):= ( k A - k A^2 - 2k A B e^{-nt} + k B e^{-nt} - k B^2 e^{-2nt} )So, the equation becomes:[ -n B e^{-nt} = k A - k A^2 - 2k A B e^{-nt} + k B e^{-nt} - k B^2 e^{-2nt} - m + m e^{-nt} ]Now, collect like terms.Left-hand side: ( -n B e^{-nt} )Right-hand side:- Constant terms: ( k A - k A^2 - m )- Terms with ( e^{-nt} ): ( (-2k A B + k B + m) e^{-nt} )- Terms with ( e^{-2nt} ): ( -k B^2 e^{-2nt} )So, equate coefficients:1. For ( e^{-2nt} ):Left-hand side: 0Right-hand side: ( -k B^2 )Thus, ( -k B^2 = 0 ) => ( B = 0 )But if ( B = 0 ), then ( S_p(t) = A ), a constant. Again, we end up with the same issue as before. So, this approach doesn't seem to work either.Hmm, maybe I need a different approach. Perhaps I can use the method of integrating factors for the linear part and handle the non-linear term as a perturbation. Let me try that.Rewrite the equation:[ frac{dS}{dt} + (-k)S = k S^2 - m(1 - e^{-nt}) ]This is a Bernoulli equation. The standard form is:[ frac{dS}{dt} + P(t) S = Q(t) S^n ]In our case, ( P(t) = -k ), ( Q(t) = k ), and ( n = 2 ). The substitution for Bernoulli is ( y = S^{1 - n} = S^{-1} ). So, let me set ( y = 1/S ).Then, ( frac{dy}{dt} = -frac{1}{S^2} frac{dS}{dt} )Substitute into the equation:[ -frac{1}{S^2} frac{dS}{dt} + (-k) frac{1}{S} = k ]Multiply both sides by ( -S^2 ):[ frac{dS}{dt} + k S = -k S^2 ]Wait, that's not helpful because we're back to the original equation. Maybe I need to rearrange differently.Wait, let's substitute ( y = 1/S ) into the Bernoulli equation:[ frac{dS}{dt} + (-k) S = k S^2 - m(1 - e^{-nt}) ]Divide both sides by ( S^2 ):[ frac{1}{S^2} frac{dS}{dt} - frac{k}{S} = k - frac{m}{S^2}(1 - e^{-nt}) ]But ( frac{1}{S^2} frac{dS}{dt} = -frac{dy}{dt} ), so:[ -frac{dy}{dt} - k y = k - m(1 - e^{-nt}) y^2 ]Hmm, that still has a ( y^2 ) term, which complicates things. Maybe this substitution isn't helping.Alternatively, perhaps I can use the integrating factor method on the linear part and then deal with the non-linear term separately. Let me try that.The equation is:[ frac{dS}{dt} + (-k) S = k S^2 - m(1 - e^{-nt}) ]Let me write it as:[ frac{dS}{dt} + P(t) S = Q(t) ]Where ( P(t) = -k ) and ( Q(t) = k S^2 - m(1 - e^{-nt}) ). Wait, but ( Q(t) ) still depends on ( S ), so it's not linear. That's the issue.Alternatively, maybe I can consider the equation as:[ frac{dS}{dt} = k S (1 - S) - m(1 - e^{-nt}) ]And try to solve it numerically or look for an equilibrium solution.Wait, for the long-term behavior, maybe I don't need the exact solution. Let me think about part 2 first, which asks for the long-term behavior as ( t to infty ).As ( t to infty ), ( e^{-nt} to 0 ), so the external stress term becomes ( -m(1 - 0) = -m ). So, the equation approaches:[ frac{dS}{dt} = k S (1 - S) - m ]This is a logistic equation with a constant term. The equilibrium points are found by setting ( frac{dS}{dt} = 0 ):[ k S (1 - S) - m = 0 ][ k S - k S^2 - m = 0 ][ k S^2 - k S + m = 0 ]Wait, that's a quadratic equation in ( S ):[ k S^2 - k S + m = 0 ]The discriminant is ( D = k^2 - 4 k m ).So, if ( D > 0 ), two real roots:[ S = frac{k pm sqrt{k^2 - 4 k m}}{2k} = frac{1 pm sqrt{1 - 4 m / k}}{2} ]If ( D = 0 ), one real root:[ S = frac{1}{2} ]If ( D < 0 ), no real roots, so the solution doesn't stabilize to a real number, which might mean it oscillates or tends to infinity, but since ( S ) is job satisfaction, it's bounded between 0 and 1, so maybe it approaches a limit cycle or something else.But wait, in the long-term, the external stress term becomes constant, so the equation is autonomous. The behavior depends on the roots of the quadratic.So, for the job satisfaction to stabilize, we need real roots, so ( D geq 0 ), i.e., ( k^2 - 4 k m geq 0 ) => ( k geq 4 m ).Wait, ( k^2 - 4 k m geq 0 ) => ( k(k - 4m) geq 0 ). Since ( k > 0 ), this implies ( k - 4m geq 0 ) => ( k geq 4m ).So, if ( k geq 4m ), there are real equilibrium points. Let's find them:[ S = frac{1 pm sqrt{1 - 4m/k}}{2} ]Since ( S ) must be between 0 and 1, let's check the roots.The roots are:[ S_1 = frac{1 + sqrt{1 - 4m/k}}{2} ][ S_2 = frac{1 - sqrt{1 - 4m/k}}{2} ]Since ( sqrt{1 - 4m/k} leq 1 ), both roots are between 0 and 1.Now, to determine stability, we can look at the derivative of ( frac{dS}{dt} ) with respect to ( S ) at the equilibrium points.The derivative is:[ frac{d}{dS} left( k S(1 - S) - m right) = k(1 - 2S) ]At ( S_1 ):[ k(1 - 2 S_1) = k left(1 - 2 cdot frac{1 + sqrt{1 - 4m/k}}{2}right) = k left(1 - (1 + sqrt{1 - 4m/k})right) = -k sqrt{1 - 4m/k} ]Since ( k > 0 ) and ( sqrt{1 - 4m/k} ) is real and positive (because ( k geq 4m )), this derivative is negative, so ( S_1 ) is a stable equilibrium.At ( S_2 ):[ k(1 - 2 S_2) = k left(1 - 2 cdot frac{1 - sqrt{1 - 4m/k}}{2}right) = k left(1 - (1 - sqrt{1 - 4m/k})right) = k sqrt{1 - 4m/k} ]This is positive, so ( S_2 ) is an unstable equilibrium.Therefore, if ( k geq 4m ), the job satisfaction will stabilize at ( S_1 = frac{1 + sqrt{1 - 4m/k}}{2} ).If ( k < 4m ), then ( D < 0 ), so no real equilibrium points. In this case, the solution might not stabilize to a fixed point but could approach a limit cycle or oscillate. However, since the original equation is a logistic growth with a decaying external stress, it's possible that as ( t to infty ), the solution approaches a fixed point even if ( k < 4m ), but I'm not sure. Maybe I need to analyze it more carefully.Wait, when ( t to infty ), the external stress term becomes constant, so the equation is autonomous. If there are no real equilibrium points, the solution might diverge, but since ( S ) is job satisfaction, it's bounded between 0 and 1. So, perhaps it approaches a fixed point regardless, but I'm not certain. Maybe I need to consider the behavior of the solution.Alternatively, perhaps the solution will approach the equilibrium point if it exists, and if not, it might oscillate around the carrying capacity. But I'm not sure. Maybe I should look back at the original equation.Wait, another approach: as ( t to infty ), the external stress term becomes ( -m ), so the equation is:[ frac{dS}{dt} = k S(1 - S) - m ]This is a logistic equation with a constant harvesting term. The behavior of this equation is well-known. If the harvesting term ( m ) is less than the maximum growth rate divided by 4, then there are two equilibrium points, and the solution can stabilize at the higher one. If ( m ) is too large, the population (or job satisfaction) can collapse to zero.In our case, job satisfaction ( S ) is analogous to population. So, if ( m ) is too large, ( S ) might decrease to zero. But let's see.The maximum of ( k S(1 - S) ) occurs at ( S = 1/2 ), and the maximum value is ( k cdot 1/2 cdot 1/2 = k/4 ). So, if ( m > k/4 ), then the harvesting term is too large, and the equation ( k S(1 - S) - m = 0 ) has no real roots, meaning the solution will tend to negative infinity, but since ( S ) can't be negative, it might approach zero. However, in reality, job satisfaction can't be negative, so perhaps it stabilizes at zero.But wait, in the logistic equation with harvesting, if ( m > k/4 ), the population can go extinct. So, in our case, if ( m > k/4 ), job satisfaction might decrease to zero. If ( m leq k/4 ), it stabilizes at ( S_1 ).So, putting it together, the long-term behavior is:- If ( m leq k/4 ), ( S(t) ) approaches ( S_1 = frac{1 + sqrt{1 - 4m/k}}{2} ).- If ( m > k/4 ), ( S(t) ) approaches zero.Therefore, the job satisfaction stabilizes at ( S_1 ) if ( m leq k/4 ), otherwise, it decreases to zero.But wait, in the original equation, the external stress term is ( -m(1 - e^{-nt}) ), which starts at ( -m ) when ( t = 0 ) and approaches ( 0 ) as ( t to infty ). So, initially, the stress is high, but it decreases over time. So, the long-term behavior is influenced by the limit as ( t to infty ), where the stress term is zero. Wait, no, earlier I thought it approaches ( -m ), but let me check.Wait, ( 1 - e^{-nt} ) approaches 1 as ( t to infty ), so ( -m(1 - e^{-nt}) ) approaches ( -m ). Wait, no, ( e^{-nt} ) approaches 0, so ( 1 - e^{-nt} ) approaches 1, so ( -m(1 - e^{-nt}) ) approaches ( -m ). So, the external stress term approaches ( -m ), not zero. So, my earlier analysis was correct.So, as ( t to infty ), the equation approaches ( frac{dS}{dt} = k S(1 - S) - m ), which has equilibrium points as discussed.Therefore, the long-term behavior depends on whether ( m leq k/4 ) or not.So, to summarize part 2:- If ( m leq k/4 ), job satisfaction stabilizes at ( S = frac{1 + sqrt{1 - 4m/k}}{2} ).- If ( m > k/4 ), job satisfaction decreases to zero.Now, going back to part 1, solving the differential equation.Given the complexity of the equation, maybe I can use an integrating factor or look for an exact solution.Wait, another approach: since the equation is ( frac{dS}{dt} = k S(1 - S) - m(1 - e^{-nt}) ), perhaps I can write it as:[ frac{dS}{dt} + k S^2 - k S = -m(1 - e^{-nt}) ]This is a Riccati equation. Riccati equations can sometimes be solved if we know a particular solution. Let me see if I can find a particular solution.Suppose ( S_p(t) ) is a particular solution. Let me assume it's of the form ( S_p(t) = A + B e^{-nt} ). Let's try that.Compute ( frac{dS_p}{dt} = -n B e^{-nt} ).Substitute into the equation:[ -n B e^{-nt} + k (A + B e^{-nt})^2 - k (A + B e^{-nt}) = -m(1 - e^{-nt}) ]Expand ( (A + B e^{-nt})^2 ):= ( A^2 + 2 A B e^{-nt} + B^2 e^{-2nt} )So, substitute back:[ -n B e^{-nt} + k (A^2 + 2 A B e^{-nt} + B^2 e^{-2nt}) - k (A + B e^{-nt}) = -m + m e^{-nt} ]Now, expand:= ( -n B e^{-nt} + k A^2 + 2k A B e^{-nt} + k B^2 e^{-2nt} - k A - k B e^{-nt} = -m + m e^{-nt} )Now, collect like terms:- Constant terms: ( k A^2 - k A )- Terms with ( e^{-nt} ): ( (-n B + 2k A B - k B) e^{-nt} )- Terms with ( e^{-2nt} ): ( k B^2 e^{-2nt} )- Right-hand side: ( -m + m e^{-nt} )So, equate coefficients:1. For ( e^{-2nt} ):Left-hand side: ( k B^2 )Right-hand side: 0Thus, ( k B^2 = 0 ) => ( B = 0 )But if ( B = 0 ), then ( S_p(t) = A ), a constant. Let's see if that works.If ( B = 0 ), then the equation becomes:[ 0 + k A^2 - k A = -m + 0 ]So,[ k A^2 - k A + m = 0 ]Which is the same quadratic as before. So, ( A = frac{1 pm sqrt{1 - 4m/k}}{2} )But this is only valid if ( k geq 4m ). Otherwise, no real solution.So, if ( k geq 4m ), we have particular solutions ( S_p(t) = A ), where ( A ) is the stable equilibrium point.But if ( k < 4m ), no real particular solution, so this approach fails.Hmm, this is getting complicated. Maybe I need to use another method.Wait, perhaps I can write the equation in terms of ( u = S - S_p ), where ( S_p ) is the particular solution. But since ( S_p ) is constant, this might help linearize the equation.Let me set ( S = S_p + u ), where ( S_p ) is a constant particular solution.Then, ( frac{dS}{dt} = frac{du}{dt} )Substitute into the equation:[ frac{du}{dt} = k (S_p + u)(1 - S_p - u) - m(1 - e^{-nt}) ]Expand:= ( k (S_p (1 - S_p) - S_p u + u - S_p u - u^2) - m(1 - e^{-nt}) )But ( S_p ) is a particular solution, so ( k S_p (1 - S_p) - m = 0 ) (from setting ( frac{dS_p}{dt} = 0 )).So, the equation simplifies to:[ frac{du}{dt} = -k S_p u - k u + k u^2 ]Wait, let me compute step by step.First, expand ( (S_p + u)(1 - S_p - u) ):= ( S_p (1 - S_p) - S_p u + u (1 - S_p) - u^2 )= ( S_p (1 - S_p) - S_p u + u - S_p u - u^2 )= ( S_p (1 - S_p) - 2 S_p u + u - u^2 )So, substitute back:[ frac{du}{dt} = k [S_p (1 - S_p) - 2 S_p u + u - u^2] - m(1 - e^{-nt}) ]But since ( k S_p (1 - S_p) - m = 0 ), this simplifies to:[ frac{du}{dt} = -2 k S_p u + k u - k u^2 ]= ( (-2 k S_p + k) u - k u^2 )= ( k (1 - 2 S_p) u - k u^2 )So, the equation becomes:[ frac{du}{dt} = k (1 - 2 S_p) u - k u^2 ]This is a Bernoulli equation in ( u ). Let me write it as:[ frac{du}{dt} + (-k (1 - 2 S_p)) u = -k u^2 ]This is a Bernoulli equation with ( n = 2 ). The substitution is ( v = u^{1 - 2} = 1/u ).Then, ( frac{dv}{dt} = -frac{1}{u^2} frac{du}{dt} )Substitute into the equation:[ -frac{1}{u^2} frac{du}{dt} + (-k (1 - 2 S_p)) frac{1}{u} = -k ]Multiply both sides by ( -u^2 ):[ frac{du}{dt} + k (1 - 2 S_p) u = k u^2 ]Wait, that's the same as before. Maybe I need to proceed differently.Wait, let's use the substitution ( v = 1/u ). Then, ( frac{dv}{dt} = -frac{1}{u^2} frac{du}{dt} )From the equation:[ frac{du}{dt} = k (1 - 2 S_p) u - k u^2 ]Divide both sides by ( u^2 ):[ frac{1}{u^2} frac{du}{dt} = k (1 - 2 S_p) frac{1}{u} - k ]But ( frac{1}{u^2} frac{du}{dt} = -frac{dv}{dt} ), so:[ -frac{dv}{dt} = k (1 - 2 S_p) v - k ]Multiply both sides by -1:[ frac{dv}{dt} = -k (1 - 2 S_p) v + k ]This is a linear ODE in ( v ). The integrating factor is:[ mu(t) = e^{int -k (1 - 2 S_p) dt} = e^{-k (1 - 2 S_p) t} ]Multiply both sides by ( mu(t) ):[ e^{-k (1 - 2 S_p) t} frac{dv}{dt} + (-k (1 - 2 S_p)) e^{-k (1 - 2 S_p) t} v = k e^{-k (1 - 2 S_p) t} ]The left-hand side is the derivative of ( v mu(t) ):[ frac{d}{dt} [v e^{-k (1 - 2 S_p) t}] = k e^{-k (1 - 2 S_p) t} ]Integrate both sides:[ v e^{-k (1 - 2 S_p) t} = int k e^{-k (1 - 2 S_p) t} dt + C ]Compute the integral:Let ( a = -k (1 - 2 S_p) ), then:[ int k e^{a t} dt = frac{k}{a} e^{a t} + C ]So,[ v e^{-k (1 - 2 S_p) t} = frac{k}{-k (1 - 2 S_p)} e^{-k (1 - 2 S_p) t} + C ]Simplify:[ v e^{-k (1 - 2 S_p) t} = frac{-1}{1 - 2 S_p} e^{-k (1 - 2 S_p) t} + C ]Divide both sides by ( e^{-k (1 - 2 S_p) t} ):[ v = frac{-1}{1 - 2 S_p} + C e^{k (1 - 2 S_p) t} ]Recall that ( v = 1/u ), so:[ frac{1}{u} = frac{-1}{1 - 2 S_p} + C e^{k (1 - 2 S_p) t} ]Thus,[ u = frac{1}{frac{-1}{1 - 2 S_p} + C e^{k (1 - 2 S_p) t}} ]But ( u = S - S_p ), so:[ S - S_p = frac{1}{frac{-1}{1 - 2 S_p} + C e^{k (1 - 2 S_p) t}} ]Simplify the denominator:Let me write it as:[ S - S_p = frac{1}{frac{-1}{1 - 2 S_p} + C e^{k (1 - 2 S_p) t}} ]Let me factor out ( frac{-1}{1 - 2 S_p} ):= ( frac{1}{frac{-1}{1 - 2 S_p} left(1 - C (1 - 2 S_p) e^{k (1 - 2 S_p) t}right)} )= ( frac{1 - 2 S_p}{-1 + C (1 - 2 S_p) e^{k (1 - 2 S_p) t}} )= ( frac{2 S_p - 1}{1 - C (1 - 2 S_p) e^{k (1 - 2 S_p) t}} )So,[ S = S_p + frac{2 S_p - 1}{1 - C (1 - 2 S_p) e^{k (1 - 2 S_p) t}} ]Now, apply the initial condition ( S(0) = S_0 ):At ( t = 0 ),[ S_0 = S_p + frac{2 S_p - 1}{1 - C (1 - 2 S_p)} ]Let me solve for ( C ):Multiply both sides by the denominator:[ S_0 [1 - C (1 - 2 S_p)] = S_p [1 - C (1 - 2 S_p)] + (2 S_p - 1) ]Expand:[ S_0 - S_0 C (1 - 2 S_p) = S_p - S_p C (1 - 2 S_p) + 2 S_p - 1 ]Simplify the right-hand side:= ( S_p + 2 S_p - 1 - S_p C (1 - 2 S_p) )= ( 3 S_p - 1 - S_p C (1 - 2 S_p) )So,[ S_0 - S_0 C (1 - 2 S_p) = 3 S_p - 1 - S_p C (1 - 2 S_p) ]Bring all terms to one side:[ S_0 - 3 S_p + 1 = S_0 C (1 - 2 S_p) - S_p C (1 - 2 S_p) ]Factor out ( C (1 - 2 S_p) ):= ( C (1 - 2 S_p)(S_0 - S_p) )So,[ C = frac{S_0 - 3 S_p + 1}{(1 - 2 S_p)(S_0 - S_p)} ]But this is getting quite involved. Let me recall that ( S_p ) is a constant particular solution, which is ( S_p = frac{1 pm sqrt{1 - 4m/k}}{2} ). Let's denote ( S_p = frac{1 + sqrt{1 - 4m/k}}{2} ), the stable equilibrium.So, ( 1 - 2 S_p = 1 - 2 cdot frac{1 + sqrt{1 - 4m/k}}{2} = 1 - (1 + sqrt{1 - 4m/k}) = -sqrt{1 - 4m/k} )Thus, ( 1 - 2 S_p = -sqrt{1 - 4m/k} )Similarly, ( 2 S_p - 1 = 2 cdot frac{1 + sqrt{1 - 4m/k}}{2} - 1 = (1 + sqrt{1 - 4m/k}) - 1 = sqrt{1 - 4m/k} )So, substituting back into the expression for ( S ):[ S = S_p + frac{sqrt{1 - 4m/k}}{1 - C (-sqrt{1 - 4m/k}) e^{k (-sqrt{1 - 4m/k}) t}} ]Simplify the exponent:= ( e^{-k sqrt{1 - 4m/k} t} )Let me denote ( alpha = sqrt{1 - 4m/k} ), so ( alpha = sqrt{1 - 4m/k} )Then,[ S = S_p + frac{alpha}{1 + C alpha e^{-k alpha t}} ]Now, apply the initial condition ( S(0) = S_0 ):At ( t = 0 ),[ S_0 = S_p + frac{alpha}{1 + C alpha} ]Solve for ( C ):Multiply both sides by ( 1 + C alpha ):[ S_0 (1 + C alpha) = S_p (1 + C alpha) + alpha ]Expand:[ S_0 + S_0 C alpha = S_p + S_p C alpha + alpha ]Bring terms with ( C ) to one side:[ S_0 C alpha - S_p C alpha = S_p + alpha - S_0 ]Factor out ( C alpha ):[ C alpha (S_0 - S_p) = S_p + alpha - S_0 ]Thus,[ C = frac{S_p + alpha - S_0}{alpha (S_0 - S_p)} ]= ( frac{S_p + alpha - S_0}{alpha (S_0 - S_p)} )= ( frac{-(S_0 - S_p - alpha)}{alpha (S_0 - S_p)} )= ( -frac{S_0 - S_p - alpha}{alpha (S_0 - S_p)} )= ( -frac{1}{alpha} cdot frac{S_0 - S_p - alpha}{S_0 - S_p} )But this is getting too complicated. Maybe I can express ( C ) in terms of ( S_0 ) and ( S_p ).Alternatively, perhaps I can write the solution in terms of ( S_p ) and ( alpha ).Recall that:[ S = S_p + frac{alpha}{1 + C alpha e^{-k alpha t}} ]And from the initial condition:[ S_0 = S_p + frac{alpha}{1 + C alpha} ]Let me solve for ( C ):[ S_0 - S_p = frac{alpha}{1 + C alpha} ][ 1 + C alpha = frac{alpha}{S_0 - S_p} ][ C alpha = frac{alpha}{S_0 - S_p} - 1 ][ C = frac{1}{S_0 - S_p} - frac{1}{alpha} ]But ( alpha = sqrt{1 - 4m/k} ), and ( S_p = frac{1 + alpha}{2} )So,[ S_0 - S_p = S_0 - frac{1 + alpha}{2} ]Thus,[ C = frac{1}{S_0 - frac{1 + alpha}{2}} - frac{1}{alpha} ]This is quite involved, but perhaps we can leave it in terms of ( C ).So, the general solution is:[ S(t) = S_p + frac{alpha}{1 + C alpha e^{-k alpha t}} ]Where ( C ) is determined by the initial condition.But this is only valid when ( k geq 4m ), i.e., when ( alpha ) is real.If ( k < 4m ), then ( alpha ) is imaginary, and the solution would involve trigonometric functions, indicating oscillatory behavior. However, since ( S ) is job satisfaction, it's bounded between 0 and 1, so oscillations might not be physical unless damped.But given the complexity, perhaps the solution is best left in terms of the integrating factor and the particular solution, acknowledging that it's valid for ( k geq 4m ).Alternatively, maybe I can write the solution using the integrating factor without assuming a particular solution.Wait, another approach: since the equation is linear in ( S ) with a non-linear term, perhaps I can write it as:[ frac{dS}{dt} + k S = k S^2 - m(1 - e^{-nt}) ]This is a Bernoulli equation, so using the substitution ( y = S^{-1} ), we get:[ frac{dy}{dt} - k y = -k + m(1 - e^{-nt}) y^2 ]Wait, that's still non-linear. Maybe I need to use a different substitution.Alternatively, perhaps I can use the method of variation of parameters on the linear part.The homogeneous equation is:[ frac{dS}{dt} + k S = 0 ]Solution: ( S_h = C e^{-k t} )Now, find a particular solution ( S_p ) using variation of parameters.Let ( S_p = C(t) e^{-k t} )Then,[ frac{dS_p}{dt} = C'(t) e^{-k t} - k C(t) e^{-k t} ]Substitute into the original equation:[ C'(t) e^{-k t} - k C(t) e^{-k t} + k C(t) e^{-k t} = k S_p^2 - m(1 - e^{-nt}) ]Simplify:[ C'(t) e^{-k t} = k (C(t) e^{-k t})^2 - m(1 - e^{-nt}) ]So,[ C'(t) = k C(t)^2 e^{-k t} - m(1 - e^{-nt}) e^{k t} ]This is a Riccati equation for ( C(t) ), which is still difficult to solve.Given the time I've spent and the complexity, perhaps it's best to accept that the solution involves an integral that can't be expressed in elementary functions, or to leave it in terms of the integrating factor and particular solution.Alternatively, perhaps the solution can be expressed using the logistic function with a time-dependent term.But given the time constraints, I think I'll have to present the solution in terms of the integrating factor and the particular solution, acknowledging that it's valid for ( k geq 4m ).So, summarizing:1. The solution to the differential equation is:[ S(t) = S_p + frac{alpha}{1 + C alpha e^{-k alpha t}} ]Where ( S_p = frac{1 + sqrt{1 - 4m/k}}{2} ), ( alpha = sqrt{1 - 4m/k} ), and ( C ) is determined by the initial condition ( S(0) = S_0 ).2. The long-term behavior as ( t to infty ) is:- If ( k geq 4m ), ( S(t) ) approaches ( S_p = frac{1 + sqrt{1 - 4m/k}}{2} ).- If ( k < 4m ), ( S(t) ) approaches zero.But I need to express the solution more explicitly. Let me try to write ( C ) in terms of ( S_0 ).From the initial condition:[ S_0 = S_p + frac{alpha}{1 + C alpha} ]Solve for ( C ):[ S_0 - S_p = frac{alpha}{1 + C alpha} ][ 1 + C alpha = frac{alpha}{S_0 - S_p} ][ C = frac{1}{alpha (S_0 - S_p)} - frac{1}{alpha} ]= ( frac{1 - (S_0 - S_p)}{alpha (S_0 - S_p)} )= ( frac{S_p - S_0 + 1}{alpha (S_0 - S_p)} )But this is getting too messy. Maybe I can write the solution as:[ S(t) = frac{S_p + frac{alpha}{1 + C alpha e^{-k alpha t}}}{1} ]But I think it's better to present the solution in terms of the logistic function with a time-dependent perturbation, but I'm not sure.Alternatively, perhaps I can write the solution using the integrating factor method for the linear part and then express the non-linear term as an integral.Wait, let's try that.The equation is:[ frac{dS}{dt} + k S = k S^2 - m(1 - e^{-nt}) ]This is a Bernoulli equation. The standard form is:[ frac{dS}{dt} + P(t) S = Q(t) S^n ]Here, ( P(t) = k ), ( Q(t) = k ), ( n = 2 ).The substitution is ( y = S^{1 - 2} = S^{-1} ).Then, ( frac{dy}{dt} = -S^{-2} frac{dS}{dt} )Substitute into the equation:[ -S^{-2} frac{dS}{dt} + k S^{-1} = k ]Multiply both sides by ( -S^2 ):[ frac{dS}{dt} - k S = -k S^2 ]Wait, that's the same as before. Maybe I need to proceed differently.Alternatively, perhaps I can write the equation as:[ frac{dS}{dt} = k S(1 - S) - m(1 - e^{-nt}) ]And recognize that this is a Riccati equation with time-dependent coefficients. Riccati equations with time-dependent coefficients generally don't have closed-form solutions unless specific conditions are met.Given that, perhaps the best approach is to leave the solution in terms of an integral or acknowledge that it's complex and focus on the long-term behavior.But since the problem asks to solve the differential equation, I need to provide an explicit solution.Wait, perhaps I can use the method of integrating factors on the linear part and then express the solution as a series expansion, but that might be beyond the scope.Alternatively, perhaps I can use the substitution ( z = S - S_p ), where ( S_p ) is the particular solution, and linearize around it.But I think I've exhausted my methods here. Given the time I've spent, I'll have to conclude that the solution involves an integral that can't be expressed in elementary functions, but for the purposes of this problem, I can express it in terms of the logistic function with a time-dependent term.Alternatively, perhaps I can write the solution using the method of variation of parameters, but I'm not sure.Wait, another idea: since the equation is ( frac{dS}{dt} = k S(1 - S) - m(1 - e^{-nt}) ), perhaps I can write it as:[ frac{dS}{dt} = k S - k S^2 - m + m e^{-nt} ]This is a linear equation in ( S ) if we treat ( S^2 ) as a source term. But it's still non-linear.Alternatively, perhaps I can use the method of undetermined coefficients for the non-homogeneous part.But given the time, I think I'll have to accept that the solution is complex and present it in terms of the integrating factor and the particular solution, as I did earlier.So, to summarize:1. The solution to the differential equation is:[ S(t) = S_p + frac{alpha}{1 + C alpha e^{-k alpha t}} ]Where ( S_p = frac{1 + sqrt{1 - 4m/k}}{2} ), ( alpha = sqrt{1 - 4m/k} ), and ( C ) is determined by the initial condition ( S(0) = S_0 ).2. The long-term behavior as ( t to infty ) is:- If ( k geq 4m ), ( S(t) ) approaches ( S_p = frac{1 + sqrt{1 - 4m/k}}{2} ).- If ( k < 4m ), ( S(t) ) approaches zero.But I need to express ( C ) in terms of ( S_0 ).From the initial condition:[ S_0 = S_p + frac{alpha}{1 + C alpha} ]Solving for ( C ):[ S_0 - S_p = frac{alpha}{1 + C alpha} ][ 1 + C alpha = frac{alpha}{S_0 - S_p} ][ C = frac{1}{alpha (S_0 - S_p)} - frac{1}{alpha} ]= ( frac{1 - (S_0 - S_p)}{alpha (S_0 - S_p)} )= ( frac{S_p - S_0 + 1}{alpha (S_0 - S_p)} )But this is still complicated. Alternatively, perhaps I can write ( C ) as:[ C = frac{S_p - S_0 + 1}{alpha (S_0 - S_p)} ]= ( frac{-(S_0 - S_p - 1)}{alpha (S_0 - S_p)} )= ( -frac{S_0 - S_p - 1}{alpha (S_0 - S_p)} )But I think it's best to leave ( C ) in terms of ( S_0 ) and ( S_p ) without further simplification.Therefore, the solution is:[ S(t) = S_p + frac{alpha}{1 + left( frac{alpha}{S_0 - S_p} - 1 right) alpha e^{-k alpha t}} ]But this is quite involved. Alternatively, perhaps I can write it as:[ S(t) = frac{S_p + frac{alpha}{1 + C alpha e^{-k alpha t}}}{1} ]Where ( C ) is determined by the initial condition.In conclusion, the solution involves a logistic function with a time-dependent term, and the long-term behavior depends on the relationship between ( k ) and ( m ).</think>

🧠:<think>Okay, so I have this problem about opinion dynamics in a society modeled as a network. The researcher is using social network theory and differential equations. Let me try to break this down step by step.First, part 1a asks about the stability of the equilibrium state where dx_i/dt = 0 for all individuals i. The differential equation given is:dx_i/dt = sum_{j in N(i)} a_{ij}(x_j - x_i) - b_i(x_i)^3 + c_iSo, to find the equilibrium, we set dx_i/dt = 0 for all i. That gives us:sum_{j in N(i)} a_{ij}(x_j^* - x_i^*) - b_i(x_i^*)^3 + c_i = 0Now, to analyze the stability, I remember that for differential equations, we usually linearize around the equilibrium and check the eigenvalues of the Jacobian matrix. If all eigenvalues have negative real parts, the equilibrium is stable.So, let's consider a small perturbation around the equilibrium x_i^*. Let x_i(t) = x_i^* + ε_i(t), where ε_i is small. Then, substituting into the differential equation:dε_i/dt = sum_{j in N(i)} a_{ij}(ε_j - ε_i) - 3b_i(x_i^*)^2 ε_iBecause (x_i + ε_i)^3 ≈ x_i^3 + 3x_i^2 ε_i, so the derivative term becomes -3b_i(x_i^*)^2 ε_i.So, the linearized system is:dε/dt = [sum_{j in N(i)} a_{ij}(ε_j - ε_i)] - 3b_i(x_i^*)^2 ε_iThis can be written in matrix form as dε/dt = (A - 3B diag(x_i^*^2)) ε, where A is the adjacency matrix with weights a_{ij}, and B is a diagonal matrix with b_i on the diagonal.Wait, actually, the first term is the Laplacian matrix of the network. The Laplacian L is defined as L = D - A, where D is the degree matrix with D_{ii} = sum_{j} a_{ij}. So, the term sum_{j} a_{ij}(ε_j - ε_i) is equivalent to -L ε.Therefore, the linearized system is dε/dt = (-L - 3B diag(x_i^*^2)) ε.So, the Jacobian matrix J is -L - 3B diag(x_i^*^2). To determine stability, we need to check if all eigenvalues of J have negative real parts. Since L is a symmetric matrix (assuming a_{ij} = a_{ji}, which is often the case in undirected networks), it has real eigenvalues. The smallest eigenvalue of L is 0, corresponding to the eigenvector of all ones.Now, the term -L will have eigenvalues -λ, where λ are the eigenvalues of L. So, the eigenvalues of -L are non-positive. Then, we subtract 3B diag(x_i^*^2). This is a diagonal matrix with negative entries (since B is positive and x_i^*^2 is positive), so each diagonal entry is -3b_i(x_i^*)^2.Therefore, the Jacobian J is a negative definite matrix if all the eigenvalues are negative. Since L is positive semi-definite, -L is negative semi-definite, and subtracting another positive definite matrix (since 3B diag(x_i^*^2) is positive definite) makes J negative definite. Hence, all eigenvalues of J are negative, which means the equilibrium is stable.Wait, but I should be careful here. The Jacobian is -L - 3B diag(x_i^*^2). So, if we consider the eigenvalues, for each eigenvalue μ of L, the corresponding eigenvalue of J is -μ - 3b_i(x_i^*)^2. But actually, diag(x_i^*^2) is a diagonal matrix, so when combined with L, it's not straightforward. Maybe I need to consider the eigenvalues of the entire matrix.Alternatively, perhaps I can think of the system as a combination of the Laplacian dynamics and the cubic term. The Laplacian term tends to make opinions converge towards each other, while the cubic term acts as a resistance to extreme opinions.In any case, the linearization leads to a Jacobian matrix where each diagonal entry is -sum_{j} a_{ij} - 3b_i(x_i^*)^2, and the off-diagonal entries are a_{ij}. So, it's a symmetric matrix if the network is undirected.For stability, we can use the fact that if the Jacobian is negative definite, then the equilibrium is stable. A symmetric matrix is negative definite if all its eigenvalues are negative. Since L is positive semi-definite, -L is negative semi-definite, and subtracting a positive definite matrix (the diag term) makes J negative definite. Therefore, the equilibrium is stable.So, the conditions for stability are that b_i > 0, which they are, and the network is connected? Wait, no, the network could be disconnected, but in that case, each connected component would have its own equilibrium. But assuming the network is connected, then the equilibrium is stable.Wait, but the problem doesn't specify the network is connected. So, maybe the stability depends on the network's connectivity. If the network is disconnected, then each component has its own equilibrium, and the stability would be within each component.But in general, as long as the Jacobian is negative definite, the equilibrium is stable. So, the key condition is that the Jacobian matrix is negative definite, which is true if the Laplacian is positive semi-definite and the diag term is positive definite.Therefore, the equilibrium is stable if the network is such that the Laplacian is positive semi-definite (which it always is) and the diag term is positive definite, which it is since b_i > 0 and x_i^* is real.Wait, but x_i^* could be zero? If x_i^* = 0, then the diag term becomes zero, so the Jacobian would be -L, which is negative semi-definite, not negative definite. So, in that case, the equilibrium might be unstable or neutrally stable.So, perhaps the stability depends on whether x_i^* is zero or not. If x_i^* ≠ 0, then the diag term is positive definite, making J negative definite, hence stable. If x_i^* = 0, then J is -L, which has a zero eigenvalue, so the equilibrium is neutrally stable.But in the problem, c_i is positive, so x_i^* can't be zero because the equation is sum a_{ij}(x_j^* - x_i^*) - b_i(x_i^*)^3 + c_i = 0. If x_i^* = 0, then we have sum a_{ij}(-x_i^*) + c_i = 0, but x_i^* = 0, so sum a_{ij}(0 - 0) + c_i = c_i = 0, which contradicts c_i > 0. Therefore, x_i^* cannot be zero, so the diag term is positive definite, making J negative definite. Hence, the equilibrium is stable.So, the conditions for stability are that the network is such that the Laplacian is positive semi-definite (which it is), and c_i > 0, which ensures x_i^* ≠ 0, making the diag term positive definite. Therefore, the equilibrium is stable.Moving on to part 1b, which is part 2 of the problem. The network is a complete graph with n individuals, a_{ij} = a > 0 for all i ≠ j, b_i = b > 0, and c_i = c > 0 for all i.We need to find the equilibrium opinion x^* in terms of a, b, c, and n, and analyze how n affects x^*.Since the network is complete, each individual is connected to every other individual. So, for each i, N(i) is all other n-1 individuals.The equilibrium equation is:sum_{j in N(i)} a (x_j^* - x_i^*) - b (x_i^*)^3 + c = 0But since the network is complete and all a_{ij} are equal, and all b_i and c_i are equal, we can assume that all x_i^* are equal at equilibrium. Let's denote x_i^* = x^* for all i.Then, the equation becomes:sum_{j ≠ i} a (x^* - x^*) - b (x^*)^3 + c = 0But sum_{j ≠ i} a (x^* - x^*) = 0, because x^* - x^* = 0. So, the equation simplifies to:0 - b (x^*)^3 + c = 0Therefore, -b (x^*)^3 + c = 0 => (x^*)^3 = c / b => x^* = (c / b)^{1/3}Wait, but that seems too simple. Let me double-check.Wait, in the complete graph, each individual is connected to n-1 others. So, the sum over j in N(i) is sum_{j ≠ i} a (x_j^* - x_i^*). If all x_j^* = x^*, then this sum is sum_{j ≠ i} a (x^* - x^*) = 0. So, yes, the equation reduces to -b (x^*)^3 + c = 0, so x^* = (c / b)^{1/3}.But wait, that seems independent of n. Is that correct? Because in the complete graph, each individual is influenced by all others, but if all opinions are the same, the influence cancels out, leaving only the cubic term and the external influence.So, the equilibrium opinion is x^* = (c / b)^{1/3}, which is independent of n. That's interesting. So, the network size n doesn't affect the equilibrium opinion in this case.But let me think again. If n increases, does that affect the equilibrium? In the equation, the sum term is zero because all x_j^* = x_i^*, so n doesn't appear in the equation. Therefore, x^* is indeed independent of n.Wait, but in the original differential equation, the sum term is over all neighbors. In a complete graph, each individual has n-1 neighbors, so the sum is (n-1)a (x_j^* - x_i^*). But if all x_j^* = x_i^*, then the sum is zero. So, the equation is -b (x^*)^3 + c = 0, regardless of n.Therefore, the equilibrium opinion is x^* = (c / b)^{1/3}, and it doesn't depend on n. So, increasing n doesn't change the equilibrium opinion.But wait, is that realistic? Intuitively, in a larger network, each individual is influenced by more people, but if all opinions are the same, the influence cancels out. So, the external influence c and the resistance b determine the equilibrium, regardless of the network size.So, the expression for x^* is (c / b)^{1/3}, and n doesn't affect it.Wait, but let me check if I made a mistake in the equilibrium equation. The original equation is:sum_{j in N(i)} a_{ij} (x_j - x_i) - b_i x_i^3 + c_i = 0In the complete graph, a_{ij} = a for all j ≠ i, so sum_{j in N(i)} a_{ij} (x_j - x_i) = a sum_{j ≠ i} (x_j - x_i)If all x_j = x, then this sum is a sum_{j ≠ i} (x - x) = 0. So, yes, the equation becomes -b x^3 + c = 0, so x = (c / b)^{1/3}.Therefore, the equilibrium opinion is indeed independent of n.So, to summarize:a. The equilibrium is stable because the Jacobian matrix is negative definite due to the positive semi-definite Laplacian and the positive definite diagonal term from the cubic resistance.b. The equilibrium opinion is x^* = (c / b)^{1/3}, and it doesn't depend on the network size n.</think>

🧠:<think>Okay, so I have this problem about a multinational corporation's communication security, and it's divided into two sub-problems. Let me tackle them one by one.Starting with Sub-problem 1: They're using RSA encryption with a public key (e, N) where e is 65537 and N is 589,723. I need to find the private key d. I remember that in RSA, the private key d is the modular inverse of e modulo φ(N), where φ(N) is Euler's totient function. Since N is the product of two primes p and q, φ(N) is (p-1)(q-1). So, my first task is to factor N into p and q.The problem mentions that p and q are close to each other in value. That should help because if they're close, I can try to find them by checking around the square root of N. Let me calculate the square root of 589,723. Hmm, sqrt(589,723) is approximately 768 because 768 squared is 590,000, which is a bit higher. So, maybe p and q are around 767 or 769.Let me check if 589,723 is divisible by 767. Dividing 589,723 by 767: 767 times 767 is 588,289. Subtract that from 589,723: 589,723 - 588,289 = 1,434. Hmm, that's not a multiple of 767. Maybe I need to try a different approach.Alternatively, I can use the fact that N = p*q and that p and q are close. So, let me set p = q + k where k is a small number. Then N = q*(q + k) = q² + k*q. Since k is small, maybe I can approximate q as sqrt(N). Let's compute sqrt(589,723) more accurately. 768² is 589,824, which is just a bit higher than N. So, 768² is 589,824, so N is 589,723, which is 101 less than 768². So, maybe p and q are 767 and 769? Let me check: 767 * 769. Let's compute that.767 * 769: I can write this as (768 - 1)(768 + 1) = 768² - 1 = 589,824 - 1 = 589,823. But N is 589,723, which is 100 less. Hmm, that's not matching. Maybe I need to try a different pair.Wait, perhaps I made a mistake in my initial assumption. Let me try dividing N by 767. 589,723 divided by 767. Let me compute 767 * 767: 767*700=536,900; 767*60=46,020; 767*7=5,369. Adding those together: 536,900 + 46,020 = 582,920; 582,920 + 5,369 = 588,289. So, 767*767=588,289. Then, 589,723 - 588,289 = 1,434. So, 1,434 divided by 767 is 1.866. Not an integer. So, 767 is not a factor.How about 769? Let me try dividing 589,723 by 769. 769 * 700 = 538,300. 769 * 60 = 46,140. 769 * 6 = 4,614. Adding those: 538,300 + 46,140 = 584,440; 584,440 + 4,614 = 589,054. Then, 589,723 - 589,054 = 669. 669 divided by 769 is less than 1, so not a factor. Hmm.Maybe I need to try a different approach. Let me use the fact that N is 589,723 and try to find factors around 768. Let me check 768 - 1 = 767, which we saw didn't work. How about 768 + 2 = 770? Let me see if 770 divides N. 770 * 767: Wait, 770 is 70*11, so maybe it's not a prime. Alternatively, let me compute 589,723 divided by 770. 770 * 760 = 585,200. 589,723 - 585,200 = 4,523. 4,523 divided by 770 is about 5.87, not an integer. So, not a factor.Alternatively, maybe I should use Pollard's Rho algorithm or some other factoring method, but since I'm doing this manually, perhaps I can try another approach. Let me compute N modulo small primes to see if it's divisible by any. Let's try 3: 5+8+9+7+2+3 = 34, which is not divisible by 3. Next, 5: last digit is 3, so no. 7: Let's see, 589,723 divided by 7. 7*84,246=589,722, so 589,723 is 589,722 +1, so remainder 1. Not divisible by 7. 11: Alternating sum: 5 -8 +9 -7 +2 -3 = 5 -8= -3; -3 +9=6; 6 -7= -1; -1 +2=1; 1 -3= -2. Not divisible by 11.13: Let's see, 13*45,363=589,719. 589,723 -589,719=4. Not divisible by 13. 17: 17*34,689=589,713. 589,723 -589,713=10. Not divisible by 17. 19: 19*31,038=589,722. 589,723 -589,722=1. Not divisible by 19. 23: Let's see, 23*25,640=589,720. 589,723 -589,720=3. Not divisible by 23.29: 29*20,335=589,715. 589,723 -589,715=8. Not divisible by 29. 31: 31*19,023=589,713. 589,723 -589,713=10. Not divisible by 31. 37: Let me try 37*15,938=589,706. 589,723 -589,706=17. Not divisible by 37.Hmm, this is taking too long. Maybe I should try a different strategy. Since p and q are close, maybe I can approximate φ(N) as (sqrt(N) -1)^2, but that might not be precise. Alternatively, I can use the fact that φ(N) = (p-1)(q-1) = pq - p - q +1 = N - (p + q) +1. So, if I can find p + q, I can compute φ(N).Let me denote s = p + q and t = p*q = N. Then, φ(N) = t - s +1. So, if I can find s, I can compute φ(N). But how? I know that s = p + q, and since p and q are close, maybe I can find s by trying values around 2*sqrt(N). Let me compute 2*sqrt(589,723). sqrt(589,723) is approximately 768, so 2*768=1536. So, s is around 1536.Let me try s=1536. Then φ(N)=589,723 -1536 +1=588,188. Then, I need to find d such that e*d ≡1 mod φ(N). So, 65537*d ≡1 mod 588,188. To find d, I can use the extended Euclidean algorithm.Let me compute gcd(65537, 588188) and express it as a linear combination.First, divide 588188 by 65537:588188 ÷ 65537 ≈ 8.97, so 8*65537=524,296. Subtract: 588,188 - 524,296 = 63,892.Now, gcd(65537, 63,892). Compute 65537 ÷ 63,892 = 1 with remainder 1,645.Now, gcd(63,892, 1,645). Compute 63,892 ÷ 1,645 ≈38.8, so 38*1,645=62,510. Subtract: 63,892 -62,510=1,382.Now, gcd(1,645, 1,382). Compute 1,645 ÷1,382=1 with remainder 263.gcd(1,382,263). 1,382 ÷263=5 with remainder 67 (5*263=1,315; 1,382-1,315=67).gcd(263,67). 263 ÷67=3 with remainder 62.gcd(67,62). 67 ÷62=1 with remainder 5.gcd(62,5). 62 ÷5=12 with remainder 2.gcd(5,2). 5 ÷2=2 with remainder 1.gcd(2,1). 2 ÷1=2 with remainder 0. So, gcd is 1, which is expected.Now, working backwards to express 1 as a combination:1 = 5 - 2*2But 2 = 62 - 5*12, so:1 = 5 - 2*(62 -5*12) = 5 -2*62 +24*5 =25*5 -2*62But 5 =67 -62*1, so:1=25*(67 -62) -2*62=25*67 -25*62 -2*62=25*67 -27*62But 62=263 -67*3, so:1=25*67 -27*(263 -67*3)=25*67 -27*263 +81*67=106*67 -27*263But 67=1,382 -263*5, so:1=106*(1,382 -263*5) -27*263=106*1,382 -530*263 -27*263=106*1,382 -557*263But 263=1,645 -1,382*1, so:1=106*1,382 -557*(1,645 -1,382)=106*1,382 -557*1,645 +557*1,382=663*1,382 -557*1,645But 1,382=63,892 -1,645*38, so:1=663*(63,892 -1,645*38) -557*1,645=663*63,892 -663*1,645*38 -557*1,645=663*63,892 - (663*38 +557)*1,645Compute 663*38: 600*38=22,800; 63*38=2,394; total=22,800+2,394=25,194. So, 25,194 +557=25,751.Thus, 1=663*63,892 -25,751*1,645But 1,645=65537 -63,892*1, so:1=663*63,892 -25,751*(65537 -63,892)=663*63,892 -25,751*65537 +25,751*63,892= (663 +25,751)*63,892 -25,751*65537=26,414*63,892 -25,751*65537But 63,892=588,188 -65537*8, so:1=26,414*(588,188 -65537*8) -25,751*65537=26,414*588,188 -26,414*65537*8 -25,751*65537=26,414*588,188 - (26,414*8 +25,751)*65537Compute 26,414*8: 20,000*8=160,000; 6,414*8=51,312; total=160,000+51,312=211,312. So, 211,312 +25,751=237,063.Thus, 1=26,414*588,188 -237,063*65537Therefore, -237,063*65537 ≡1 mod 588,188. So, d ≡ -237,063 mod 588,188.Compute -237,063 mod 588,188: 588,188 -237,063=351,125. So, d=351,125.Wait, let me verify this. Compute 65537 *351,125 mod 588,188. Let me compute 65537*351,125. That's a big number, but I can compute it modulo 588,188.Alternatively, since we know that 65537*d ≡1 mod φ(N), and we've found d=351,125, let's check if 65537*351,125 mod 588,188 is 1.But maybe it's easier to check using the extended Euclidean steps. From above, we have 1=26,414*588,188 -237,063*65537. So, rearranged, 1= -237,063*65537 +26,414*588,188. Therefore, -237,063*65537 ≡1 mod 588,188, so d= -237,063 mod 588,188=351,125.So, the private key d is 351,125.Wait, but let me double-check. Let me compute 65537 *351,125 mod 588,188. Let me compute 65537 *351,125 first. That's a huge number, but perhaps I can compute it modulo 588,188.Alternatively, since 65537 *351,125 = (65537 *351,125) mod 588,188. Let me compute 65537 mod 588,188 is 65537. 351,125 mod 588,188 is 351,125. So, compute 65537 *351,125 mod 588,188.Alternatively, note that 65537 *351,125 = (65537 * (351,125 mod 588,188)) mod 588,188. But 351,125 is less than 588,188, so we can proceed.But perhaps it's easier to compute 65537 *351,125 mod 588,188 by breaking it down.Let me compute 65537 *351,125:First, note that 65537 *351,125 = 65537 * (350,000 +1,125) = 65537*350,000 +65537*1,125.Compute 65537*350,000: 65537*350,000=65537*35*10,000=2,293,795*10,000=22,937,950,000.Compute 65537*1,125: 65537*1,000=65,537,000; 65537*125=8,192,125. So total is 65,537,000 +8,192,125=73,729,125.So total is 22,937,950,000 +73,729,125=23,011,679,125.Now, compute 23,011,679,125 mod 588,188.To compute this, divide 23,011,679,125 by 588,188 and find the remainder.But this is a bit tedious. Alternatively, note that 588,188 *39,000=588,188*30,000=17,645,640,000; 588,188*9,000=5,293,692,000. So total 17,645,640,000 +5,293,692,000=22,939,332,000.Subtract this from 23,011,679,125: 23,011,679,125 -22,939,332,000=72,347,125.Now, compute 72,347,125 mod 588,188.Divide 72,347,125 by 588,188:588,188 *122=588,188*100=58,818,800; 588,188*20=11,763,760; 588,188*2=1,176,376. So total 58,818,800 +11,763,760=70,582,560 +1,176,376=71,758,936.Subtract from 72,347,125: 72,347,125 -71,758,936=588,189.Wait, 588,189 is just 1 more than 588,188, so 588,189 mod 588,188=1.Therefore, 23,011,679,125 mod 588,188=1. So, yes, 65537*351,125 ≡1 mod 588,188. Therefore, d=351,125 is correct.So, the private key d is 351,125.Now, moving on to Sub-problem 2: The corporation uses a symmetric key system with an LFSR generating a maximum length sequence. The period required is 2^100 -1 bits. I need to find the minimum degree n of the primitive polynomial required.I recall that the period of an LFSR with a primitive polynomial of degree n is 2^n -1. So, to get a period of 2^100 -1, the degree n must be 100. Because the maximum period is 2^n -1, so n=100 gives period 2^100 -1.Wait, but let me make sure. The problem says the LFSR produces a maximum length sequence, which requires that the polynomial is primitive. So, yes, the period is 2^n -1. Therefore, to have period 2^100 -1, n must be 100.So, the minimum degree n is 100.Wait, but let me think again. Is there a possibility of a smaller n? No, because the period is 2^n -1, so to get 2^100 -1, n must be 100. There's no smaller n that would give a period of 2^100 -1 because 2^n -1 grows exponentially with n. So, n=100 is the minimum.Therefore, the answers are:Sub-problem 1: d=351,125Sub-problem 2: n=100</think>

🧠:<think>Okay, so I have this problem where a parent is studying the effects of different disciplinary methods on children's academic performance. The methods are divided into non-corporal and corporal, and the academic performance is measured by GPA. The parent used a logistic growth model to model the GPA over time for each group. First, I need to calculate the GPA for both groups after 10 years. The logistic growth model is given by:[ text{GPA}_{text{group}}(t) = frac{K_{text{group}}}{1 + frac{K_{text{group}} - text{GPA}_{0, text{group}}}{text{GPA}_{0, text{group}}} e^{-r_{text{group}} t}} ]For the non-corporal group, the parameters are ( K_{text{non}} = 4 ), ( text{GPA}_{0, text{non}} = 1.8 ), and ( r_{text{non}} = 0.3 ). For the corporal group, they are ( K_{text{corp}} = 3.5 ), ( text{GPA}_{0, text{corp}} = 1.7 ), and ( r_{text{corp}} = 0.25 ).Alright, so I need to plug in t = 10 for both groups.Starting with the non-corporal group:First, let's write down the formula again with the given values:[ text{GPA}_{text{non}}(10) = frac{4}{1 + frac{4 - 1.8}{1.8} e^{-0.3 times 10}} ]Let me compute each part step by step.Compute ( 4 - 1.8 ): that's 2.2.Then, divide that by 1.8: 2.2 / 1.8. Let me calculate that. 2.2 divided by 1.8 is approximately 1.2222.So, the denominator becomes ( 1 + 1.2222 e^{-0.3 times 10} ).Compute the exponent: -0.3 * 10 = -3. So, e^{-3} is approximately... e is about 2.71828, so e^{-3} is roughly 1 / e^3 ≈ 1 / 20.0855 ≈ 0.0498.Multiply that by 1.2222: 1.2222 * 0.0498 ≈ 0.0608.So, the denominator is 1 + 0.0608 ≈ 1.0608.Therefore, GPA_non(10) ≈ 4 / 1.0608 ≈ let's compute that. 4 divided by 1.0608. Well, 1.0608 * 3.77 ≈ 4, because 1.0608 * 3 = 3.1824, 1.0608 * 4 = 4.2432. So, 4 / 1.0608 is approximately 3.77.Wait, let me compute it more accurately.1.0608 * 3.77: 1 * 3.77 = 3.77, 0.0608 * 3.77 ≈ 0.229. So total ≈ 3.77 + 0.229 ≈ 3.999, which is almost 4. So, 3.77 is the approximate value.But let me do it more precisely.Compute 4 / 1.0608:1.0608 goes into 4 how many times?1.0608 * 3 = 3.1824Subtract that from 4: 4 - 3.1824 = 0.8176Bring down a zero: 8.1761.0608 goes into 8.176 about 7.7 times because 1.0608 * 7 = 7.4256, 1.0608 * 7.7 ≈ 8.174 (since 1.0608 * 0.7 = 0.74256, so 7 * 1.0608 + 0.7 * 1.0608 = 7.4256 + 0.74256 ≈ 8.16816). So, 7.7 gives approximately 8.16816, which is very close to 8.176.So, 3.77 is accurate to two decimal places.Thus, GPA_non(10) ≈ 3.77.Now, moving on to the corporal group.The formula is:[ text{GPA}_{text{corp}}(10) = frac{3.5}{1 + frac{3.5 - 1.7}{1.7} e^{-0.25 times 10}} ]Compute each part:3.5 - 1.7 = 1.8Divide by 1.7: 1.8 / 1.7 ≈ 1.0588So, the denominator is 1 + 1.0588 e^{-0.25 * 10}Compute the exponent: -0.25 * 10 = -2.5e^{-2.5} is approximately 1 / e^{2.5} ≈ 1 / 12.1825 ≈ 0.0821.Multiply by 1.0588: 1.0588 * 0.0821 ≈ 0.087.So, denominator is 1 + 0.087 ≈ 1.087.Therefore, GPA_corp(10) ≈ 3.5 / 1.087 ≈ let's compute that.3.5 divided by 1.087.1.087 * 3 = 3.261Subtract from 3.5: 3.5 - 3.261 = 0.239Bring down a zero: 2.391.087 goes into 2.39 about 2.2 times because 1.087 * 2 = 2.174, which is less than 2.39. 1.087 * 2.2 ≈ 2.3914.So, 3.22 is the approximate value.Wait, let me verify:1.087 * 3.22:1.087 * 3 = 3.2611.087 * 0.22 = 0.23914So, total is 3.261 + 0.23914 ≈ 3.50014, which is very close to 3.5.So, 3.22 is accurate.Thus, GPA_corp(10) ≈ 3.22.So, after 10 years, the non-corporal group has a GPA of approximately 3.77, and the corporal group has a GPA of approximately 3.22.Moving on to the second part: determining the time t at which the GPA growth rate is maximized for each group. The differential equation given is:[ frac{d(text{GPA}_{text{group}})}{dt} = r_{text{group}} text{GPA}_{text{group}} left( 1 - frac{text{GPA}_{text{group}}}{K_{text{group}}} right) ]So, the growth rate is a function of GPA, which itself is a function of time. To find the maximum growth rate, we can take the derivative of this growth rate with respect to time and set it to zero. However, since the growth rate is a function of GPA, which follows a logistic curve, the maximum growth rate occurs when GPA is at half the carrying capacity. That is, when GPA = K/2.Alternatively, since the logistic growth model's derivative is maximized at the inflection point, which is when GPA = K/2. So, we can set GPA(t) = K/2 and solve for t.So, for each group, set GPA(t) = K/2 and solve for t.Starting with the non-corporal group:K_non = 4, so K/2 = 2.So, set GPA_non(t) = 2.Using the logistic model:2 = 4 / [1 + (4 - 1.8)/1.8 * e^{-0.3 t}]Simplify:2 = 4 / [1 + (2.2 / 1.8) e^{-0.3 t}]Compute 2.2 / 1.8 ≈ 1.2222.So,2 = 4 / [1 + 1.2222 e^{-0.3 t}]Multiply both sides by denominator:2 [1 + 1.2222 e^{-0.3 t}] = 4Divide both sides by 2:1 + 1.2222 e^{-0.3 t} = 2Subtract 1:1.2222 e^{-0.3 t} = 1Divide both sides by 1.2222:e^{-0.3 t} = 1 / 1.2222 ≈ 0.8182Take natural logarithm:-0.3 t = ln(0.8182)Compute ln(0.8182): approximately -0.2007.So,-0.3 t ≈ -0.2007Divide both sides by -0.3:t ≈ (-0.2007) / (-0.3) ≈ 0.669 years.So, approximately 0.67 years, or about 8 months.Now, for the corporal group:K_corp = 3.5, so K/2 = 1.75.Set GPA_corp(t) = 1.75.Using the logistic model:1.75 = 3.5 / [1 + (3.5 - 1.7)/1.7 * e^{-0.25 t}]Simplify:1.75 = 3.5 / [1 + (1.8 / 1.7) e^{-0.25 t}]Compute 1.8 / 1.7 ≈ 1.0588.So,1.75 = 3.5 / [1 + 1.0588 e^{-0.25 t}]Multiply both sides by denominator:1.75 [1 + 1.0588 e^{-0.25 t}] = 3.5Divide both sides by 1.75:1 + 1.0588 e^{-0.25 t} = 2Subtract 1:1.0588 e^{-0.25 t} = 1Divide both sides by 1.0588:e^{-0.25 t} ≈ 1 / 1.0588 ≈ 0.9449Take natural logarithm:-0.25 t = ln(0.9449) ≈ -0.0563So,-0.25 t ≈ -0.0563Divide both sides by -0.25:t ≈ (-0.0563) / (-0.25) ≈ 0.225 years.Which is approximately 0.225 years, or about 2.7 months.So, the growth rate is maximized at approximately 0.67 years for the non-corporal group and 0.225 years for the corporal group.Wait, that seems a bit counterintuitive. The non-corporal group has a higher growth rate parameter (r=0.3 vs. r=0.25), but their maximum growth occurs later? Hmm, let me think.Actually, the time to maximum growth rate is when GPA = K/2. Since the non-corporal group starts at a higher GPA (1.8 vs. 1.7), they might reach K/2 faster? Wait, but in our calculations, non-corporal group reaches K/2 at ~0.67 years, while corporal group reaches K/2 at ~0.225 years. So, the corporal group reaches half their carrying capacity faster, but their overall growth rate is lower.But let's verify the calculations.For non-corporal group:We set GPA = 2. The equation was:2 = 4 / [1 + 1.2222 e^{-0.3 t}]So, 1 + 1.2222 e^{-0.3 t} = 21.2222 e^{-0.3 t} = 1e^{-0.3 t} = 0.8182ln(0.8182) ≈ -0.2007So, t ≈ 0.2007 / 0.3 ≈ 0.669, which is correct.For corporal group:Set GPA = 1.75.1.75 = 3.5 / [1 + 1.0588 e^{-0.25 t}]So, 1 + 1.0588 e^{-0.25 t} = 21.0588 e^{-0.25 t} = 1e^{-0.25 t} ≈ 0.9449ln(0.9449) ≈ -0.0563t ≈ 0.0563 / 0.25 ≈ 0.225, which is correct.So, even though the non-corporal group has a higher growth rate, they start at a higher GPA, so it takes them longer to reach half their carrying capacity. The corporal group, starting lower, reaches half their K faster, even with a lower r.That makes sense because the time to reach K/2 is influenced by both the initial GPA and the growth rate. The formula for t when GPA = K/2 is:t = (1/r) * ln[(K - GPA0)/GPA0 * (K/(2) - GPA0)/(K - GPA0))]Wait, maybe another way to think about it is that the time to reach K/2 is t = (1/r) * ln[(K - GPA0)/(GPA0) * (K/(2) - GPA0)/(K - GPA0))]Wait, perhaps it's better to recall that for logistic growth, the time to reach K/2 is t = (1/r) * ln[(K - GPA0)/(GPA0)].Wait, no, actually, when solving for t when GPA = K/2, we get:K/2 = K / [1 + ((K - GPA0)/GPA0) e^{-rt}]Multiply both sides by denominator:K/2 [1 + ((K - GPA0)/GPA0) e^{-rt}] = KDivide both sides by K:1/2 [1 + ((K - GPA0)/GPA0) e^{-rt}] = 1Multiply both sides by 2:1 + ((K - GPA0)/GPA0) e^{-rt} = 2Subtract 1:((K - GPA0)/GPA0) e^{-rt} = 1So,e^{-rt} = GPA0 / (K - GPA0)Take natural log:-rt = ln[GPA0 / (K - GPA0)]So,t = (1/r) * ln[(K - GPA0)/GPA0]Ah, so the time to reach K/2 is t = (1/r) * ln[(K - GPA0)/GPA0]So, for non-corporal group:t_non = (1/0.3) * ln[(4 - 1.8)/1.8] = (1/0.3) * ln[2.2 / 1.8] ≈ (1/0.3) * ln(1.2222) ≈ (3.3333) * 0.2007 ≈ 0.669 years.For corporal group:t_corp = (1/0.25) * ln[(3.5 - 1.7)/1.7] = (4) * ln[1.8 / 1.7] ≈ 4 * ln(1.0588) ≈ 4 * 0.0572 ≈ 0.2288 years.So, that's consistent with our earlier calculations.Therefore, the maximum growth rate occurs at approximately 0.67 years for non-corporal and 0.23 years for corporal.So, summarizing:1. After 10 years, non-corporal group has GPA ≈ 3.77, corporal group ≈ 3.22.2. The maximum growth rate occurs at ~0.67 years for non-corporal and ~0.23 years for corporal.Final Answer1. After 10 years, the non-corporal group has a GPA of boxed{3.77} and the corporal group has a GPA of boxed{3.22}.2. The maximum GPA growth rate occurs at approximately boxed{0.67} years for the non-corporal group and boxed{0.23} years for the corporal group.</think>