🧠:<think>Alright, so I have this differential equation to solve: ( frac{d^2E}{dt^2} + 3frac{dE}{dt} - 4E = 0 ). It's a second-order linear homogeneous differential equation with constant coefficients. I remember that to solve these, I need to find the characteristic equation first. The characteristic equation for a differential equation like ( ay'' + by' + cy = 0 ) is ( ar^2 + br + c = 0 ). So in this case, the equation becomes ( r^2 + 3r - 4 = 0 ). Let me solve this quadratic equation. Using the quadratic formula, ( r = frac{-b pm sqrt{b^2 - 4ac}}{2a} ). Plugging in the values, ( a = 1 ), ( b = 3 ), ( c = -4 ). So,( r = frac{-3 pm sqrt{9 + 16}}{2} = frac{-3 pm sqrt{25}}{2} = frac{-3 pm 5}{2} ).So the roots are ( r = frac{-3 + 5}{2} = 1 ) and ( r = frac{-3 - 5}{2} = -4 ). Since we have two real and distinct roots, the general solution to the differential equation is ( E(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t} ), where ( r_1 ) and ( r_2 ) are the roots we found. So,( E(t) = C_1 e^{t} + C_2 e^{-4t} ).Now, I need to find the constants ( C_1 ) and ( C_2 ) using the initial conditions. The initial conditions are ( E(0) = 5 ) and ( frac{dE}{dt}(0) = 2 ).First, let's compute ( E(0) ):( E(0) = C_1 e^{0} + C_2 e^{0} = C_1 + C_2 = 5 ).So, equation (1): ( C_1 + C_2 = 5 ).Next, compute the first derivative ( frac{dE}{dt} ):( frac{dE}{dt} = C_1 e^{t} - 4C_2 e^{-4t} ).Then, evaluate at ( t = 0 ):( frac{dE}{dt}(0) = C_1 e^{0} - 4C_2 e^{0} = C_1 - 4C_2 = 2 ).So, equation (2): ( C_1 - 4C_2 = 2 ).Now, I have a system of two equations:1. ( C_1 + C_2 = 5 )2. ( C_1 - 4C_2 = 2 )I can solve this system using substitution or elimination. Let's use elimination. Subtract equation (2) from equation (1):( (C_1 + C_2) - (C_1 - 4C_2) = 5 - 2 )Simplify:( C_1 + C_2 - C_1 + 4C_2 = 3 )Which simplifies to:( 5C_2 = 3 )So, ( C_2 = frac{3}{5} ).Now, substitute ( C_2 ) back into equation (1):( C_1 + frac{3}{5} = 5 )So, ( C_1 = 5 - frac{3}{5} = frac{25}{5} - frac{3}{5} = frac{22}{5} ).Therefore, the particular solution is:( E(t) = frac{22}{5} e^{t} + frac{3}{5} e^{-4t} ).Okay, that was part 1. Now, part 2 asks for the total number of significant events during the first 10 years. That means I need to compute the integral of ( E(t) ) from 0 to 10.So, total events ( = int_{0}^{10} E(t) dt = int_{0}^{10} left( frac{22}{5} e^{t} + frac{3}{5} e^{-4t} right) dt ).Let me compute this integral term by term.First, integrate ( frac{22}{5} e^{t} ):The integral of ( e^{t} ) is ( e^{t} ), so:( frac{22}{5} int e^{t} dt = frac{22}{5} e^{t} ).Second, integrate ( frac{3}{5} e^{-4t} ):The integral of ( e^{kt} ) is ( frac{1}{k} e^{kt} ). So here, ( k = -4 ), so:( frac{3}{5} int e^{-4t} dt = frac{3}{5} times left( frac{1}{-4} e^{-4t} right) = -frac{3}{20} e^{-4t} ).Putting it all together, the integral is:( frac{22}{5} e^{t} - frac{3}{20} e^{-4t} ) evaluated from 0 to 10.So, compute at 10:( frac{22}{5} e^{10} - frac{3}{20} e^{-40} ).And compute at 0:( frac{22}{5} e^{0} - frac{3}{20} e^{0} = frac{22}{5} - frac{3}{20} ).Subtract the lower limit from the upper limit:Total events = ( left( frac{22}{5} e^{10} - frac{3}{20} e^{-40} right) - left( frac{22}{5} - frac{3}{20} right) ).Simplify this expression:First, let's compute the constants:( frac{22}{5} - frac{3}{20} = frac{88}{20} - frac{3}{20} = frac{85}{20} = frac{17}{4} ).So, total events = ( frac{22}{5} e^{10} - frac{3}{20} e^{-40} - frac{17}{4} ).Hmm, that seems correct. Let me double-check the integral steps.Yes, integrating ( e^{t} ) gives ( e^{t} ), and integrating ( e^{-4t} ) gives ( -frac{1}{4} e^{-4t} ). So, the coefficients are correct.So, the total number of significant events is ( frac{22}{5} e^{10} - frac{3}{20} e^{-40} - frac{17}{4} ).But since ( e^{-40} ) is an extremely small number (because 40 is a large exponent in the negative), it can be approximated as nearly zero. So, for practical purposes, the term ( - frac{3}{20} e^{-40} ) is negligible.Therefore, the total number of significant events is approximately ( frac{22}{5} e^{10} - frac{17}{4} ).To get a numerical value, I can compute ( e^{10} ) approximately. I remember that ( e^{10} ) is about 22026.4658. Let me verify that.Yes, ( e^{10} approx 22026.4658 ).So, compute ( frac{22}{5} times 22026.4658 ):( frac{22}{5} = 4.4 ), so 4.4 * 22026.4658 ≈ 4.4 * 22026.4658.Let me compute 4 * 22026.4658 = 88105.8632.Then, 0.4 * 22026.4658 = 8810.58632.Adding together: 88105.8632 + 8810.58632 ≈ 96916.4495.Now, subtract ( frac{17}{4} ), which is 4.25:96916.4495 - 4.25 ≈ 96912.1995.So, approximately 96,912.2 significant events.But wait, let me make sure I didn't make a calculation error.Wait, 4.4 * 22026.4658:Compute 22026.4658 * 4 = 88105.863222026.4658 * 0.4 = 8810.58632Adding them: 88105.8632 + 8810.58632 = 96916.44952Yes, that's correct.Subtract 4.25: 96916.44952 - 4.25 = 96912.19952.So, approximately 96,912.2.But since we're dealing with the number of events, which should be an integer, but the model might allow for fractional events? Hmm, but in reality, events are discrete, but since this is a model, it's okay to have a continuous function.But perhaps the question expects an exact expression rather than a numerical approximation. Let me check the question again.It says, "calculate the total number of significant events... Use the integral of E(t) over [0,10] to find the total." It doesn't specify whether to leave it in terms of exponentials or compute numerically. Since in part 1, they asked for E(t), which we expressed in exact terms, maybe they want the exact expression here as well.So, the exact total is ( frac{22}{5} e^{10} - frac{3}{20} e^{-40} - frac{17}{4} ). Alternatively, we can write it as:( frac{22}{5} e^{10} - frac{17}{4} - frac{3}{20} e^{-40} ).But since ( e^{-40} ) is negligible, it's approximately ( frac{22}{5} e^{10} - frac{17}{4} ).But perhaps the question expects an exact answer, so maybe we should leave it as is.Alternatively, if they want a numerical value, we can compute it as approximately 96,912.2.But let me see if I can represent it more neatly.First, let's write all terms with denominator 20:( frac{22}{5} = frac{88}{20} ), ( frac{17}{4} = frac{85}{20} ).So,Total events = ( frac{88}{20} e^{10} - frac{85}{20} - frac{3}{20} e^{-40} ).Factor out ( frac{1}{20} ):Total events = ( frac{1}{20} (88 e^{10} - 85 - 3 e^{-40}) ).That might be a cleaner way to write it.Alternatively, we can write it as:( frac{88 e^{10} - 85 - 3 e^{-40}}{20} ).Either way is acceptable, but perhaps the first form is better.So, to recap:1. The solution to the differential equation is ( E(t) = frac{22}{5} e^{t} + frac{3}{5} e^{-4t} ).2. The total number of significant events over the first 10 years is ( frac{22}{5} e^{10} - frac{3}{20} e^{-40} - frac{17}{4} ), which is approximately 96,912.2.I think that's all. Let me just double-check my steps.- Solved the characteristic equation correctly: roots at 1 and -4. Correct.- Formulated the general solution: correct.- Applied initial conditions:At t=0, E(0)=5: ( C1 + C2 =5 ). Correct.First derivative: ( C1 e^{t} -4 C2 e^{-4t} ). At t=0: ( C1 -4 C2 =2 ). Correct.Solving the system: C1=22/5, C2=3/5. Correct.Integral computation:Integrated each term correctly: yes.Evaluated at 10 and 0, subtracted: correct.Simplified the constants: yes.Approximated ( e^{-40} ) as negligible: yes.Numerical approximation: correct.So, I think everything checks out.Final Answer1. The solution is ( boxed{E(t) = frac{22}{5} e^{t} + frac{3}{5} e^{-4t}} ).2. The total number of significant events is approximately ( boxed{96912.2} ).</think>

🧠:<think>Okay, so I have this problem about an NGO executive analyzing donor data. They have a function C(t) that models donor contributions over time, and it's a piecewise function. The first part is a quadratic from t=0 to t=2, and then it switches to an exponential function from t=2 to t=5. The executive wants to understand the trends and patterns to optimize fundraising. First, the problem is asking me to determine the conditions on the parameters a1, b1, c1, a2, b2, and c2 so that the function C(t) is continuous and differentiable at t=2. That makes sense because if the function isn't continuous or differentiable at the point where it changes from quadratic to exponential, it might not accurately model the donor behavior.So, for continuity at t=2, the value of the quadratic function at t=2 should equal the value of the exponential function at t=2. That means:a1*(2)^2 + b1*(2) + c1 = a2*e^(b2*2) + c2Simplifying that, it's:4a1 + 2b1 + c1 = a2*e^(2b2) + c2That's the first condition.Next, for differentiability at t=2, the derivatives of both pieces should be equal at that point. The derivative of the quadratic function is 2a1*t + b1, and the derivative of the exponential function is a2*b2*e^(b2*t). So, evaluating both at t=2:2a1*(2) + b1 = a2*b2*e^(b2*2)Simplifying:4a1 + b1 = a2*b2*e^(2b2)So, that's the second condition.Therefore, the two conditions are:1. 4a1 + 2b1 + c1 = a2*e^(2b2) + c22. 4a1 + b1 = a2*b2*e^(2b2)These conditions ensure that the function is both continuous and smooth at the transition point t=2.Moving on to the second part, the executive wants to forecast C(t) for the next year, from t=5 to t=6. They assume that the growth rate follows the pattern observed in the interval t=2 to t=5, which is the exponential part. So, I think that means we should extend the exponential function beyond t=5.Given that the exponential function is a2*e^(b2*t) + c2, we can model the forecast for t >5 as the same function. So, for 5 < t ≤6, C(t) = a2*e^(b2*t) + c2.But wait, do we need to adjust the parameters? Since the exponential function is already defined up to t=5, maybe we can just continue it as is. Alternatively, perhaps we should consider the growth rate from t=2 to t=5 and project that forward.Wait, the growth rate is given by the derivative. The derivative of the exponential function is a2*b2*e^(b2*t). So, if we want to model the growth rate for t >5, it would be the same as the growth rate at t=5, which is a2*b2*e^(b2*5).But actually, since the exponential function is already capturing the growth, if we assume the same growth pattern, we can just continue the exponential function beyond t=5. So, the model for t >5 would be C(t) = a2*e^(b2*t) + c2.But wait, at t=5, the function is already a2*e^(b2*5) + c2. So, for t=6, it would be a2*e^(b2*6) + c2.Therefore, the expected increase from t=5 to t=6 is C(6) - C(5) = [a2*e^(6b2) + c2] - [a2*e^(5b2) + c2] = a2*e^(5b2)*(e^(b2) -1).So, that's the expected increase.But hold on, is there a way to express this in terms of the growth rate? The growth rate is the derivative, which is a2*b2*e^(b2*t). At t=5, the growth rate is a2*b2*e^(5b2). So, the increase from t=5 to t=6 can also be approximated by the growth rate at t=5 multiplied by the time interval, which is 1 year. So, approximately, the increase would be a2*b2*e^(5b2)*1.But since the function is exponential, the exact increase is a2*e^(5b2)*(e^(b2) -1). So, the exact increase is a2*e^(5b2)*(e^(b2) -1), and the approximate increase using the growth rate is a2*b2*e^(5b2). Depending on how precise we need to be, we can use either.But since the problem says to develop a suitable mathematical model to forecast C(t) for the next year, I think just extending the exponential function is sufficient. So, the forecast model is C(t) = a2*e^(b2*t) + c2 for 5 < t ≤6.And the expected increase is C(6) - C(5) = a2*e^(6b2) + c2 - (a2*e^(5b2) + c2) = a2*e^(5b2)*(e^(b2) -1).So, summarizing:1. The conditions for continuity and differentiability at t=2 are: - 4a1 + 2b1 + c1 = a2*e^(2b2) + c2 - 4a1 + b1 = a2*b2*e^(2b2)2. The forecast model for t between 5 and 6 is C(t) = a2*e^(b2*t) + c2, and the expected increase from t=5 to t=6 is a2*e^(5b2)*(e^(b2) -1).I think that's it. I don't see any mistakes in the reasoning. The key was ensuring continuity and differentiability at the switch point, and then extending the exponential model for forecasting.</think>

🧠:<think>Alright, so I've got these two problems about horse racing, and I need to solve them. Let me take them one at a time.Starting with Race A. The problem says that the speed of the horse is given by the function v(t) = at² + bt + c. They give me three points: at t=0, v=0; at t=5, v=20; and at t=10, v=50. So I need to find the constants a, b, and c.Okay, so since v(0) = 0, let me plug t=0 into the equation. That gives v(0) = a*(0)² + b*(0) + c = c. So c must be 0. That simplifies things a bit.Now the equation becomes v(t) = at² + bt. Got that. Now, at t=5, v=20. Plugging that in: 20 = a*(5)² + b*(5) => 20 = 25a + 5b. Let me write that as equation (1): 25a + 5b = 20.Similarly, at t=10, v=50. Plugging that in: 50 = a*(10)² + b*(10) => 50 = 100a + 10b. Let me write that as equation (2): 100a + 10b = 50.Now I have a system of two equations:1) 25a + 5b = 202) 100a + 10b = 50Hmm, maybe I can solve this system. Let me see. Maybe I can simplify equation (1) first. If I divide equation (1) by 5, I get 5a + b = 4. Let's call this equation (1a): 5a + b = 4.Similarly, equation (2) can be simplified. If I divide equation (2) by 10, I get 10a + b = 5. Let's call this equation (2a): 10a + b = 5.Now, I have:(1a): 5a + b = 4(2a): 10a + b = 5If I subtract equation (1a) from equation (2a), I can eliminate b.(10a + b) - (5a + b) = 5 - 410a + b - 5a - b = 15a = 1So, a = 1/5. That's 0.2.Now, plug a back into equation (1a): 5*(1/5) + b = 4 => 1 + b = 4 => b = 3.So, a = 1/5, b = 3, c = 0.Wait, let me double-check with equation (2) to make sure.100a + 10b = 50100*(1/5) + 10*3 = 20 + 30 = 50. Yep, that works. So the constants are a=1/5, b=3, c=0.Alright, moving on to Race B. The problem says that the horse's acceleration is given by a(t) = 3t + 2. The horse starts from rest, and the race lasts for 10 seconds. I need to find the total distance covered.Hmm, okay. So acceleration is the derivative of velocity, right? So if I have acceleration, I can integrate it to get velocity, and then integrate velocity to get distance.Since the horse starts from rest, initial velocity v(0) = 0. Let me write that down.First, find velocity as a function of time. The acceleration a(t) = 3t + 2. So velocity v(t) is the integral of a(t) dt.So, integrating a(t):v(t) = ∫ (3t + 2) dt = (3/2)t² + 2t + CWe know that at t=0, v(0) = 0. So plugging t=0 in:0 = (3/2)*(0)² + 2*(0) + C => C = 0.So, v(t) = (3/2)t² + 2t.Now, to find the total distance, I need to integrate velocity from t=0 to t=10.So, distance D = ∫₀¹⁰ v(t) dt = ∫₀¹⁰ [(3/2)t² + 2t] dtLet me compute that integral.First, integrate term by term:∫ (3/2)t² dt = (3/2)*(t³/3) = (1/2)t³∫ 2t dt = 2*(t²/2) = t²So, putting it together:D = [ (1/2)t³ + t² ] evaluated from 0 to 10.Compute at t=10:(1/2)*(10)³ + (10)² = (1/2)*1000 + 100 = 500 + 100 = 600Compute at t=0:(1/2)*(0)³ + (0)² = 0So, D = 600 - 0 = 600 meters.Wait, that seems straightforward. Let me double-check the integration steps.Starting with a(t) = 3t + 2.Integrate once to get v(t):v(t) = ∫ (3t + 2) dt = (3/2)t² + 2t + C. Since v(0)=0, C=0.Integrate v(t) to get distance:D = ∫₀¹⁰ [(3/2)t² + 2t] dtCompute integral:(3/2)*(t³/3) + 2*(t²/2) = (1/2)t³ + t². Evaluated from 0 to 10.At 10: (1/2)*1000 + 100 = 500 + 100 = 600.Yep, that's correct. So the total distance is 600 meters.Wait, just to make sure, let me compute the integral another way.Alternatively, I can compute the definite integral directly.∫₀¹⁰ (3/2)t² + 2t dt = [ (3/2)*(t³/3) + 2*(t²/2) ] from 0 to 10Simplify:(3/2)*(1/3)t³ + 2*(1/2)t² = (1/2)t³ + t²Same as before. So, yeah, 600 meters.I think that's solid.Final Answer1. The constants are ( a = boxed{dfrac{1}{5}} ), ( b = boxed{3} ), and ( c = boxed{0} ).2. The total distance covered is ( boxed{600} ) meters.</think>

🧠:<think>First, I need to determine the number of retirees and students in the seating arrangement. Since the participants are forming a perfect circle with students seated between two retirees, the number of students must equal the number of retirees. Given that there are 30 participants in total, I can set up the equation R + S = 30, where R represents the number of retirees and S represents the number of students. Since S = R, the equation becomes 2R = 30, leading to R = 15 and S = 15.Next, I need to calculate the minimum total time required for all students to learn all the strategies. It's stated that there are twice as many students as retirees, so if there are 15 retirees, there must be 30 students. Each teaching strategy takes 15 minutes, and each student can only attend one session at a time. Since there are 15 unique strategies, each student needs to attend 15 sessions. Therefore, the total time required for one student to learn all strategies is 15 sessions multiplied by 15 minutes per session, which equals 225 minutes. Since all students need to learn all strategies simultaneously, the minimum total time required is 225 minutes.</think>