🧠:<think>Okay, so I'm trying to figure out the volume of an oil slick over time and when it peaks. Let me start by understanding the problem step by step.First, the oil spill is spreading radially, forming a circular slick. The area of this slick is given by the function A(t) = k * t², where k is 0.5 km²/hr². So, the area increases quadratically over time. That makes sense because as time goes on, the radius increases, and the area of a circle is πr², which would also be quadratic in time if r is linear in time.Next, the thickness of the oil slick is given by h(t) = h₀ * e^(-λt). Here, h₀ is 0.01 meters, and λ is 0.1 per hour. So, the thickness decreases exponentially over time. That means as time goes on, the oil slick becomes thinner.Now, the first sub-problem is to determine the volume V(t) as a function of time. Volume is generally area multiplied by thickness, right? So, V(t) should be A(t) multiplied by h(t). Let me write that down:V(t) = A(t) * h(t)Substituting the given functions:V(t) = (0.5 * t²) * (0.01 * e^(-0.1t))Let me compute the constants first. 0.5 multiplied by 0.01 is 0.005. So,V(t) = 0.005 * t² * e^(-0.1t)Hmm, that seems straightforward. So, V(t) is 0.005 t² e^(-0.1t). I should make sure the units are consistent. A(t) is in km², h(t) is in meters. But wait, km² is square kilometers, and meters is a linear measure. To get volume in cubic meters, I need to convert km² to m².1 km² is (1000 m)^2 = 1,000,000 m². So, A(t) is in km², which is 1e6 m² per km². So, actually, V(t) in m³ would be:V(t) = A(t) * h(t) = (0.5 t² km²) * (0.01 m) * (1e6 m²/km²)Let me compute that:0.5 * 0.01 * 1e6 = 0.5 * 0.01 * 1,000,000 = 0.5 * 10,000 = 5,000.Wait, that's 5,000 m³? Wait, no, that's not right. Wait, let's do it step by step.A(t) is 0.5 t² km². So, in m², that's 0.5 t² * 1e6 m². Then, h(t) is 0.01 meters. So, V(t) is:V(t) = (0.5 * t² * 1e6 m²) * (0.01 m) = 0.5 * 1e6 * 0.01 * t² * m³Calculating the constants: 0.5 * 1e6 = 5e5, then 5e5 * 0.01 = 5e3. So, V(t) = 5,000 t² e^(-0.1t) m³.Wait, that seems a lot. Let me double-check.0.5 km²/hr² is the rate of area increase. So, at time t, A(t) = 0.5 t² km². To convert km² to m², multiply by 1e6. So, 0.5 t² * 1e6 m². Then, h(t) is 0.01 meters. So, volume is area times thickness:V(t) = (0.5 t² * 1e6 m²) * (0.01 m) = 0.5 * 1e6 * 0.01 * t² * m³.Calculating 0.5 * 1e6 = 5e5, 5e5 * 0.01 = 5e3, so 5,000 t² e^(-0.1t) m³.Wait, but in the initial substitution, I didn't convert units and got V(t) = 0.005 t² e^(-0.1t). But that would be in km² * meters, which is km²·m. To convert that to m³, since 1 km = 1e3 m, 1 km² = (1e3 m)^2 = 1e6 m². So, 1 km²·m = 1e6 m³.So, V(t) in km²·m is 0.005 t² e^(-0.1t). To convert to m³, multiply by 1e6:V(t) = 0.005 * 1e6 * t² e^(-0.1t) = 5,000 t² e^(-0.1t) m³.Yes, that matches. So, the volume function is V(t) = 5,000 t² e^(-0.1t) m³.Okay, so that's the first part. Now, the second sub-problem is to find the time t at which the volume reaches a peak before it starts to decrease.So, we need to find the maximum of V(t). To find the maximum, we can take the derivative of V(t) with respect to t, set it equal to zero, and solve for t.Let me write V(t) as:V(t) = 5,000 t² e^(-0.1t)To find dV/dt, we'll use the product rule. Let me denote u = t² and v = e^(-0.1t). Then, du/dt = 2t and dv/dt = -0.1 e^(-0.1t).So, dV/dt = 5,000 (u'v + uv') = 5,000 [2t e^(-0.1t) + t² (-0.1) e^(-0.1t)]Factor out e^(-0.1t):dV/dt = 5,000 e^(-0.1t) [2t - 0.1 t²]Set this equal to zero to find critical points:5,000 e^(-0.1t) [2t - 0.1 t²] = 0Since 5,000 is positive and e^(-0.1t) is always positive, the equation equals zero when 2t - 0.1 t² = 0.So, 2t - 0.1 t² = 0Factor out t:t (2 - 0.1 t) = 0So, t = 0 or 2 - 0.1 t = 0Solving 2 - 0.1 t = 0:0.1 t = 2 => t = 2 / 0.1 = 20 hours.So, critical points at t = 0 and t = 20 hours.Now, we need to determine whether t = 20 is a maximum. Since the volume starts at zero, increases to a peak, and then decreases, t = 20 should be the time of maximum volume.To confirm, we can check the second derivative or analyze the behavior around t = 20.Alternatively, consider the sign of dV/dt before and after t = 20.For t < 20, say t = 10:2t - 0.1 t² = 20 - 10 = 10 > 0, so dV/dt > 0, volume increasing.For t > 20, say t = 30:2t - 0.1 t² = 60 - 90 = -30 < 0, so dV/dt < 0, volume decreasing.Therefore, t = 20 hours is indeed the time when the volume reaches its maximum.So, summarizing:1. V(t) = 5,000 t² e^(-0.1t) m³2. The volume peaks at t = 20 hours.I think that's it. Let me just double-check the calculations.For the volume, converting units correctly was important. Initially, I thought maybe I missed a step, but breaking it down, it seems correct.For the derivative, applying the product rule correctly, factoring, solving for t, and confirming with the sign change all make sense. So, I'm confident in these answers.</think>

🧠:<think>Okay, so I have this problem about optimizing the wear resistance of a composite material for a kitchen countertop. The function given is W(x, y, z) = 4x² + 3y² + 5z, and the constraint is that x + y + z = 1. I need to find the proportions x, y, z that maximize W. Then, I also have to calculate the total cost per unit proportion based on the costs of each component.First, I remember that when dealing with optimization problems with constraints, the method of Lagrange multipliers is useful. So, maybe I should set that up. The idea is to find the maximum of W subject to the constraint x + y + z = 1.Let me write down the Lagrangian function. It should be the original function minus a multiplier times the constraint. So:L(x, y, z, λ) = 4x² + 3y² + 5z - λ(x + y + z - 1)Now, to find the critical points, I need to take the partial derivatives of L with respect to x, y, z, and λ, and set them equal to zero.Let's compute each partial derivative:∂L/∂x = 8x - λ = 0 --> 8x = λ∂L/∂y = 6y - λ = 0 --> 6y = λ∂L/∂z = 5 - λ = 0 --> λ = 5∂L/∂λ = -(x + y + z - 1) = 0 --> x + y + z = 1Okay, so from ∂L/∂z, we get λ = 5. Then, plugging that into the other equations:From ∂L/∂x: 8x = 5 --> x = 5/8From ∂L/∂y: 6y = 5 --> y = 5/6Wait, hold on. If x = 5/8 and y = 5/6, then let's check if x + y + z = 1.x + y = 5/8 + 5/6. Let me compute that. The common denominator is 24.5/8 = 15/24, 5/6 = 20/24. So, 15/24 + 20/24 = 35/24.But 35/24 is greater than 1, which is 24/24. So, x + y = 35/24, which is already more than 1, meaning z would have to be negative to satisfy x + y + z = 1. But proportions can't be negative. That doesn't make sense.Hmm, so maybe I made a mistake here. Let me double-check my partial derivatives.Wait, ∂L/∂x is derivative of 4x², which is 8x, minus λ, correct.∂L/∂y is derivative of 3y², which is 6y, minus λ, correct.∂L/∂z is derivative of 5z, which is 5, minus λ, correct.So, the partial derivatives seem right. So, if λ = 5, then x = 5/8, y = 5/6, but that gives x + y > 1, which is impossible.So, this suggests that the maximum occurs at a boundary of the domain, since the critical point found is not feasible.In optimization problems with constraints, if the critical point found via Lagrange multipliers is not within the feasible region, the maximum must occur on the boundary.So, the feasible region is the simplex x + y + z = 1, with x, y, z ≥ 0.Therefore, the maximum must occur when one or more of the variables are zero.So, perhaps I need to check the boundaries where one of x, y, or z is zero.Let me consider each case:Case 1: z = 0. Then, x + y = 1. The function becomes W = 4x² + 3y².We can express y as 1 - x, so W = 4x² + 3(1 - x)².Let me expand that: 4x² + 3(1 - 2x + x²) = 4x² + 3 - 6x + 3x² = 7x² - 6x + 3.To find the maximum of this quadratic function, since the coefficient of x² is positive, it opens upwards, so the minimum is at the vertex, but we are looking for the maximum on the interval x ∈ [0,1].So, the maximum will occur at one of the endpoints.Compute W at x=0: W = 0 + 3(1)² = 3.Compute W at x=1: W = 4(1)² + 3(0)² = 4.So, in this case, maximum is 4 at x=1, y=0, z=0.Case 2: y = 0. Then, x + z = 1. The function becomes W = 4x² + 5z.Express z as 1 - x, so W = 4x² + 5(1 - x) = 4x² + 5 - 5x.This is a quadratic in x: 4x² -5x +5.Again, since the coefficient of x² is positive, it opens upwards, so the minimum is at the vertex, but maximum occurs at endpoints.Compute W at x=0: W = 0 +5(1) =5.Compute W at x=1: W=4(1) +5(0)=4.Thus, maximum is 5 at x=0, z=1, y=0.Case 3: x = 0. Then, y + z =1. The function becomes W=3y² +5z.Express z as 1 - y, so W=3y² +5(1 - y)=3y² +5 -5y.This is quadratic in y: 3y² -5y +5.Again, opens upwards, so maximum at endpoints.Compute W at y=0: W=0 +5(1)=5.Compute W at y=1: W=3(1) +5(0)=3.Thus, maximum is 5 at y=0, z=1, x=0.So, from the three cases, the maximum wear resistance is 5, achieved when either x=0, y=0, z=1 or when x=0, z=1, y=0. Wait, actually, in Case 2, x=0, z=1, y=0, and in Case 3, same thing. So, the maximum is 5 when z=1, x=0, y=0.But wait, in Case 1, the maximum was 4, which is less than 5. So, the overall maximum is 5, achieved when z=1, x=0, y=0.But hold on, earlier when I tried using Lagrange multipliers, I got x=5/8, y=5/6, which is not feasible because x + y >1. So, that suggests that the maximum is indeed on the boundary, specifically at z=1.Therefore, the proportions that maximize wear resistance are x=0, y=0, z=1.But let me check if that makes sense. If I have all the wear-resistant additive, then the wear resistance is 5*1=5. If I have some other combination, like in Case 1, putting all in quartz gives W=4, which is less than 5. Similarly, putting all in resin gives W=3, which is even less. So, indeed, putting all in z gives the highest wear resistance.So, the maximum wear resistance is 5, achieved when z=1, x=0, y=0.Now, moving on to the second part: calculating the total cost per unit proportion when x, y, z are chosen to maximize W.The cost per unit proportion is given as 20 for quartz, 15 for resin, and 50 for the additive.So, the total cost per unit proportion is 20x + 15y + 50z.Since we have x=0, y=0, z=1, plugging in:Total cost = 20*0 + 15*0 + 50*1 = 50.Wait, that seems straightforward. But let me make sure I didn't miss anything.Alternatively, is the cost per unit proportion referring to the cost per unit volume or something else? The problem says "cost per unit proportion," so I think it's per unit of the mixture. So, if x, y, z are proportions, then the total cost is 20x +15y +50z, yes.So, with x=0, y=0, z=1, total cost is 50.But let me think again: is the maximum wear resistance achieved at z=1, so all the mixture is the additive. That seems a bit counterintuitive because usually, you want a mix of materials for other properties, but in this case, since the wear resistance function is linear in z and quadratic in x and y, with coefficients 4 and 3, which are less than 5. So, z has a higher coefficient, so increasing z gives higher wear resistance.But since the coefficients for x² and y² are positive, meaning that increasing x or y also increases wear resistance, but quadratically. So, perhaps there is a point where increasing x or y could give higher W than just putting all into z.Wait, but in the interior critical point, we saw that x=5/8, y=5/6, which is x≈0.625, y≈0.833, which is more than 1 when summed, which is impossible. So, that suggests that the maximum is indeed on the boundary.But let me test another approach. Maybe using substitution.Given x + y + z =1, so z =1 -x -y.Substitute into W: W=4x² +3y² +5(1 -x -y)=4x² +3y² +5 -5x -5y.So, W=4x² +3y² -5x -5y +5.Now, to find the maximum of this function over x, y ≥0, and x + y ≤1.To find critical points, take partial derivatives with respect to x and y.∂W/∂x =8x -5.Set to zero: 8x -5=0 --> x=5/8≈0.625.Similarly, ∂W/∂y=6y -5.Set to zero: 6y -5=0 --> y=5/6≈0.833.So, the critical point is at x=5/8, y=5/6, z=1 -5/8 -5/6.Compute z: 1 - (15/24 + 20/24)=1 -35/24= -11/24≈-0.458.Negative, which is not feasible.Therefore, as before, the maximum must be on the boundary.So, on the boundaries, which are the edges of the simplex.So, the boundaries are when one variable is zero.So, as before, when z=0, x + y=1, W=4x² +3y².We saw that maximum is 4 at x=1, y=0.When y=0, x + z=1, W=4x² +5z.Maximum is 5 at x=0, z=1.When x=0, y + z=1, W=3y² +5z.Maximum is 5 at y=0, z=1.So, same conclusion.Therefore, the maximum wear resistance is 5, achieved when z=1, x=0, y=0.Thus, the proportions are x=0, y=0, z=1.Therefore, the total cost is 20*0 +15*0 +50*1=50.So, the total cost per unit proportion is 50.Wait, but let me think again: if all the mixture is the additive, is that practical? Maybe in reality, you can't have 100% additive, but in the problem, it's allowed since proportions can be zero or one.So, the answer is x=0, y=0, z=1, with total cost 50.But just to make sure, let me check the function W at z=1: W=5*1=5.If I take another point, say, z=0.9, x=0.1, y=0, then W=4*(0.1)^2 +3*(0)^2 +5*(0.9)=0.04 +0 +4.5=4.54, which is less than 5.Similarly, z=0.95, x=0.05, y=0: W=4*(0.05)^2 +5*0.95=0.01 +4.75=4.76 <5.So, yes, increasing z beyond 1 is not possible, but as z approaches 1, W approaches 5.Therefore, the maximum is indeed at z=1.So, I think that's solid.Final Answer1. The proportions that maximize wear resistance are ( x = 0 ), ( y = 0 ), and ( z = 1 ). 2. The total cost per unit proportion is boxed{50} dollars.</think>

🧠:<think>Okay, so I have this problem about a football player who scored goals in a sequence defined by a recursive formula. The sequence is G(n) = G(n-1) + G(n-2) + 1, with initial conditions G(1) = 2 and G(2) = 3. I need to find a closed-form expression for G(n) and then use it to compute the total number of goals over the first 10 matches.Alright, let's start with part 1. I remember that for linear recursions, especially second-order ones, we can use characteristic equations or generating functions. Since the recursion here is linear and has constant coefficients, I think the characteristic equation method might work. But there's a constant term in the recursion, the "+1", so that might complicate things a bit.First, let me write down the recursion:G(n) = G(n-1) + G(n-2) + 1Hmm, this is a nonhomogeneous linear recurrence relation because of the constant term. To solve this, I think I need to find the general solution, which is the sum of the homogeneous solution and a particular solution.So, let's first solve the homogeneous part:G(n) - G(n-1) - G(n-2) = 0The characteristic equation for this would be:r^2 - r - 1 = 0Let me solve this quadratic equation. The discriminant is D = (1)^2 - 4*1*(-1) = 1 + 4 = 5.So, the roots are r = [1 ± sqrt(5)] / 2.Let me denote them as r1 = (1 + sqrt(5))/2 and r2 = (1 - sqrt(5))/2.Therefore, the general solution to the homogeneous equation is:G_h(n) = A*(r1)^n + B*(r2)^nWhere A and B are constants to be determined by initial conditions.Now, since the nonhomogeneous term is a constant (1), I can look for a particular solution. Let's assume the particular solution is a constant, say G_p(n) = C.Plugging this into the recursion:C = C + C + 1Wait, that would be:C = C + C + 1 => C = 2C + 1 => -C = 1 => C = -1So, the particular solution is G_p(n) = -1.Therefore, the general solution is:G(n) = G_h(n) + G_p(n) = A*(r1)^n + B*(r2)^n - 1Now, we can use the initial conditions to find A and B.Given G(1) = 2 and G(2) = 3.Let's plug in n=1:G(1) = A*r1 + B*r2 - 1 = 2So,A*r1 + B*r2 = 3 ...(1)Similarly, for n=2:G(2) = A*(r1)^2 + B*(r2)^2 - 1 = 3So,A*(r1)^2 + B*(r2)^2 = 4 ...(2)Now, I need to compute (r1)^2 and (r2)^2.We know that r1 and r2 are roots of r^2 - r -1 =0, so r1^2 = r1 + 1 and r2^2 = r2 +1.Therefore, equation (2) becomes:A*(r1 +1) + B*(r2 +1) = 4Which simplifies to:A*r1 + A + B*r2 + B = 4But from equation (1), A*r1 + B*r2 = 3, so substituting:3 + A + B = 4 => A + B = 1 ...(3)So now, we have two equations:From equation (1): A*r1 + B*r2 = 3From equation (3): A + B = 1We can solve these two equations for A and B.Let me write r1 = (1 + sqrt(5))/2 and r2 = (1 - sqrt(5))/2.So, let's denote sqrt(5) as s for simplicity.Then, r1 = (1 + s)/2, r2 = (1 - s)/2.So, equation (1) becomes:A*(1 + s)/2 + B*(1 - s)/2 = 3Multiply both sides by 2:A*(1 + s) + B*(1 - s) = 6Equation (3): A + B = 1Let me write these two equations:1. A*(1 + s) + B*(1 - s) = 62. A + B = 1Let me solve equation 2 for B: B = 1 - ASubstitute into equation 1:A*(1 + s) + (1 - A)*(1 - s) = 6Expand:A*(1 + s) + (1 - s) - A*(1 - s) = 6Factor A:A*(1 + s - 1 + s) + (1 - s) = 6Simplify:A*(2s) + (1 - s) = 6So,2s*A = 6 - (1 - s) = 5 + sThus,A = (5 + s)/(2s)Similarly, since s = sqrt(5), we can write:A = (5 + sqrt(5))/(2*sqrt(5))Let me rationalize the denominator:Multiply numerator and denominator by sqrt(5):A = (5 + sqrt(5)) * sqrt(5) / (2*5) = (5*sqrt(5) + 5)/10 = (5(sqrt(5) + 1))/10 = (sqrt(5) + 1)/2So, A = (sqrt(5) + 1)/2Then, from equation (3), B = 1 - A = 1 - (sqrt(5) + 1)/2 = (2 - sqrt(5) - 1)/2 = (1 - sqrt(5))/2So, B = (1 - sqrt(5))/2Therefore, the general solution is:G(n) = A*r1^n + B*r2^n - 1Plugging in A and B:G(n) = [(sqrt(5) + 1)/2]*[(1 + sqrt(5))/2]^n + [(1 - sqrt(5))/2]*[(1 - sqrt(5))/2]^n - 1Wait, hold on. Let me check that.Wait, A is (sqrt(5) + 1)/2, which is actually r1. Similarly, B is (1 - sqrt(5))/2, which is r2.So, A = r1 and B = r2.Therefore, G(n) = r1*(r1)^n + r2*(r2)^n - 1 = r1^{n+1} + r2^{n+1} - 1Wait, that seems a bit off. Let me see:Wait, no. Wait, A is (sqrt(5) + 1)/2, which is r1, and B is (1 - sqrt(5))/2, which is r2.So, G(n) = A*r1^n + B*r2^n - 1 = r1^{n+1} + r2^{n+1} - 1?Wait, no, because A is r1, so A*r1^n = r1^{n+1}, similarly B*r2^n = r2^{n+1}So, G(n) = r1^{n+1} + r2^{n+1} - 1Wait, that seems correct.But let me verify with n=1 and n=2.For n=1:G(1) = r1^{2} + r2^{2} -1We know r1^2 = r1 +1, similarly r2^2 = r2 +1So, G(1) = (r1 +1) + (r2 +1) -1 = r1 + r2 +1But r1 + r2 = [ (1 + sqrt(5))/2 + (1 - sqrt(5))/2 ] = (2)/2 =1So, G(1) = 1 +1 =2, which matches.Similarly, for n=2:G(2) = r1^{3} + r2^{3} -1Compute r1^3 and r2^3.We can use the recurrence relation for Fibonacci numbers, since r1 and r2 are the roots related to Fibonacci.Wait, actually, r1^n = r1^{n-1} + r1^{n-2}, similarly for r2.But maybe it's easier to compute r1^3:r1^3 = r1*(r1^2) = r1*(r1 +1) = r1^2 + r1 = (r1 +1) + r1 = 2*r1 +1Similarly, r2^3 = 2*r2 +1Therefore, G(2) = (2*r1 +1) + (2*r2 +1) -1 = 2*(r1 + r2) +1Again, r1 + r2 =1, so 2*1 +1 =3, which matches G(2)=3.Good, so the formula seems correct.Therefore, the closed-form expression is:G(n) = r1^{n+1} + r2^{n+1} -1Where r1 = (1 + sqrt(5))/2 and r2 = (1 - sqrt(5))/2.Alternatively, since r1 and r2 are the golden ratio and its conjugate, we can express this in terms of Fibonacci numbers, but I think this is sufficient.Alternatively, since r1 = phi and r2 = psi, where phi is the golden ratio and psi is its conjugate, we can write:G(n) = phi^{n+1} + psi^{n+1} -1But perhaps it's better to write it in terms of the roots as above.So, that's the closed-form expression.Now, moving on to part 2: Compute the total number of goals over the first 10 matches using this closed-form expression.So, the total T = G(1) + G(2) + ... + G(10)Given that G(n) = r1^{n+1} + r2^{n+1} -1, so:T = sum_{n=1 to 10} [ r1^{n+1} + r2^{n+1} -1 ] = sum_{n=1 to 10} r1^{n+1} + sum_{n=1 to 10} r2^{n+1} - sum_{n=1 to 10} 1Compute each sum separately.First, sum_{n=1 to 10} r1^{n+1} = sum_{k=2 to 11} r1^k = sum_{k=0 to 11} r1^k - r1^0 - r1^1Similarly, sum_{n=1 to 10} r2^{n+1} = sum_{k=2 to 11} r2^k = sum_{k=0 to 11} r2^k - r2^0 - r2^1And sum_{n=1 to 10} 1 =10So, let's compute these.First, sum_{k=0 to 11} r1^k is a geometric series with ratio r1, so sum = (r1^{12} -1)/(r1 -1)Similarly, sum_{k=0 to 11} r2^k = (r2^{12} -1)/(r2 -1)Therefore,sum_{n=1 to 10} r1^{n+1} = (r1^{12} -1)/(r1 -1) -1 - r1Similarly,sum_{n=1 to 10} r2^{n+1} = (r2^{12} -1)/(r2 -1) -1 - r2So, let's compute these.First, let's compute (r1^{12} -1)/(r1 -1) -1 - r1Similarly for r2.But this might get complicated. Alternatively, perhaps we can find a recursive formula for the sum.Alternatively, since G(n) = r1^{n+1} + r2^{n+1} -1, then sum_{n=1 to 10} G(n) = sum_{n=1 to10} [r1^{n+1} + r2^{n+1}] -10So, sum_{n=1 to10} [r1^{n+1} + r2^{n+1}] = sum_{n=2 to11} [r1^n + r2^n]Which is equal to sum_{n=0 to11} [r1^n + r2^n] - [r1^0 + r2^0] - [r1^1 + r2^1]Compute sum_{n=0 to11} [r1^n + r2^n] = sum_{n=0 to11} r1^n + sum_{n=0 to11} r2^nEach is a geometric series.Sum_{n=0 to11} r1^n = (r1^{12} -1)/(r1 -1)Similarly, sum_{n=0 to11} r2^n = (r2^{12} -1)/(r2 -1)So, sum_{n=0 to11} [r1^n + r2^n] = (r1^{12} -1)/(r1 -1) + (r2^{12} -1)/(r2 -1)Then, subtract [r1^0 + r2^0] =1 +1=2 and [r1^1 + r2^1] = r1 + r2 =1So, total sum is:[(r1^{12} -1)/(r1 -1) + (r2^{12} -1)/(r2 -1)] -2 -1 = [(r1^{12} -1)/(r1 -1) + (r2^{12} -1)/(r2 -1)] -3Therefore, the total T is:[(r1^{12} -1)/(r1 -1) + (r2^{12} -1)/(r2 -1) -3] -10Wait, no. Wait, the total T is:sum_{n=1 to10} G(n) = [sum_{n=2 to11} (r1^n + r2^n)] -10Which is equal to [sum_{n=0 to11} (r1^n + r2^n) - (r1^0 + r2^0) - (r1^1 + r2^1)] -10Which is [sum_{n=0 to11} (r1^n + r2^n) -2 -1] -10 = sum_{n=0 to11} (r1^n + r2^n) -13So, T = [ (r1^{12} -1)/(r1 -1) + (r2^{12} -1)/(r2 -1) ] -13Hmm, this seems complicated, but maybe we can compute it numerically.Alternatively, perhaps we can find a recursive relation for the sum.Wait, let me think differently.Given that G(n) = G(n-1) + G(n-2) +1, with G(1)=2, G(2)=3.Let me compute the first 10 terms manually to find the total.Compute G(1) to G(10):G(1) =2G(2)=3G(3)=G(2)+G(1)+1=3+2+1=6G(4)=G(3)+G(2)+1=6+3+1=10G(5)=G(4)+G(3)+1=10+6+1=17G(6)=G(5)+G(4)+1=17+10+1=28G(7)=G(6)+G(5)+1=28+17+1=46G(8)=G(7)+G(6)+1=46+28+1=75G(9)=G(8)+G(7)+1=75+46+1=122G(10)=G(9)+G(8)+1=122+75+1=198So, the sequence is: 2,3,6,10,17,28,46,75,122,198Now, let's compute the total:2 +3=55+6=1111+10=2121+17=3838+28=6666+46=112112+75=187187+122=309309+198=507So, total is 507.Wait, that's straightforward. But the question says to compute it using the closed-form expression. So, maybe I should compute it using the formula.Alternatively, perhaps the closed-form expression can help us find a formula for the sum.Alternatively, since G(n) = r1^{n+1} + r2^{n+1} -1, then sum_{n=1 to10} G(n) = sum_{n=1 to10} [r1^{n+1} + r2^{n+1}] -10Which is sum_{k=2 to11} [r1^k + r2^k] -10So, sum_{k=2 to11} [r1^k + r2^k] = sum_{k=0 to11} [r1^k + r2^k] - (r1^0 + r2^0) - (r1^1 + r2^1)Which is sum_{k=0 to11} [r1^k + r2^k] -2 -1 = sum_{k=0 to11} [r1^k + r2^k] -3So, total T = [sum_{k=0 to11} (r1^k + r2^k) -3] -10 = sum_{k=0 to11} (r1^k + r2^k) -13Now, compute sum_{k=0 to11} (r1^k + r2^k)We can compute this as sum_{k=0 to11} r1^k + sum_{k=0 to11} r2^kEach is a geometric series.Sum_{k=0 to n} r^k = (r^{n+1} -1)/(r -1)So, sum_{k=0 to11} r1^k = (r1^{12} -1)/(r1 -1)Similarly, sum_{k=0 to11} r2^k = (r2^{12} -1)/(r2 -1)Therefore, sum_{k=0 to11} (r1^k + r2^k) = (r1^{12} -1)/(r1 -1) + (r2^{12} -1)/(r2 -1)Now, let's compute these terms.First, compute r1^{12} and r2^{12}.But r1 and r2 are roots of r^2 = r +1, so we can use this to find higher powers.Alternatively, we can use the fact that r1^n + r2^n is related to Fibonacci numbers.Wait, actually, r1^n + r2^n = F(n+1)*sqrt(5), where F(n) is the nth Fibonacci number. Wait, no, actually, it's known that r1^n + r2^n = L(n), where L(n) is the nth Lucas number.Yes, Lucas numbers satisfy L(n) = r1^n + r2^n.And Lucas numbers start with L(0)=2, L(1)=1, L(2)=3, L(3)=4, L(4)=7, etc.So, if that's the case, then sum_{k=0 to11} (r1^k + r2^k) = sum_{k=0 to11} L(k)So, sum_{k=0 to11} L(k)I can compute this sum.Lucas numbers:L(0)=2L(1)=1L(2)=3L(3)=4L(4)=7L(5)=11L(6)=18L(7)=29L(8)=47L(9)=76L(10)=123L(11)=199So, sum from k=0 to11: 2 +1 +3 +4 +7 +11 +18 +29 +47 +76 +123 +199Let me compute this step by step:Start with 2.2 +1=33 +3=66 +4=1010 +7=1717 +11=2828 +18=4646 +29=7575 +47=122122 +76=198198 +123=321321 +199=520So, sum_{k=0 to11} L(k) =520Therefore, sum_{k=0 to11} (r1^k + r2^k) =520Therefore, total T =520 -13=507Which matches the manual computation.So, the total number of goals is 507.Therefore, the closed-form expression is G(n) = r1^{n+1} + r2^{n+1} -1, where r1=(1+sqrt(5))/2 and r2=(1-sqrt(5))/2, and the total over the first 10 matches is 507.Final AnswerThe total number of goals scored in the first 10 matches is boxed{507}.</think>

🧠:<think>Okay, so I have this problem about modeling voter support dynamics in Dominica using nonlinear differential equations. It's divided into two parts. Let me try to tackle them one by one.Starting with part 1: I need to analyze the stability of the fixed points of the given system and discuss when the opposition party's support increases. The system is:[begin{cases}frac{dV_{DLP}}{dt} = a V_{Undecided}(t) - b V_{DLP}(t) V_{Opp}(t) frac{dV_{Opp}}{dt} = c V_{Undecided}(t) - d V_{Opp}(t) V_{DLP}(t) frac{dV_{Undecided}}{dt} = -a V_{Undecided}(t) - c V_{Undecided}(t)end{cases}]First, since the total population ( P(t) ) is constant, I can express ( V_{Undecided}(t) ) as ( P - V_{DLP}(t) - V_{Opp}(t) ). That might simplify things later on.But for fixed points, I need to find the points where all derivatives are zero. So, set each derivative to zero:1. ( a V_{Undecided} - b V_{DLP} V_{Opp} = 0 )2. ( c V_{Undecided} - d V_{Opp} V_{DLP} = 0 )3. ( -a V_{Undecided} - c V_{Undecided} = 0 )Looking at equation 3: ( - (a + c) V_{Undecided} = 0 ). So, ( V_{Undecided} = 0 ) at fixed points.Plugging ( V_{Undecided} = 0 ) into equations 1 and 2:1. ( - b V_{DLP} V_{Opp} = 0 )2. ( - d V_{Opp} V_{DLP} = 0 )So, either ( V_{DLP} = 0 ) or ( V_{Opp} = 0 ).Therefore, the fixed points are:1. ( V_{DLP} = 0 ), ( V_{Opp} = 0 ), ( V_{Undecided} = P )2. ( V_{DLP} = P ), ( V_{Opp} = 0 ), ( V_{Undecided} = 0 )3. ( V_{DLP} = 0 ), ( V_{Opp} = P ), ( V_{Undecided} = 0 )Wait, but if ( V_{DLP} = 0 ) and ( V_{Opp} = 0 ), then all voters are undecided, but that's only possible if ( P = 0 ), which isn't the case. So maybe that fixed point isn't feasible unless ( P = 0 ). So the relevant fixed points are when one party has all the support, and the other has none.So, fixed points are:- All undecided: ( V_{DLP} = 0 ), ( V_{Opp} = 0 ), ( V_{Undecided} = P )- DLP wins: ( V_{DLP} = P ), ( V_{Opp} = 0 ), ( V_{Undecided} = 0 )- Opp wins: ( V_{DLP} = 0 ), ( V_{Opp} = P ), ( V_{Undecided} = 0 )But the all undecided point is only stable if... Hmm, let's analyze the stability.To do that, I need to linearize the system around each fixed point and find the eigenvalues.Starting with the all undecided fixed point: ( V_{DLP} = 0 ), ( V_{Opp} = 0 ), ( V_{Undecided} = P ).Let me denote the variables as ( x = V_{DLP} ), ( y = V_{Opp} ), ( z = V_{Undecided} ). Then the system is:[begin{cases}frac{dx}{dt} = a z - b x y frac{dy}{dt} = c z - d x y frac{dz}{dt} = - (a + c) zend{cases}]At the fixed point ( (0, 0, P) ), let's compute the Jacobian matrix.The Jacobian J is:[J = begin{bmatrix}frac{partial dot{x}}{partial x} & frac{partial dot{x}}{partial y} & frac{partial dot{x}}{partial z} frac{partial dot{y}}{partial x} & frac{partial dot{y}}{partial y} & frac{partial dot{y}}{partial z} frac{partial dot{z}}{partial x} & frac{partial dot{z}}{partial y} & frac{partial dot{z}}{partial z}end{bmatrix}]Calculating each derivative:For ( dot{x} = a z - b x y ):- ( frac{partial dot{x}}{partial x} = -b y )- ( frac{partial dot{x}}{partial y} = -b x )- ( frac{partial dot{x}}{partial z} = a )For ( dot{y} = c z - d x y ):- ( frac{partial dot{y}}{partial x} = -d y )- ( frac{partial dot{y}}{partial y} = -d x )- ( frac{partial dot{y}}{partial z} = c )For ( dot{z} = - (a + c) z ):- ( frac{partial dot{z}}{partial x} = 0 )- ( frac{partial dot{z}}{partial y} = 0 )- ( frac{partial dot{z}}{partial z} = - (a + c) )Evaluating at ( (0, 0, P) ):J becomes:[J = begin{bmatrix}0 & 0 & a 0 & 0 & c 0 & 0 & - (a + c)end{bmatrix}]The eigenvalues are the diagonal elements since it's a triangular matrix. So eigenvalues are 0, 0, and ( - (a + c) ). Since one eigenvalue is negative and the others are zero, the fixed point is a saddle point. So it's unstable.Next, consider the fixed point where DLP has all the support: ( (P, 0, 0) ).Compute the Jacobian at ( (P, 0, 0) ):Again, using the partial derivatives:For ( dot{x} ):- ( frac{partial dot{x}}{partial x} = -b y = 0 )- ( frac{partial dot{x}}{partial y} = -b x = -b P )- ( frac{partial dot{x}}{partial z} = a )For ( dot{y} ):- ( frac{partial dot{y}}{partial x} = -d y = 0 )- ( frac{partial dot{y}}{partial y} = -d x = -d P )- ( frac{partial dot{y}}{partial z} = c )For ( dot{z} ):- All partials are 0 except ( frac{partial dot{z}}{partial z} = - (a + c) )So J at ( (P, 0, 0) ):[J = begin{bmatrix}0 & -b P & a 0 & -d P & c 0 & 0 & - (a + c)end{bmatrix}]To find eigenvalues, solve det(J - λI) = 0.The matrix is upper triangular except for the first two rows. Let's write it out:[begin{bmatrix}-λ & -b P & a 0 & -d P - λ & c 0 & 0 & - (a + c) - λend{bmatrix}]The eigenvalues are the diagonal elements because the matrix is upper triangular. So eigenvalues are:1. ( -λ = 0 ) => λ = 02. ( -d P - λ = 0 ) => λ = -d P3. ( - (a + c) - λ = 0 ) => λ = - (a + c)So eigenvalues are 0, -d P, - (a + c). Since two eigenvalues are negative and one is zero, the fixed point is a saddle point. So it's also unstable.Similarly, for the fixed point where Opp has all support: ( (0, P, 0) ).Compute the Jacobian at ( (0, P, 0) ):For ( dot{x} ):- ( frac{partial dot{x}}{partial x} = -b y = -b P )- ( frac{partial dot{x}}{partial y} = -b x = 0 )- ( frac{partial dot{x}}{partial z} = a )For ( dot{y} ):- ( frac{partial dot{y}}{partial x} = -d y = -d P )- ( frac{partial dot{y}}{partial y} = -d x = 0 )- ( frac{partial dot{y}}{partial z} = c )For ( dot{z} ):- All partials are 0 except ( frac{partial dot{z}}{partial z} = - (a + c) )So J at ( (0, P, 0) ):[J = begin{bmatrix}- b P & 0 & a - d P & 0 & c 0 & 0 & - (a + c)end{bmatrix}]Again, the eigenvalues are the diagonal elements because it's upper triangular:1. ( - b P - λ = 0 ) => λ = -b P2. ( 0 - λ = 0 ) => λ = 03. ( - (a + c) - λ = 0 ) => λ = - (a + c)So eigenvalues are -b P, 0, - (a + c). Again, two negative eigenvalues and one zero. So this fixed point is also a saddle point, hence unstable.Hmm, so all the fixed points are saddle points? That suggests that the system doesn't settle into any of these fixed points, but maybe oscillates or something else.But wait, maybe I missed some fixed points. Because in nonlinear systems, sometimes fixed points can be non-trivial, not just the corners. Let me check.Suppose ( V_{Undecided} = 0 ), but both ( V_{DLP} ) and ( V_{Opp} ) are positive. Then from equations 1 and 2:1. ( - b V_{DLP} V_{Opp} = 0 )2. ( - d V_{DLP} V_{Opp} = 0 )Which implies either ( V_{DLP} = 0 ) or ( V_{Opp} = 0 ). So no, there are no fixed points with both parties having positive support and no undecided voters.What about fixed points with ( V_{Undecided} neq 0 )?Wait, but from equation 3, ( frac{dV_{Undecided}}{dt} = - (a + c) V_{Undecided} ). So unless ( V_{Undecided} = 0 ), it's decreasing. So the only fixed points are when ( V_{Undecided} = 0 ). So the only fixed points are the ones we found earlier.So, all fixed points are unstable saddle points. That suggests that the system might not have stable fixed points, but perhaps limit cycles or other behaviors.But the question is about the stability of fixed points and when the opposition support increases.So, if all fixed points are unstable, maybe the system can have oscillatory behavior or approach some other kind of attractor.But to analyze when ( V_{Opp} ) increases, let's look at the derivative ( frac{dV_{Opp}}{dt} = c V_{Undecided} - d V_{Opp} V_{DLP} ).So, ( V_{Opp} ) increases when ( c V_{Undecided} > d V_{Opp} V_{DLP} ).Given that ( V_{Undecided} = P - V_{DLP} - V_{Opp} ), substitute:( c (P - V_{DLP} - V_{Opp}) > d V_{Opp} V_{DLP} )So, ( c P - c V_{DLP} - c V_{Opp} > d V_{Opp} V_{DLP} )Rearranged:( c P > c V_{DLP} + c V_{Opp} + d V_{Opp} V_{DLP} )Hmm, not sure if that helps.Alternatively, maybe consider the ratio of the derivatives.Alternatively, think about the system in terms of fractions. Let me define ( x = V_{DLP}/P ), ( y = V_{Opp}/P ), ( z = V_{Undecided}/P ). Then ( x + y + z = 1 ).Then, the system becomes:[begin{cases}frac{dx}{dt} = a z - b P x y frac{dy}{dt} = c z - d P x y frac{dz}{dt} = - (a + c) zend{cases}]But since ( z = 1 - x - y ), we can write:[begin{cases}frac{dx}{dt} = a (1 - x - y) - b P x y frac{dy}{dt} = c (1 - x - y) - d P x y frac{dz}{dt} = - (a + c) (1 - x - y)end{cases}]But maybe that complicates things more.Alternatively, perhaps subtract the two equations for ( dot{x} ) and ( dot{y} ):( frac{dx}{dt} - frac{dy}{dt} = (a - c) z - (b - d) x y )But not sure.Alternatively, consider the difference ( w = x - y ). Then:( frac{dw}{dt} = frac{dx}{dt} - frac{dy}{dt} = (a - c) z - (b - d) x y )But again, not sure.Alternatively, think about when ( V_{Opp} ) is increasing. So, ( frac{dy}{dt} > 0 ). So:( c z > d x y )But ( z = 1 - x - y ), so:( c (1 - x - y) > d x y )So, ( c (1 - x - y) > d x y )Which can be written as:( c > (d x y) / (1 - x - y) )Hmm, not sure.Alternatively, think about the initial conditions. Suppose at t=0, ( V_{DLP} = V_{DLP0} ), ( V_{Opp} = V_{Opp0} ), so ( V_{Undecided} = P - V_{DLP0} - V_{Opp0} ).If initially, the undecided voters are high, then both ( V_{DLP} ) and ( V_{Opp} ) can increase if they convert undecided voters faster than they lose support to each other.But the terms ( -b V_{DLP} V_{Opp} ) and ( -d V_{DLP} V_{Opp} ) represent the interaction between the two parties, where each party loses support proportional to the product of their supporters.So, if ( a > c ), then DLP converts undecided voters faster, which might help DLP grow more. Similarly, if ( c > a ), Opp might grow faster.But the interaction terms complicate things.Alternatively, maybe consider symmetric case where ( a = c ) and ( b = d ). Then the system is symmetric, and perhaps the difference ( w = x - y ) can be analyzed.But since the problem doesn't specify symmetry, maybe we need a different approach.Alternatively, consider the total change in ( V_{Opp} ):( frac{dV_{Opp}}{dt} = c V_{Undecided} - d V_{Opp} V_{DLP} )So, for ( V_{Opp} ) to increase, ( c V_{Undecided} > d V_{Opp} V_{DLP} ).Given that ( V_{Undecided} = P - V_{DLP} - V_{Opp} ), substitute:( c (P - V_{DLP} - V_{Opp}) > d V_{Opp} V_{DLP} )So, ( c P > c V_{DLP} + c V_{Opp} + d V_{Opp} V_{DLP} )Not sure.Alternatively, maybe think about the ratio ( frac{dV_{Opp}}{dt} / frac{dV_{DLP}}{dt} ):( frac{c V_{Undecided} - d V_{Opp} V_{DLP}}{a V_{Undecided} - b V_{Opp} V_{DLP}} )If this ratio is positive, both are increasing or decreasing together. If negative, one is increasing while the other is decreasing.But perhaps more straightforward: to find when ( V_{Opp} ) increases, we need ( c V_{Undecided} > d V_{Opp} V_{DLP} ).So, conditions under which this holds.Given that ( V_{Undecided} = P - V_{DLP} - V_{Opp} ), perhaps when the undecided pool is large enough relative to the product ( V_{Opp} V_{DLP} ).Alternatively, maybe consider the initial phase when ( V_{DLP} ) and ( V_{Opp} ) are small, so ( V_{Undecided} ) is close to P. Then, ( c V_{Undecided} ) is significant, while ( d V_{Opp} V_{DLP} ) is small. So, initially, both parties might gain support.But as they gain support, ( V_{Undecided} ) decreases, and the interaction terms become more significant.So, perhaps if ( c > d V_{Opp} V_{DLP} / V_{Undecided} ), then ( V_{Opp} ) increases.But without specific values, it's hard to say.Alternatively, maybe consider the system as a competition between two species with mutual interference terms.In ecology, similar models have been studied. The stability and growth depend on the parameters.But perhaps, to find when ( V_{Opp} ) can grow, we need that the inflow from undecided is greater than the outflow due to interaction with DLP.So, ( c V_{Undecided} > d V_{Opp} V_{DLP} ).Given that ( V_{Undecided} = P - V_{DLP} - V_{Opp} ), maybe we can write:( c (P - V_{DLP} - V_{Opp}) > d V_{Opp} V_{DLP} )Which can be rearranged as:( c P > c V_{DLP} + c V_{Opp} + d V_{Opp} V_{DLP} )Not sure if that helps.Alternatively, maybe consider the ratio ( frac{c}{d} > frac{V_{Opp} V_{DLP}}{V_{Undecided}} ).So, if ( frac{c}{d} ) is large enough, then ( V_{Opp} ) can increase even if ( V_{DLP} ) is large.Alternatively, think about the initial conditions. If initially, ( V_{Opp0} ) is small, then ( V_{Opp} ) can increase if ( c V_{Undecided0} > d V_{Opp0} V_{DLP0} ).Since ( V_{Undecided0} = P - V_{DLP0} - V_{Opp0} ), which is likely large if ( V_{DLP0} ) and ( V_{Opp0} ) are small.So, in the beginning, ( V_{Opp} ) can increase if ( c ) is sufficiently large relative to ( d ).But as ( V_{Opp} ) and ( V_{DLP} ) grow, the term ( d V_{Opp} V_{DLP} ) can become significant, potentially causing ( V_{Opp} ) to stop growing or even decrease.So, the opposition's support increases when the rate at which they convert undecided voters ( c V_{Undecided} ) is greater than the rate at which they lose support to DLP ( d V_{Opp} V_{DLP} ).Therefore, conditions for ( V_{Opp} ) to increase over time are when ( c V_{Undecided} > d V_{Opp} V_{DLP} ), which depends on the parameters ( c ) and ( d ), as well as the current values of ( V_{Undecided} ), ( V_{Opp} ), and ( V_{DLP} ).Moving on to part 2: Express ( V_{Undecided}(t) ) in terms of ( V_{DLP} ) and ( V_{Opp} ), and find the general solution assuming initial conditions.Given that ( P(t) = P ) is constant, ( V_{Undecided}(t) = P - V_{DLP}(t) - V_{Opp}(t) ).So, substitute ( V_{Undecided} = P - x - y ) into the system:[begin{cases}frac{dx}{dt} = a (P - x - y) - b x y frac{dy}{dt} = c (P - x - y) - d x y frac{dz}{dt} = - (a + c) zend{cases}]But since ( z = P - x - y ), we can write the system in terms of x and y only.Let me write the first two equations:1. ( frac{dx}{dt} = a (P - x - y) - b x y )2. ( frac{dy}{dt} = c (P - x - y) - d x y )This is a system of two nonlinear ODEs. Solving such systems analytically is generally difficult, but maybe we can find some relation between x and y.Let me subtract the two equations:( frac{dx}{dt} - frac{dy}{dt} = (a - c)(P - x - y) - (b - d) x y )Let me denote ( w = x - y ). Then, ( frac{dw}{dt} = frac{dx}{dt} - frac{dy}{dt} = (a - c)(P - x - y) - (b - d) x y )But ( P - x - y = z ), so:( frac{dw}{dt} = (a - c) z - (b - d) x y )But ( z = P - x - y ), so:( frac{dw}{dt} = (a - c)(P - x - y) - (b - d) x y )Not sure if that helps.Alternatively, maybe consider the ratio ( frac{dy}{dx} ).From the two equations:( frac{dy}{dt} = c (P - x - y) - d x y )( frac{dx}{dt} = a (P - x - y) - b x y )So,( frac{dy}{dx} = frac{c (P - x - y) - d x y}{a (P - x - y) - b x y} )This is a first-order ODE in terms of y and x. It might be separable or have an integrating factor.Let me denote ( u = P - x - y ). Then, ( u = V_{Undecided} ).So, ( frac{du}{dt} = - frac{dx}{dt} - frac{dy}{dt} = - [a u - b x y] - [c u - d x y] = - a u + b x y - c u + d x y = - (a + c) u + (b + d) x y )But from the original equation, ( frac{du}{dt} = - (a + c) u ). Wait, that's a contradiction unless ( (b + d) x y = 0 ). But that's only when either x or y is zero, which is not generally the case.Wait, no, actually, from the original system, ( frac{dz}{dt} = - (a + c) z ), which is ( frac{du}{dt} = - (a + c) u ). So, that equation is separate from the others.So, we can solve ( frac{du}{dt} = - (a + c) u ) independently.The solution is:( u(t) = u(0) e^{ - (a + c) t } )Where ( u(0) = P - V_{DLP0} - V_{Opp0} ).So, ( V_{Undecided}(t) = (P - V_{DLP0} - V_{Opp0}) e^{ - (a + c) t } )That's useful. So, the undecided voters decay exponentially.Now, knowing that, we can substitute ( u(t) ) back into the equations for ( frac{dx}{dt} ) and ( frac{dy}{dt} ).So, ( frac{dx}{dt} = a u(t) - b x y )( frac{dy}{dt} = c u(t) - d x y )But since ( u(t) ) is known, we can write:( frac{dx}{dt} + b x y = a u(t) )( frac{dy}{dt} + d x y = c u(t) )This is a system of linear ODEs with variable coefficients because ( u(t) ) is known but x and y are functions of t.But solving this system is still non-trivial because of the ( x y ) terms.Alternatively, maybe consider dividing the two equations:( frac{frac{dx}{dt} + b x y}{frac{dy}{dt} + d x y} = frac{a u(t)}{c u(t)} = frac{a}{c} )So,( frac{dx}{dt} + b x y = frac{a}{c} (frac{dy}{dt} + d x y) )Simplify:( frac{dx}{dt} + b x y = frac{a}{c} frac{dy}{dt} + frac{a d}{c} x y )Rearrange:( frac{dx}{dt} - frac{a}{c} frac{dy}{dt} = left( frac{a d}{c} - b right) x y )Hmm, not sure if that helps.Alternatively, maybe express ( frac{dx}{dt} ) and ( frac{dy}{dt} ) in terms of u(t):From ( frac{dx}{dt} = a u(t) - b x y )From ( frac{dy}{dt} = c u(t) - d x y )Let me denote ( frac{dx}{dt} = a u - b x y ) as equation (1)and ( frac{dy}{dt} = c u - d x y ) as equation (2)Let me solve equation (1) for ( x y ):( x y = frac{a u - frac{dx}{dt}}{b} )Similarly, from equation (2):( x y = frac{c u - frac{dy}{dt}}{d} )Set them equal:( frac{a u - frac{dx}{dt}}{b} = frac{c u - frac{dy}{dt}}{d} )Multiply both sides by b d:( d (a u - frac{dx}{dt}) = b (c u - frac{dy}{dt}) )Expand:( a d u - d frac{dx}{dt} = b c u - b frac{dy}{dt} )Rearrange:( - d frac{dx}{dt} + b frac{dy}{dt} = (b c - a d) u )But from earlier, we have:( frac{dx}{dt} - frac{a}{c} frac{dy}{dt} = left( frac{a d}{c} - b right) x y )Not sure if combining these helps.Alternatively, maybe consider the ratio ( frac{dy}{dx} ) again.From equation (1) and (2):( frac{dy}{dx} = frac{c u - d x y}{a u - b x y} )Let me denote ( v = y / x ). Then, ( y = v x ), and ( frac{dy}{dx} = v + x frac{dv}{dx} )Substitute into the ratio:( v + x frac{dv}{dx} = frac{c u - d x (v x)}{a u - b x (v x)} = frac{c u - d v x^2}{a u - b v x^2} )But this seems complicated because u is a function of t, which is related to x and y.Alternatively, maybe consider that u(t) is known, so express everything in terms of t.Given that u(t) = (P - V_{DLP0} - V_{Opp0}) e^{ - (a + c) t }, we can write:( frac{dx}{dt} = a (P - V_{DLP0} - V_{Opp0}) e^{ - (a + c) t } - b x y )( frac{dy}{dt} = c (P - V_{DLP0} - V_{Opp0}) e^{ - (a + c) t } - d x y )This is a system of nonlinear ODEs with forcing terms decaying exponentially. Solving this analytically might not be straightforward, but perhaps we can look for an integrating factor or some substitution.Alternatively, consider that as t increases, u(t) approaches zero, so the system approaches:( frac{dx}{dt} = - b x y )( frac{dy}{dt} = - d x y )Which is a simpler system. The solution to this would be:From ( frac{dx}{dt} = - b x y ), and ( frac{dy}{dt} = - d x y )Divide the two equations:( frac{dx}{dy} = frac{b}{d} )So, ( x = frac{b}{d} y + C )But with initial conditions, maybe we can find C.But this is only for the limit as t approaches infinity, which might not help for finite t.Alternatively, perhaps consider the system when u(t) is small, but that might not be helpful either.Given the complexity, maybe the general solution isn't expressible in closed form, and we have to rely on qualitative analysis or numerical methods.But the problem asks for the general solution assuming initial conditions. Maybe it's expecting us to express the solution in terms of integrals or something.Alternatively, perhaps consider that the system can be decoupled by some substitution.Let me try adding the two equations:( frac{dx}{dt} + frac{dy}{dt} = (a + c) u(t) - (b + d) x y )But since ( u = P - x - y ), we have:( frac{d}{dt}(x + y) = (a + c)(P - x - y) - (b + d) x y )Let me denote ( s = x + y ). Then,( frac{ds}{dt} = (a + c)(P - s) - (b + d) x y )But we still have the term ( x y ), which complicates things.Alternatively, maybe consider the product ( x y ). Let me denote ( p = x y ). Then,( frac{dp}{dt} = x frac{dy}{dt} + y frac{dx}{dt} )From the original equations:( frac{dx}{dt} = a u - b p )( frac{dy}{dt} = c u - d p )So,( frac{dp}{dt} = x (c u - d p) + y (a u - b p) = (c x + a y) u - (d x + b y) p )But ( x + y = s ), so ( c x + a y = c (s - y) + a y = c s + (a - c) y )Not sure if that helps.Alternatively, maybe express in terms of s and p.But this seems getting too involved. Maybe the problem expects us to recognize that the system is complex and perhaps only the undecided voters can be solved explicitly, while the other variables require more involved methods.Given that, perhaps the general solution isn't expressible in a simple closed form, and we can only provide the expression for ( V_{Undecided}(t) ) and note that ( V_{DLP}(t) ) and ( V_{Opp}(t) ) satisfy the given ODEs with the known u(t).Alternatively, maybe consider that the system can be transformed into a Bernoulli equation or something similar.Wait, let me think differently. Since u(t) is known, maybe we can write the equations for x and y as:( frac{dx}{dt} + b x y = a u(t) )( frac{dy}{dt} + d x y = c u(t) )This is a system of linear ODEs with variable coefficients (because of the x y term). But I don't think standard linear ODE techniques apply here because of the nonlinearity.Alternatively, maybe consider a substitution where we let ( x y = q ). Then,From the first equation: ( frac{dx}{dt} = a u - b q )From the second equation: ( frac{dy}{dt} = c u - d q )Also, ( frac{dq}{dt} = x frac{dy}{dt} + y frac{dx}{dt} = x (c u - d q) + y (a u - b q) = (c x + a y) u - (d x + b y) q )But ( x + y = s ), so ( c x + a y = c (s - y) + a y = c s + (a - c) y ). Not helpful.Alternatively, express ( c x + a y = c x + a (s - x) = (c - a) x + a s ). Still not helpful.I think I'm stuck here. Maybe the problem expects us to recognize that the system is complex and only provide the expression for ( V_{Undecided}(t) ), and note that ( V_{DLP} ) and ( V_{Opp} ) satisfy the given ODEs with the known u(t), but their solutions aren't expressible in closed form without further assumptions.Alternatively, maybe consider that if ( b = d ), the system simplifies. Let me check.If ( b = d ), then the equations become:( frac{dx}{dt} = a u - b x y )( frac{dy}{dt} = c u - b x y )Subtracting them:( frac{dx}{dt} - frac{dy}{dt} = (a - c) u )Integrate both sides:( x(t) - y(t) = (a - c) int_{0}^{t} u(tau) dtau + C )Given initial conditions ( x(0) = V_{DLP0} ), ( y(0) = V_{Opp0} ), so ( C = V_{DLP0} - V_{Opp0} ).Thus,( x(t) - y(t) = (a - c) int_{0}^{t} u(tau) dtau + V_{DLP0} - V_{Opp0} )But ( u(tau) = (P - V_{DLP0} - V_{Opp0}) e^{ - (a + c) tau } ), so the integral is:( int_{0}^{t} u(tau) dtau = frac{P - V_{DLP0} - V_{Opp0}}{a + c} (1 - e^{ - (a + c) t }) )So,( x(t) - y(t) = (a - c) frac{P - V_{DLP0} - V_{Opp0}}{a + c} (1 - e^{ - (a + c) t }) + V_{DLP0} - V_{Opp0} )This gives a relation between x and y. Then, we can use one of the original equations to solve for x or y.For example, from ( frac{dx}{dt} = a u - b x y ), and knowing that ( y = x - [ (a - c) frac{P - V_{DLP0} - V_{Opp0}}{a + c} (1 - e^{ - (a + c) t }) + V_{DLP0} - V_{Opp0} ] ), we can substitute y in terms of x and solve.But this still leads to a nonlinear ODE for x(t), which might not have a closed-form solution.So, unless there's a specific substitution or assumption, I think the general solution for ( V_{DLP}(t) ) and ( V_{Opp}(t) ) isn't expressible in a simple closed form. Therefore, the answer might involve expressing ( V_{Undecided}(t) ) explicitly and noting that ( V_{DLP} ) and ( V_{Opp} ) satisfy the given ODEs with the known u(t).In summary:1. The fixed points are all unstable saddle points, suggesting no stable equilibrium where one party dominates. The opposition's support increases when ( c V_{Undecided} > d V_{Opp} V_{DLP} ).2. ( V_{Undecided}(t) = (P - V_{DLP0} - V_{Opp0}) e^{ - (a + c) t } ). The general solution for ( V_{DLP}(t) ) and ( V_{Opp}(t) ) requires solving a system of nonlinear ODEs, which may not have a closed-form solution without additional assumptions.</think>