🧠:<think>First, I need to determine the maximum radius of the circular fountain that John can install in the center of his 50 feet by 30 feet garden while adhering to the local building code requirement of being at least 5 feet away from the property boundaries.Since the fountain is circular and centered, the radius is limited by the smallest distance from the center to any boundary. The garden is 50 feet long and 30 feet wide. The center of the garden is at (25 feet, 15 feet). The distance from the center to the nearest boundary on the length side is 25 feet, and on the width side, it's 15 feet. To comply with the 5 feet setback, the maximum radius cannot exceed 15 feet minus 5 feet, which equals 10 feet.Next, for the square gazebo, John wants it to be as large as possible while maintaining a 5 feet distance from all property boundaries and keeping its sides parallel to the garden edges. The garden measures 50 feet by 30 feet. To ensure the 5 feet setback, the available space for the gazebo is reduced by 10 feet on both the length and the width (5 feet on each side). This results in a usable area of 40 feet by 20 feet. Since the gazebo must be square, the maximum side length is determined by the smaller dimension of the usable area, which is 20 feet.</think>

🧠:<think>First, I need to determine the maximum number of devices that can be repaired in one day by the current team. There are 5 expert technicians, each repairing an average of 10 devices per day, and 10 junior technicians, each repairing an average of 3 devices per day.Calculating the total devices repaired by experts in one day:5 experts * 10 devices/expert = 50 devices.Calculating the total devices repaired by juniors in one day:10 juniors * 3 devices/junior = 30 devices.Adding these together gives the total devices repaired per day:50 devices + 30 devices = 80 devices per day.Next, to determine if the team can meet the goal of repairing all 150 devices within 7 days, I'll multiply the daily capacity by 7:80 devices/day * 7 days = 560 devices.Since 560 devices exceed the required 150 devices, the team can easily meet the goal within the week.Now, if one expert technician falls ill for 3 days, the number of available expert technicians during those days reduces to 4. Recalculating the daily repair capacity:4 experts * 10 devices/expert = 40 devices.10 juniors * 3 devices/junior = 30 devices.Total devices repaired per day during the illness: 40 + 30 = 70 devices.Over the 3 days, the team can repair:70 devices/day * 3 days = 210 devices.For the remaining 4 days, with all 5 experts available:80 devices/day * 4 days = 320 devices.Total devices repaired in the week: 210 + 320 = 530 devices, which still exceeds the 150 devices needed. Therefore, the team can still meet the goal without needing additional repairs per day.</think>

🧠:<think>Okay, so I have this problem about a soccer defender improving their performance through a fitness regimen. There are two parts: one involving a differential equation for speed and another for endurance. Let me try to tackle them one by one.Starting with part 1: The defender's speed S(t) is given by the differential equation dS/dt = kS(10 - S) - aE(t), where E(t) is 3 + 0.5t. The initial condition is S(0) = 5 m/s. I need to solve this differential equation for S(t).Hmm, so this is a first-order nonlinear ordinary differential equation because of the S(10 - S) term. It looks like a logistic growth model but with an additional term subtracted, which is aE(t). Since E(t) is given explicitly as 3 + 0.5t, I can substitute that into the equation.So, substituting E(t), the equation becomes:dS/dt = kS(10 - S) - a(3 + 0.5t)This is a Riccati equation because it's of the form dS/dt = P(t) + Q(t)S + R(t)S^2. In this case, P(t) = -a(3 + 0.5t), Q(t) = 10k, and R(t) = -k.Riccati equations are generally difficult to solve unless we can find a particular solution. Maybe I can assume a particular solution of the form S_p(t) = At + B, where A and B are constants to be determined. Let's try that.So, if S_p(t) = At + B, then dS_p/dt = A.Substituting into the differential equation:A = k(At + B)(10 - At - B) - a(3 + 0.5t)Let me expand the right-hand side:First, compute (At + B)(10 - At - B):= At*10 - (At)^2 - At*B + B*10 - B*At - B^2= 10At - A^2t^2 - ABt + 10B - ABt - B^2Combine like terms:= -A^2t^2 + (10A - 2AB)t + (10B - B^2)Multiply by k:= -kA^2t^2 + k(10A - 2AB)t + k(10B - B^2)Now subtract a(3 + 0.5t):= -kA^2t^2 + k(10A - 2AB)t + k(10B - B^2) - 3a - 0.5a tSo, putting it all together, the equation is:A = -kA^2t^2 + [k(10A - 2AB) - 0.5a]t + [k(10B - B^2) - 3a]Since the left-hand side is a constant A, the coefficients of t^2, t, and the constant term on the right must all be zero except for the constant term, which should equal A.So, we can set up the following equations:1. Coefficient of t^2: -kA^2 = 02. Coefficient of t: k(10A - 2AB) - 0.5a = 03. Constant term: k(10B - B^2) - 3a = AFrom equation 1: -kA^2 = 0. Since k is a constant and presumably non-zero (otherwise the equation would be linear), this implies A^2 = 0, so A = 0.Plugging A = 0 into equation 2: k(10*0 - 2*0*B) - 0.5a = 0 => -0.5a = 0 => a = 0.But a is a constant given in the problem, so unless a is zero, this approach might not work. Hmm, maybe assuming a linear particular solution isn't the right way to go.Alternatively, perhaps I can use an integrating factor or another substitution. Let me think.The equation is dS/dt + ( -k(10 - S) )S = -a(3 + 0.5t)Wait, actually, it's a Bernoulli equation because of the S^2 term. Bernoulli equations can be linearized by substituting u = 1/S.Let me try that substitution.Let u = 1/S, then du/dt = -1/S^2 dS/dt.So, rearranging the original equation:dS/dt = kS(10 - S) - a(3 + 0.5t)Multiply both sides by -1/S^2:-1/S^2 dS/dt = -k(10 - S)/S + a(3 + 0.5t)/S^2But du/dt = -1/S^2 dS/dt, so:du/dt = -k(10 - S)/S + a(3 + 0.5t)/S^2But S = 1/u, so substitute that in:du/dt = -k(10 - 1/u)/(1/u) + a(3 + 0.5t)/(1/u)^2Simplify each term:First term: -k(10 - 1/u)/(1/u) = -k(10u - 1)Second term: a(3 + 0.5t)/(1/u)^2 = a(3 + 0.5t)u^2So, putting it together:du/dt = -k(10u - 1) + a(3 + 0.5t)u^2Hmm, this seems more complicated. Maybe this substitution isn't helpful. Let me think of another approach.Alternatively, perhaps I can write the equation as:dS/dt + (a(3 + 0.5t) - kS(10 - S)) = 0But I don't see an obvious integrating factor here. Maybe it's better to consider numerical methods, but since this is a problem-solving question, perhaps I need to find an analytical solution.Wait, maybe I can rearrange the equation:dS/dt = kS(10 - S) - a(3 + 0.5t)Let me write it as:dS/dt + a(3 + 0.5t) = kS(10 - S)This is a Bernoulli equation in terms of S. Let me write it in standard Bernoulli form:dS/dt + P(t)S = Q(t)S^nIn this case, rearranging:dS/dt - k(10 - S)S = -a(3 + 0.5t)Wait, that's not the standard Bernoulli form. Let me see:Actually, the standard Bernoulli equation is dS/dt + P(t)S = Q(t)S^n. Comparing, we have:dS/dt - k(10 - S)S = -a(3 + 0.5t)Which can be written as:dS/dt + (-k(10 - S))S = -a(3 + 0.5t)But that's not quite the standard form because the coefficient of S is not just a function of t, but also depends on S. Hmm, maybe that's not helpful.Alternatively, perhaps I can consider this as a Riccati equation and see if I can find a particular solution. But earlier, assuming a linear particular solution led to a contradiction unless a=0, which isn't necessarily the case.Alternatively, maybe I can use an integrating factor if I can manipulate the equation into a linear form. Let me try rearranging:dS/dt = kS(10 - S) - a(3 + 0.5t)Let me expand the right-hand side:= 10kS - kS^2 - 3a - 0.5a tSo, the equation is:dS/dt + kS^2 - 10kS + 3a + 0.5a t = 0This is a quadratic in S, making it a Riccati equation. As such, it's generally difficult to solve without a known particular solution.Perhaps I can look for a particular solution of the form S_p(t) = mt + n, a linear function. Let's try that.Let S_p(t) = mt + n, then dS_p/dt = m.Substitute into the equation:m = k(mt + n)(10 - mt - n) - a(3 + 0.5t)Expand the right-hand side:First, compute (mt + n)(10 - mt - n):= mt*10 - (mt)^2 - mt*n + n*10 - n*mt - n^2= 10mt - m^2t^2 - mnt + 10n - mnt - n^2Combine like terms:= -m^2t^2 + (10m - 2mn)t + (10n - n^2)Multiply by k:= -k m^2 t^2 + k(10m - 2mn)t + k(10n - n^2)Subtract a(3 + 0.5t):= -k m^2 t^2 + k(10m - 2mn)t + k(10n - n^2) - 3a - 0.5a tSo, the equation becomes:m = -k m^2 t^2 + [k(10m - 2mn) - 0.5a] t + [k(10n - n^2) - 3a]Since the left-hand side is a constant m, the coefficients of t^2, t, and the constant term on the right must satisfy:1. Coefficient of t^2: -k m^2 = 02. Coefficient of t: k(10m - 2mn) - 0.5a = 03. Constant term: k(10n - n^2) - 3a = mFrom equation 1: -k m^2 = 0. Since k ≠ 0, this implies m^2 = 0, so m = 0.Plugging m = 0 into equation 2: k(0 - 0) - 0.5a = 0 => -0.5a = 0 => a = 0.But a is a constant given in the problem, so unless a = 0, this approach doesn't work. Therefore, assuming a linear particular solution only works if a = 0, which isn't necessarily the case here.Hmm, maybe I need to try a different approach. Perhaps I can consider this as a nonhomogeneous logistic equation and use an integrating factor or variation of parameters.Alternatively, maybe I can use substitution to make it linear. Let me try dividing both sides by S^2:dS/dt / S^2 = k(10 - S)/S - a(3 + 0.5t)/S^2Let me set u = 1/S, then du/dt = -1/S^2 dS/dtSo, -du/dt = k(10 - S)/S - a(3 + 0.5t)/S^2Substitute S = 1/u:- du/dt = k(10 - 1/u)/(1/u) - a(3 + 0.5t)u^2Simplify:- du/dt = k(10u - 1) - a(3 + 0.5t)u^2Multiply both sides by -1:du/dt = -k(10u - 1) + a(3 + 0.5t)u^2This still looks complicated, but maybe I can rearrange it:du/dt + k(10u - 1) = a(3 + 0.5t)u^2This is a Bernoulli equation in terms of u, with n = 2. The standard form is du/dt + P(t)u = Q(t)u^n.Here, P(t) = 10k, Q(t) = a(3 + 0.5t), and n = 2.To solve this, we can use the substitution v = u^{1 - n} = u^{-1}.So, let v = 1/u, then dv/dt = -1/u^2 du/dt.From the equation:du/dt = a(3 + 0.5t)u^2 - k(10u - 1)Multiply both sides by -1/u^2:-1/u^2 du/dt = -a(3 + 0.5t) + k(10u - 1)/u^2But dv/dt = -1/u^2 du/dt, so:dv/dt = -a(3 + 0.5t) + k(10u - 1)/u^2But u = 1/v, so substitute:= -a(3 + 0.5t) + k(10*(1/v) - 1)/(1/v)^2Simplify each term:First term: -a(3 + 0.5t)Second term: k(10/v - 1)/(1/v^2) = k(10/v - 1) * v^2 = k(10v - v^2)So, putting it together:dv/dt = -a(3 + 0.5t) + k(10v - v^2)This is a linear differential equation in terms of v:dv/dt - k(10v - v^2) = -a(3 + 0.5t)Wait, no, actually, it's still nonlinear because of the v^2 term. Hmm, maybe I made a miscalculation.Wait, let's go back to the substitution step. After substitution, we had:dv/dt = -a(3 + 0.5t) + k(10v - v^2)Wait, that's still nonlinear because of the v^2 term. So, this substitution didn't linearize the equation. Maybe I need to try a different substitution or approach.Alternatively, perhaps I can consider this as a Riccati equation and look for a particular solution. But without knowing a particular solution, it's difficult.Wait, maybe I can assume that the particular solution is of the form v_p(t) = pt + q, a linear function. Let's try that.Let v_p(t) = pt + q, then dv_p/dt = p.Substitute into the equation:p = -a(3 + 0.5t) + k(10(pt + q) - (pt + q)^2)Expand the right-hand side:= -3a - 0.5a t + k(10pt + 10q - p^2t^2 - 2pqt - q^2)So, grouping terms:= -3a - 0.5a t + 10k p t + 10k q - k p^2 t^2 - 2k p q t - k q^2Now, equate coefficients of like powers of t:Coefficient of t^2: -k p^2 = 0 => p^2 = 0 => p = 0Coefficient of t: -0.5a + 10k p - 2k p q = 0But p = 0, so this becomes -0.5a = 0 => a = 0, which again is a problem unless a = 0.So, this approach also leads to a contradiction unless a = 0. Therefore, perhaps assuming a linear particular solution isn't working.Hmm, maybe I need to consider that this equation doesn't have an analytical solution and instead use numerical methods. But since this is a problem-solving question, perhaps I need to find an integrating factor or another substitution.Wait, going back to the original equation:dS/dt = kS(10 - S) - a(3 + 0.5t)Let me rearrange it as:dS/dt + a(3 + 0.5t) = kS(10 - S)This is a Bernoulli equation with n = 2. Let me write it in standard form:dS/dt + P(t)S = Q(t)S^nHere, P(t) = -k(10 - S), but that still depends on S, so it's not helpful.Wait, perhaps I can write it as:dS/dt = kS(10 - S) - a(3 + 0.5t)Let me divide both sides by S(10 - S):dS/dt / [S(10 - S)] = k - a(3 + 0.5t)/[S(10 - S)]This might not help directly, but perhaps I can use partial fractions on the left-hand side.Let me set:1/[S(10 - S)] = A/S + B/(10 - S)Solving for A and B:1 = A(10 - S) + B SLet S = 0: 1 = 10A => A = 1/10Let S = 10: 1 = 10B => B = 1/10So, 1/[S(10 - S)] = (1/10)(1/S + 1/(10 - S))Therefore, the equation becomes:(1/10)(1/S + 1/(10 - S)) dS/dt = k - a(3 + 0.5t)/[S(10 - S)]But this seems more complicated. Maybe integrating both sides with respect to t.Wait, let me write it as:(1/10)(1/S + 1/(10 - S)) dS = [k - a(3 + 0.5t)/[S(10 - S)]] dtBut integrating the right-hand side is still difficult because of the S terms.Alternatively, perhaps I can consider this as a separable equation. Let me try rearranging:dS / [kS(10 - S) - a(3 + 0.5t)] = dtBut this is not separable because S is a function of t, so we can't separate variables directly.Hmm, this is getting complicated. Maybe I need to consider that without specific values for k and a, it's difficult to solve analytically. But the problem doesn't provide specific values for k and a, so perhaps the solution is left in terms of integrals or expressed implicitly.Alternatively, maybe I can write the solution using an integrating factor after some manipulation.Wait, let me consider the equation again:dS/dt = kS(10 - S) - a(3 + 0.5t)This is a Riccati equation, and generally, Riccati equations don't have closed-form solutions unless a particular solution is known. Since I couldn't find a particular solution earlier, maybe I need to express the solution in terms of integrals.Alternatively, perhaps I can use the substitution z = S - 5, since the initial condition is S(0) = 5. Maybe this centers the equation around the initial value.Let z = S - 5, so S = z + 5, and dS/dt = dz/dt.Substitute into the equation:dz/dt = k(z + 5)(10 - (z + 5)) - a(3 + 0.5t)Simplify the terms inside the parentheses:10 - z - 5 = 5 - zSo, the equation becomes:dz/dt = k(z + 5)(5 - z) - a(3 + 0.5t)Multiply out (z + 5)(5 - z):= (5 + z)(5 - z) = 25 - z^2So, dz/dt = k(25 - z^2) - a(3 + 0.5t)This is still a Riccati equation but perhaps in a simpler form. Let me write it as:dz/dt + a(3 + 0.5t) = k(25 - z^2)This is a Bernoulli equation with n = 2. Let me write it in standard form:dz/dt + P(t)z = Q(t)z^nHere, P(t) = 0, Q(t) = -k, and n = 2. Wait, no, because the equation is:dz/dt = -k z^2 + 25k - a(3 + 0.5t)So, rearranged:dz/dt + k z^2 = 25k - a(3 + 0.5t)This is a Riccati equation, but again, without a particular solution, it's difficult to solve.Alternatively, perhaps I can write it as:dz/dt = -k z^2 + (25k - 3a - 0.5a t)This is a Riccati equation of the form dz/dt = A z^2 + B(t) z + C(t), where A = -k, B(t) = 0, and C(t) = 25k - 3a - 0.5a t.Since B(t) = 0, it's a special case. The general solution for Riccati equations when B(t) = 0 is:z(t) = sqrt(C(t)/A) * tan( sqrt(C(t)/A) * (t - t0) + phi )But this only works if C(t) is a constant, which it isn't here because C(t) = 25k - 3a - 0.5a t is linear in t. Therefore, this approach doesn't apply.Hmm, perhaps I need to consider that this equation doesn't have an elementary closed-form solution and instead express the solution in terms of integrals or use a series expansion. But since this is a problem-solving question, maybe I'm missing a trick.Wait, let me go back to the original equation:dS/dt = kS(10 - S) - a(3 + 0.5t)Let me consider this as a nonhomogeneous logistic equation. The homogeneous part is dS/dt = kS(10 - S), which has the solution S(t) = 10 / (1 + 9 e^{-10kt}), given S(0) = 5.But the nonhomogeneous term is -a(3 + 0.5t), which complicates things. Maybe I can use variation of parameters or another method.Alternatively, perhaps I can write the equation as:dS/dt + a(3 + 0.5t) = kS(10 - S)Let me consider this as a Bernoulli equation and use the substitution u = 1/S.Wait, I tried that earlier, but it led to a more complicated equation. Maybe I need to proceed differently.Alternatively, perhaps I can write the equation in terms of t as the independent variable and integrate both sides. Let me try separating variables.But the equation is:dS/dt = kS(10 - S) - a(3 + 0.5t)This isn't separable because S is a function of t, so we can't directly separate variables. Therefore, perhaps I need to use an integrating factor or another technique.Wait, maybe I can write this as:dS/dt + a(3 + 0.5t) = kS(10 - S)Let me rearrange it as:dS/dt - kS(10 - S) = -a(3 + 0.5t)This is a Bernoulli equation with n = 2. Let me use the substitution u = S^{1 - 2} = 1/S.Then, du/dt = -1/S^2 dS/dtSo, from the equation:dS/dt = kS(10 - S) - a(3 + 0.5t)Multiply both sides by -1/S^2:-1/S^2 dS/dt = -k(10 - S)/S + a(3 + 0.5t)/S^2But du/dt = -1/S^2 dS/dt, so:du/dt = -k(10 - S)/S + a(3 + 0.5t)/S^2But S = 1/u, so substitute:du/dt = -k(10 - 1/u)/(1/u) + a(3 + 0.5t)u^2Simplify each term:First term: -k(10 - 1/u)/(1/u) = -k(10u - 1)Second term: a(3 + 0.5t)u^2So, the equation becomes:du/dt = -k(10u - 1) + a(3 + 0.5t)u^2This is still a Riccati equation, which is nonlinear. Therefore, without a particular solution, it's difficult to proceed.At this point, I think it's best to consider that this differential equation doesn't have a closed-form solution in terms of elementary functions, and the solution would need to be expressed implicitly or found numerically. However, since the problem asks to solve the differential equation, perhaps I need to present the solution in terms of integrals or another form.Alternatively, maybe I can write the solution using the integrating factor method for Bernoulli equations. Let me recall that for a Bernoulli equation du/dt + P(t)u = Q(t)u^n, the substitution v = u^{1 - n} linearizes the equation.In our case, after substitution, we have:du/dt = -k(10u - 1) + a(3 + 0.5t)u^2Let me write this as:du/dt + k(10u - 1) = a(3 + 0.5t)u^2This is a Bernoulli equation with n = 2, so let me set v = u^{-1}.Then, dv/dt = -u^{-2} du/dtFrom the equation:du/dt = a(3 + 0.5t)u^2 - k(10u - 1)Multiply both sides by -u^{-2}:- u^{-2} du/dt = -a(3 + 0.5t) + k(10u - 1)u^{-2}But dv/dt = -u^{-2} du/dt, so:dv/dt = -a(3 + 0.5t) + k(10u - 1)u^{-2}But u = 1/v, so substitute:= -a(3 + 0.5t) + k(10*(1/v) - 1) * v^2Simplify:= -a(3 + 0.5t) + k(10v - v^2)So, the equation becomes:dv/dt = -a(3 + 0.5t) + 10k v - k v^2This is still a Riccati equation, which is nonlinear. Therefore, without a particular solution, it's difficult to proceed.Given that I'm stuck here, perhaps I need to accept that the solution can't be expressed in a simple closed form and instead present it as an implicit solution or leave it in terms of integrals.Alternatively, maybe I can write the solution using the method of integrating factors for the Bernoulli equation. Let me try that.For the Bernoulli equation:dv/dt + P(t)v = Q(t)v^nHere, after substitution, we have:dv/dt - 10k v + k v^2 = -a(3 + 0.5t)Wait, that's still nonlinear. I think I'm going in circles here.Perhaps I need to consider that without specific values for k and a, the solution remains in terms of integrals. Alternatively, maybe the problem expects a qualitative analysis rather than an explicit solution.But the problem says "solve the differential equation for S(t)", so perhaps I need to proceed differently.Wait, maybe I can write the equation as:dS/dt + a(3 + 0.5t) = kS(10 - S)Let me consider this as a nonhomogeneous equation and use the method of variation of parameters. First, solve the homogeneous equation:dS/dt = kS(10 - S)This is the logistic equation, whose solution is:S(t) = 10 / (1 + 9 e^{-10kt})Given S(0) = 5, which fits because 10 / (1 + 9) = 5.Now, for the nonhomogeneous part, we can use variation of parameters. Let me denote the homogeneous solution as S_h(t) = 10 / (1 + 9 e^{-10kt}).Then, the particular solution S_p(t) can be found by assuming S_p(t) = S_h(t) * u(t), where u(t) is a function to be determined.So, S_p(t) = [10 / (1 + 9 e^{-10kt})] * u(t)Compute dS_p/dt:dS_p/dt = [d/dt (10 / (1 + 9 e^{-10kt}))] * u(t) + [10 / (1 + 9 e^{-10kt})] * du/dtBut d/dt (10 / (1 + 9 e^{-10kt})) = k * 10 * 9 e^{-10kt} / (1 + 9 e^{-10kt})^2 = k * 90 e^{-10kt} / (1 + 9 e^{-10kt})^2But 10 / (1 + 9 e^{-10kt}) = S_h(t), so:d/dt S_h(t) = k S_h(t) (10 - S_h(t))Which is the homogeneous equation.Therefore, dS_p/dt = k S_h(t) (10 - S_h(t)) u(t) + S_h(t) du/dtSubstitute S_p(t) into the original equation:dS_p/dt = k S_p(t) (10 - S_p(t)) - a(3 + 0.5t)Substitute S_p(t) and dS_p/dt:k S_h(t) (10 - S_h(t)) u(t) + S_h(t) du/dt = k [S_h(t) u(t)] (10 - S_h(t) u(t)) - a(3 + 0.5t)This looks messy, but let's try to simplify.First, expand the right-hand side:= k S_h(t) u(t) (10 - S_h(t) u(t)) - a(3 + 0.5t)= 10k S_h(t) u(t) - k S_h(t)^2 u(t)^2 - a(3 + 0.5t)Now, equate to the left-hand side:k S_h(t) (10 - S_h(t)) u(t) + S_h(t) du/dt = 10k S_h(t) u(t) - k S_h(t)^2 u(t)^2 - a(3 + 0.5t)Subtract 10k S_h(t) u(t) from both sides:k S_h(t) (10 - S_h(t)) u(t) - 10k S_h(t) u(t) + S_h(t) du/dt = - k S_h(t)^2 u(t)^2 - a(3 + 0.5t)Simplify the left-hand side:k S_h(t) (10 - S_h(t) - 10) u(t) + S_h(t) du/dt = - k S_h(t)^2 u(t)^2 - a(3 + 0.5t)Which simplifies to:k S_h(t) (- S_h(t)) u(t) + S_h(t) du/dt = - k S_h(t)^2 u(t)^2 - a(3 + 0.5t)So:- k S_h(t)^2 u(t) + S_h(t) du/dt = - k S_h(t)^2 u(t)^2 - a(3 + 0.5t)Rearrange terms:S_h(t) du/dt = - k S_h(t)^2 u(t)^2 + k S_h(t)^2 u(t) - a(3 + 0.5t)Divide both sides by S_h(t):du/dt = - k S_h(t) u(t)^2 + k S_h(t) u(t) - a(3 + 0.5t)/S_h(t)This is still a nonlinear equation in u(t), so it doesn't help much. Therefore, variation of parameters doesn't seem to simplify the problem.At this point, I think it's best to conclude that the differential equation doesn't have a closed-form solution in terms of elementary functions and that the solution would need to be found numerically or expressed implicitly. However, since the problem asks to solve it, perhaps I need to present the solution in terms of integrals.Alternatively, maybe I can write the solution using the method of integrating factors for the Bernoulli equation. Let me try that again.For the Bernoulli equation:du/dt + P(t)u = Q(t)u^nWe can use the substitution v = u^{1 - n}, which linearizes the equation. In our case, after substitution, we have:dv/dt = -a(3 + 0.5t) + 10k v - k v^2Wait, that's still a Riccati equation. I think I'm stuck here.Given the time I've spent and the lack of progress, I think I need to accept that without specific values for k and a, or a particular solution, the equation can't be solved analytically in a simple form. Therefore, the solution would need to be expressed implicitly or found numerically.However, since the problem is part of a training plan, maybe the coach expects a qualitative answer rather than an explicit formula. But the problem specifically says to solve the differential equation, so perhaps I'm missing something.Wait, maybe I can consider that the equation is linear in S if I rearrange it. Let me try:dS/dt + a(3 + 0.5t) = kS(10 - S)But this is still nonlinear because of the S^2 term.Alternatively, perhaps I can write it as:dS/dt = -k S^2 + 10k S - a(3 + 0.5t)This is a Riccati equation, and as such, it's generally difficult to solve without a particular solution. However, if I can find a particular solution, I can reduce it to a linear equation.Let me assume that the particular solution is of the form S_p(t) = At + B, a linear function. Then, dS_p/dt = A.Substitute into the equation:A = -k (At + B)^2 + 10k (At + B) - a(3 + 0.5t)Expand the right-hand side:= -k (A^2 t^2 + 2AB t + B^2) + 10k At + 10k B - 3a - 0.5a t= -k A^2 t^2 - 2k AB t - k B^2 + 10k A t + 10k B - 3a - 0.5a tGroup like terms:= (-k A^2) t^2 + (-2k AB + 10k A - 0.5a) t + (-k B^2 + 10k B - 3a)Set this equal to A:So, we have:- k A^2 t^2 + (-2k AB + 10k A - 0.5a) t + (-k B^2 + 10k B - 3a) = AThis must hold for all t, so the coefficients of t^2, t, and the constant term must be zero, and the constant term must equal A.Therefore, we have the following system of equations:1. Coefficient of t^2: -k A^2 = 0 => A^2 = 0 => A = 02. Coefficient of t: -2k AB + 10k A - 0.5a = 03. Constant term: -k B^2 + 10k B - 3a = AFrom equation 1, A = 0. Plugging into equation 2:-2k * 0 * B + 10k * 0 - 0.5a = 0 => -0.5a = 0 => a = 0Again, this implies a = 0, which isn't necessarily the case. Therefore, assuming a linear particular solution only works if a = 0, which isn't helpful here.Given that I can't find a particular solution, I think I need to conclude that the differential equation doesn't have an analytical solution in terms of elementary functions and that the solution must be found numerically or expressed implicitly.However, since the problem asks to solve it, perhaps I need to present the solution in terms of integrals. Let me try that.Rewriting the original equation:dS/dt = kS(10 - S) - a(3 + 0.5t)Let me separate variables:dS / [kS(10 - S) - a(3 + 0.5t)] = dtBut this isn't separable because S is a function of t. Therefore, I can't directly integrate both sides.Alternatively, perhaps I can write the equation as:dS / [kS(10 - S) - a(3 + 0.5t)] = dtBut this doesn't help because the left-hand side is a function of S and t, making it impossible to separate variables.At this point, I think the best I can do is to express the solution implicitly. Let me integrate both sides:∫ [1 / (kS(10 - S) - a(3 + 0.5t))] dS = ∫ dtBut this integral is not straightforward and likely doesn't have an elementary antiderivative. Therefore, the solution would need to be expressed in terms of integrals or solved numerically.Given that, I think the answer for part 1 is that the solution can't be expressed in a simple closed form and would require numerical methods or an implicit solution.Moving on to part 2: The endurance E(t) is given by E(t) = E0 + ∫₀ᵗ [c / (1 + e^{-bt})] dt, with E0 = 2, b = 0.3, c = 5. We need to calculate E(t) at t = 8 weeks.So, E(t) = 2 + ∫₀⁸ [5 / (1 + e^{-0.3t})] dtI need to compute this integral. Let me focus on the integral ∫ [5 / (1 + e^{-0.3t})] dt from 0 to 8.First, let me simplify the integrand. Let me set u = -0.3t, then du = -0.3 dt, so dt = -du/0.3.But perhaps a better substitution is to let u = e^{-0.3t}, then du/dt = -0.3 e^{-0.3t} => du = -0.3 u dt => dt = -du/(0.3 u)But let's try that substitution.Let u = e^{-0.3t}, then when t = 0, u = 1, and when t = 8, u = e^{-2.4} ≈ e^{-2.4} ≈ 0.0907.So, the integral becomes:∫ [5 / (1 + u)] * (-du)/(0.3 u)But the limits change from t=0 (u=1) to t=8 (u≈0.0907). So, the integral becomes:5 / 0.3 ∫_{u=1}^{u≈0.0907} [1 / (1 + u)] * (-1/u) duThe negative sign flips the limits:= (5/0.3) ∫_{0.0907}^{1} [1 / (u(1 + u))] duNow, let's decompose the integrand using partial fractions:1 / [u(1 + u)] = A/u + B/(1 + u)Multiply both sides by u(1 + u):1 = A(1 + u) + B uLet u = 0: 1 = A(1) + 0 => A = 1Let u = -1: 1 = A(0) + B(-1) => 1 = -B => B = -1So, 1 / [u(1 + u)] = 1/u - 1/(1 + u)Therefore, the integral becomes:(5/0.3) ∫_{0.0907}^{1} [1/u - 1/(1 + u)] duIntegrate term by term:= (5/0.3) [ ln|u| - ln|1 + u| ] evaluated from 0.0907 to 1Compute at upper limit (1):ln(1) - ln(2) = 0 - ln(2) = -ln(2)Compute at lower limit (0.0907):ln(0.0907) - ln(1 + 0.0907) = ln(0.0907) - ln(1.0907)So, the integral is:(5/0.3) [ (-ln(2)) - (ln(0.0907) - ln(1.0907)) ]Simplify:= (5/0.3) [ -ln(2) - ln(0.0907) + ln(1.0907) ]Combine the logarithms:= (5/0.3) [ -ln(2) - ln(0.0907) + ln(1.0907) ]= (5/0.3) [ -ln(2) - ln(0.0907) + ln(1.0907) ]Let me compute each logarithm:ln(2) ≈ 0.6931ln(0.0907) ≈ ln(0.09) ≈ -2.4079 (since 0.0907 ≈ 0.09)ln(1.0907) ≈ 0.0870 (since ln(1.09) ≈ 0.0862)So, plugging in:= (5/0.3) [ -0.6931 - (-2.4079) + 0.0870 ]Simplify inside the brackets:= -0.6931 + 2.4079 + 0.0870 ≈ (2.4079 - 0.6931) + 0.0870 ≈ 1.7148 + 0.0870 ≈ 1.8018Therefore, the integral is:(5/0.3) * 1.8018 ≈ (5 / 0.3) * 1.8018 ≈ (5 * 1.8018) / 0.3 ≈ 9.009 / 0.3 ≈ 30.03So, the integral ∫₀⁸ [5 / (1 + e^{-0.3t})] dt ≈ 30.03Therefore, E(8) = 2 + 30.03 ≈ 32.03But let me check my calculations because the integral result seems quite large. Let me recalculate the integral more accurately.First, let's compute the integral ∫ [5 / (1 + e^{-0.3t})] dt from 0 to 8.We can also note that ∫ [1 / (1 + e^{-bt})] dt = (1/b) ln(1 + e^{-bt}) + CWait, let me check that derivative:d/dt [ (1/b) ln(1 + e^{-bt}) ] = (1/b) * [ -b e^{-bt} / (1 + e^{-bt}) ] = - e^{-bt} / (1 + e^{-bt}) = -1 / (e^{bt} + 1)But our integrand is 1 / (1 + e^{-bt}) = e^{bt} / (1 + e^{bt}) = 1 - 1 / (1 + e^{bt})Wait, actually, 1 / (1 + e^{-bt}) = e^{bt} / (1 + e^{bt}) = 1 - 1 / (1 + e^{bt})So, ∫ [1 / (1 + e^{-bt})] dt = ∫ [1 - 1 / (1 + e^{bt})] dt = t - (1/b) ln(1 + e^{bt}) + CTherefore, the integral ∫₀ᵗ [c / (1 + e^{-bt})] dt = c [ t - (1/b) ln(1 + e^{bt}) ] evaluated from 0 to t.So, for our case, c = 5, b = 0.3, t = 8.Compute:E(t) = 2 + 5 [ t - (1/0.3) ln(1 + e^{0.3t}) ] from 0 to 8= 2 + 5 [ (8 - (1/0.3) ln(1 + e^{2.4})) - (0 - (1/0.3) ln(1 + e^{0})) ]Simplify:= 2 + 5 [8 - (10/3) ln(1 + e^{2.4}) + (10/3) ln(2) ]Because ln(1 + e^0) = ln(2)Compute e^{2.4} ≈ 11.023So, ln(1 + 11.023) = ln(12.023) ≈ 2.49And ln(2) ≈ 0.6931Therefore:= 2 + 5 [8 - (10/3)(2.49) + (10/3)(0.6931) ]Compute each term:(10/3)(2.49) ≈ 8.3(10/3)(0.6931) ≈ 2.31So:= 2 + 5 [8 - 8.3 + 2.31] = 2 + 5 [2.01] = 2 + 10.05 = 12.05Wait, that's very different from my earlier result. I must have made a mistake in the substitution earlier.Wait, let's recast the integral correctly.Given that ∫ [1 / (1 + e^{-bt})] dt = t - (1/b) ln(1 + e^{bt}) + CTherefore, ∫₀ᵗ [c / (1 + e^{-bt})] dt = c [ t - (1/b) ln(1 + e^{bt}) ] from 0 to t= c [ t - (1/b) ln(1 + e^{bt}) - (0 - (1/b) ln(1 + e^{0})) ]= c [ t - (1/b) ln(1 + e^{bt}) + (1/b) ln(2) ]So, for t = 8, b = 0.3, c = 5:= 5 [8 - (1/0.3) ln(1 + e^{2.4}) + (1/0.3) ln(2) ]Compute e^{2.4} ≈ 11.023ln(1 + 11.023) = ln(12.023) ≈ 2.49ln(2) ≈ 0.6931So:= 5 [8 - (10/3)(2.49) + (10/3)(0.6931) ]Calculate each term:(10/3)(2.49) ≈ 8.3(10/3)(0.6931) ≈ 2.31So:= 5 [8 - 8.3 + 2.31] = 5 [2.01] = 10.05Therefore, E(8) = 2 + 10.05 = 12.05So, the endurance at t = 8 weeks is approximately 12.05 units.Wait, that makes more sense because the integral from 0 to 8 of 5 / (1 + e^{-0.3t}) dt is about 10.05, so E(8) = 2 + 10.05 = 12.05.Therefore, the answer for part 2 is approximately 12.05.But let me verify the integral calculation again to ensure accuracy.Compute ∫₀⁸ [5 / (1 + e^{-0.3t})] dtUsing the antiderivative:= 5 [ t - (1/0.3) ln(1 + e^{0.3t}) ] from 0 to 8At t = 8:= 5 [8 - (10/3) ln(1 + e^{2.4}) ]At t = 0:= 5 [0 - (10/3) ln(2) ]So, the integral is:5 [8 - (10/3) ln(1 + e^{2.4}) - (0 - (10/3) ln(2)) ]= 5 [8 - (10/3) ln(1 + e^{2.4}) + (10/3) ln(2) ]As before.Now, compute ln(1 + e^{2.4}):e^{2.4} ≈ 11.0231 + e^{2.4} ≈ 12.023ln(12.023) ≈ 2.49ln(2) ≈ 0.6931So:= 5 [8 - (10/3)(2.49) + (10/3)(0.6931) ]Calculate (10/3)(2.49) ≈ 8.3(10/3)(0.6931) ≈ 2.31So:= 5 [8 - 8.3 + 2.31] = 5 [2.01] = 10.05Therefore, E(8) = 2 + 10.05 = 12.05So, the endurance at 8 weeks is approximately 12.05 units.To be more precise, let's compute the integral using more accurate values.Compute e^{2.4}:2.4 is 2 + 0.4. e^2 ≈ 7.389, e^{0.4} ≈ 1.4918, so e^{2.4} ≈ 7.389 * 1.4918 ≈ 11.023ln(12.023):We know that ln(12) ≈ 2.4849, and ln(12.023) ≈ 2.49ln(2) ≈ 0.69314718056So, (10/3)(ln(12.023) - ln(2)) = (10/3)(2.49 - 0.69314718056) ≈ (10/3)(1.79685281944) ≈ 5.98950939813Therefore, the integral is:5 [8 - 5.98950939813] ≈ 5 [2.01049060187] ≈ 10.05245300935So, E(8) = 2 + 10.05245300935 ≈ 12.05245300935Rounding to two decimal places, E(8) ≈ 12.05Therefore, the endurance after 8 weeks is approximately 12.05 units.So, summarizing:1. The differential equation for S(t) doesn't have a closed-form solution in terms of elementary functions and would need to be solved numerically or expressed implicitly.2. The endurance E(8) is approximately 12.05 units.But wait, the problem for part 1 says "solve the differential equation for S(t)", so maybe I need to present it in terms of integrals or another form. Alternatively, perhaps the problem expects a qualitative answer, but given the time I've spent, I think I'll proceed with what I have.</think>

🧠:<think>Okay, so I have this problem about a table tennis enthusiast from Kraków who has been keeping track of his games. He played 200 games, won 120, lost 80. The average score difference in his wins was 4 points, and in his losses, it was 3 points. Now, there are two questions to answer.First, using Bayesian statistics, determine the probability that he will win his next game, given that he has won his last 3 games in a row against the same opponent.Second, calculate the probability that the score difference in his next game will be greater than 5 points, assuming the standard deviation is 2.5 points.Alright, let's tackle the first question. Bayesian statistics. Hmm. So, Bayesian probability is about updating our beliefs based on new evidence. In this case, the evidence is that he has won his last 3 games in a row against the same opponent. We need to find the probability that he will win the next game given this evidence.First, I need to figure out what prior information we have. The prior probability of him winning a game is 120/200, which is 0.6 or 60%. That's his overall win rate. But now, given that he's won 3 in a row against the same opponent, we need to update this probability.Wait, but Bayesian statistics usually requires a prior distribution and then updating it with likelihood. So, perhaps we can model his probability of winning as a parameter that we're estimating. Let's denote θ as the probability of winning a game. Our prior belief about θ is based on his overall performance, which is 120 wins out of 200 games. So, a Beta distribution might be appropriate as a prior because it's conjugate to the binomial likelihood.Yes, the Beta distribution is a good choice here. The Beta distribution has parameters α and β, which can be thought of as the number of prior successes and failures. So, if he has 120 wins and 80 losses, we can set α = 120 and β = 80 as our prior.Then, when we observe new data, which is 3 consecutive wins, we can update our prior to get the posterior distribution. The posterior distribution will then have parameters α' = α + number of new successes, and β' = β + number of new failures. Since he won 3 games, the number of new successes is 3, and the number of new failures is 0.So, α' = 120 + 3 = 123, and β' = 80 + 0 = 80. Therefore, the posterior distribution is Beta(123, 80). The mean of this distribution is α' / (α' + β') = 123 / (123 + 80) = 123 / 203 ≈ 0.6059 or 60.59%.So, the probability that he will win his next game, given the prior and the new evidence, is approximately 60.59%.Wait, but is this the correct approach? Because the opponent is the same, maybe we should consider opponent-specific statistics rather than overall statistics. Hmm, the problem doesn't specify whether the opponent is the same as the ones in the 200 games or a different one. It just says "against the same opponent." So, if the opponent is the same, perhaps we should look at his performance against that specific opponent.But in the data given, we don't have information about individual opponents, just the overall 120 wins and 80 losses. So, maybe we have to assume that the opponent is the same as the ones in the overall data. Therefore, the prior is still 120/200, and the update is based on 3 wins in a row.Alternatively, if the opponent is different, we might have to use a different prior, but since we don't have information on that, perhaps we stick with the overall prior.Alternatively, maybe we can model this as a Markov chain, where the probability of winning the next game depends on the previous outcomes. But the problem doesn't specify any dependence beyond the Bayesian update, so perhaps the Beta-Binomial approach is sufficient.So, I think the first answer is approximately 60.59%, which is 123/203. Let me compute that more accurately.123 divided by 203. Let's see, 203 goes into 123 zero times. 203 goes into 1230 six times (6*203=1218). Subtract 1218 from 1230, we get 12. Bring down a zero: 120. 203 goes into 120 zero times. Bring down another zero: 1200. 203 goes into 1200 five times (5*203=1015). Subtract, get 185. Bring down a zero: 1850. 203 goes into 1850 nine times (9*203=1827). Subtract, get 23. Bring down a zero: 230. 203 goes into 230 once (1*203=203). Subtract, get 27. Bring down a zero: 270. 203 goes into 270 once (1*203=203). Subtract, get 67. Bring down a zero: 670. 203 goes into 670 three times (3*203=609). Subtract, get 61. Bring down a zero: 610. 203 goes into 610 three times (3*203=609). Subtract, get 1. So, so far, we have 0.6059113...So, approximately 0.6059 or 60.59%.So, the probability is approximately 60.59%.Alternatively, if we model it as a Bayesian update with a Beta prior, the posterior predictive probability is the mean of the posterior, which is 123/203 ≈ 0.6059.Okay, that seems reasonable.Now, moving on to the second question. The score differences follow a normal distribution. We need to calculate the probability that the score difference in his next game will be greater than 5 points, assuming the standard deviation is 2.5 points.First, we need to know the mean of the score difference. The problem says that in his wins, the average score difference was 4 points, and in his losses, it was 3 points. So, overall, the average score difference is a weighted average based on the number of wins and losses.So, total score difference for wins: 120 games * 4 points = 480 points.Total score difference for losses: 80 games * 3 points = 240 points.Total score difference: 480 + 240 = 720 points.Total number of games: 200.So, overall average score difference: 720 / 200 = 3.6 points.Wait, but is that the correct way to calculate it? Because in wins, the score difference is positive (he won by 4 points), and in losses, it's negative (he lost by 3 points). So, actually, the overall average score difference would be (120*4 + 80*(-3)) / 200.Wait, that's a good point. Because in wins, the score difference is +4, and in losses, it's -3. So, the total score difference is 120*4 + 80*(-3) = 480 - 240 = 240.Therefore, the overall average score difference is 240 / 200 = 1.2 points.Wait, that's a big difference. So, which one is correct?The problem says: "the average score difference in his wins was 4 points, while the average score difference in his losses was 3 points."So, in wins, he won by an average of 4 points, so that's +4. In losses, he lost by an average of 3 points, so that's -3.Therefore, the overall average score difference is (120*4 + 80*(-3)) / 200 = (480 - 240)/200 = 240/200 = 1.2.So, the mean score difference is 1.2 points.But wait, the problem says "the score differences follow a normal distribution." So, the mean is 1.2, standard deviation is 2.5.We need to find P(X > 5), where X ~ N(1.2, 2.5^2).So, to calculate this probability, we can standardize the variable.Z = (X - μ) / σ = (5 - 1.2) / 2.5 = (3.8) / 2.5 = 1.52.So, Z = 1.52.We need to find P(Z > 1.52). Looking at the standard normal distribution table, P(Z < 1.52) is approximately 0.9357. Therefore, P(Z > 1.52) = 1 - 0.9357 = 0.0643 or 6.43%.So, the probability that the score difference will be greater than 5 points is approximately 6.43%.Wait, but let me double-check the calculations.First, the overall mean score difference: 120 wins with +4, 80 losses with -3.Total score difference: 120*4 = 480, 80*(-3) = -240. Total: 480 - 240 = 240. Divided by 200 games: 240/200 = 1.2. Correct.Standard deviation is given as 2.5.So, to find P(X > 5), standardize: (5 - 1.2)/2.5 = 3.8 / 2.5 = 1.52.Looking up 1.52 in the Z-table: the cumulative probability up to 1.52 is 0.9357, so the tail probability is 1 - 0.9357 = 0.0643.Yes, that seems correct.Alternatively, using a calculator, the exact value can be found, but 0.0643 is a good approximation.So, the probability is approximately 6.43%.Therefore, the two answers are approximately 60.59% and 6.43%.But let me think again about the first question. Is the Bayesian approach correctly applied?We used a Beta prior with α=120, β=80, then updated with 3 successes, leading to α'=123, β'=80. The posterior mean is 123/203 ≈ 0.6059.But is this the correct way to model it? Because in Bayesian terms, if we're considering the probability of winning the next game given the prior and the new evidence, the posterior predictive probability is indeed the mean of the posterior distribution.Alternatively, if we model it as a Bernoulli process with an unknown θ, and use a Beta prior, then yes, the posterior predictive probability is the mean of the posterior.But another thought: is the opponent the same as the ones in the prior data? If so, maybe we should have a different prior based on the opponent's performance. But since we don't have that information, we have to use the overall prior.Alternatively, if the opponent is different, we might have to use a different prior, but the problem doesn't specify, so we assume it's the same opponent.Therefore, I think the approach is correct.So, summarizing:1. The probability of winning the next game given 3 consecutive wins is approximately 60.59%.2. The probability that the score difference is greater than 5 points is approximately 6.43%.Final Answer1. The probability of winning the next game is boxed{0.606}.2. The probability that the score difference will be greater than 5 points is boxed{0.064}.</think>