🧠:<think>Okay, so I have this differential equation to solve: [ frac{d^2y}{dt^2} + 3frac{dy}{dt} + 2y = cos(t) ]It's a second-order linear nonhomogeneous differential equation. I remember that to solve such equations, I need to find the general solution, which is the sum of the homogeneous solution and a particular solution.First, let me tackle the homogeneous part:[ frac{d^2y}{dt^2} + 3frac{dy}{dt} + 2y = 0 ]To solve this, I should find the characteristic equation. The characteristic equation for a second-order linear homogeneous differential equation is:[ r^2 + 3r + 2 = 0 ]Let me factor this quadratic equation. Looking for two numbers that multiply to 2 and add up to 3. Hmm, 1 and 2. Yes, that works.So, factoring:[ (r + 1)(r + 2) = 0 ]Setting each factor equal to zero gives the roots:[ r = -1 quad text{and} quad r = -2 ]Since these are real and distinct roots, the homogeneous solution will be:[ y_h(t) = C_1 e^{-t} + C_2 e^{-2t} ]Where ( C_1 ) and ( C_2 ) are constants determined by initial conditions.Now, moving on to the particular solution. The nonhomogeneous term is ( cos(t) ), so I need to find a particular solution ( y_p(t) ) that satisfies the original equation.I recall that when the nonhomogeneous term is a cosine function, we can assume a particular solution of the form:[ y_p(t) = A cos(t) + B sin(t) ]Where ( A ) and ( B ) are constants to be determined.Let me compute the first and second derivatives of ( y_p(t) ):First derivative:[ y_p'(t) = -A sin(t) + B cos(t) ]Second derivative:[ y_p''(t) = -A cos(t) - B sin(t) ]Now, substitute ( y_p(t) ), ( y_p'(t) ), and ( y_p''(t) ) into the original differential equation:[ (-A cos(t) - B sin(t)) + 3(-A sin(t) + B cos(t)) + 2(A cos(t) + B sin(t)) = cos(t) ]Let me expand and collect like terms.First, distribute the coefficients:1. ( -A cos(t) - B sin(t) )2. ( + 3(-A sin(t) + B cos(t)) = -3A sin(t) + 3B cos(t) )3. ( + 2(A cos(t) + B sin(t)) = 2A cos(t) + 2B sin(t) )Now, combine all terms:- Cosine terms: - ( -A cos(t) ) - ( + 3B cos(t) ) - ( + 2A cos(t) ) Total: ( (-A + 3B + 2A) cos(t) = (A + 3B) cos(t) )- Sine terms: - ( -B sin(t) ) - ( -3A sin(t) ) - ( + 2B sin(t) ) Total: ( (-B - 3A + 2B) sin(t) = (B - 3A) sin(t) )So, putting it all together:[ (A + 3B) cos(t) + (B - 3A) sin(t) = cos(t) ]This equation must hold for all ( t ), so the coefficients of ( cos(t) ) and ( sin(t) ) on the left must equal those on the right. The right side is ( cos(t) ), which can be written as ( 1 cdot cos(t) + 0 cdot sin(t) ).Therefore, we can set up the following system of equations:1. ( A + 3B = 1 ) (coefficient of ( cos(t) ))2. ( B - 3A = 0 ) (coefficient of ( sin(t) ))Let me solve this system.From equation 2:[ B = 3A ]Substitute ( B = 3A ) into equation 1:[ A + 3(3A) = 1 ][ A + 9A = 1 ][ 10A = 1 ][ A = frac{1}{10} ]Now, substitute ( A = frac{1}{10} ) back into ( B = 3A ):[ B = 3 times frac{1}{10} = frac{3}{10} ]So, the particular solution is:[ y_p(t) = frac{1}{10} cos(t) + frac{3}{10} sin(t) ]Therefore, the general solution to the differential equation is the sum of the homogeneous and particular solutions:[ y(t) = y_h(t) + y_p(t) = C_1 e^{-t} + C_2 e^{-2t} + frac{1}{10} cos(t) + frac{3}{10} sin(t) ]That answers part 1. Now, moving on to part 2.Dr. Aria wants the blood flow to return to normal as quickly as possible after surgery. So, she needs the deviation ( y(t) ) to approach zero as ( t ) becomes large. The initial conditions are given as:- Initial deviation ( y(0) = 0 )- Initial rate of change ( y'(0) = 0 )I need to use these initial conditions to find the constants ( C_1 ) and ( C_2 ).First, let me write down the general solution again:[ y(t) = C_1 e^{-t} + C_2 e^{-2t} + frac{1}{10} cos(t) + frac{3}{10} sin(t) ]Now, compute ( y(0) ):[ y(0) = C_1 e^{0} + C_2 e^{0} + frac{1}{10} cos(0) + frac{3}{10} sin(0) ][ y(0) = C_1 + C_2 + frac{1}{10}(1) + frac{3}{10}(0) ][ y(0) = C_1 + C_2 + frac{1}{10} ]Given that ( y(0) = 0 ):[ C_1 + C_2 + frac{1}{10} = 0 ][ C_1 + C_2 = -frac{1}{10} quad text{(Equation 3)} ]Next, compute the first derivative ( y'(t) ):[ y'(t) = -C_1 e^{-t} - 2C_2 e^{-2t} - frac{1}{10} sin(t) + frac{3}{10} cos(t) ]Now, evaluate ( y'(0) ):[ y'(0) = -C_1 e^{0} - 2C_2 e^{0} - frac{1}{10} sin(0) + frac{3}{10} cos(0) ][ y'(0) = -C_1 - 2C_2 - 0 + frac{3}{10}(1) ][ y'(0) = -C_1 - 2C_2 + frac{3}{10} ]Given that ( y'(0) = 0 ):[ -C_1 - 2C_2 + frac{3}{10} = 0 ][ -C_1 - 2C_2 = -frac{3}{10} ][ C_1 + 2C_2 = frac{3}{10} quad text{(Equation 4)} ]Now, we have a system of two equations:3. ( C_1 + C_2 = -frac{1}{10} )4. ( C_1 + 2C_2 = frac{3}{10} )Let me subtract equation 3 from equation 4 to eliminate ( C_1 ):[ (C_1 + 2C_2) - (C_1 + C_2) = frac{3}{10} - (-frac{1}{10}) ][ C_1 + 2C_2 - C_1 - C_2 = frac{3}{10} + frac{1}{10} ][ C_2 = frac{4}{10} ][ C_2 = frac{2}{5} ]Now, substitute ( C_2 = frac{2}{5} ) into equation 3:[ C_1 + frac{2}{5} = -frac{1}{10} ][ C_1 = -frac{1}{10} - frac{2}{5} ]Convert ( frac{2}{5} ) to tenths: ( frac{4}{10} )[ C_1 = -frac{1}{10} - frac{4}{10} = -frac{5}{10} = -frac{1}{2} ]So, ( C_1 = -frac{1}{2} ) and ( C_2 = frac{2}{5} ).Therefore, the particular solution satisfying the initial conditions is:[ y(t) = -frac{1}{2} e^{-t} + frac{2}{5} e^{-2t} + frac{1}{10} cos(t) + frac{3}{10} sin(t) ]Now, analyzing the behavior of this solution over time.Looking at the terms:1. ( -frac{1}{2} e^{-t} ): As ( t ) increases, this term decays exponentially to zero.2. ( frac{2}{5} e^{-2t} ): Similarly, this term also decays exponentially to zero, but with a faster rate since the exponent is ( -2t ).3. ( frac{1}{10} cos(t) + frac{3}{10} sin(t) ): This is a combination of sine and cosine functions, which oscillate indefinitely with a constant amplitude.So, as ( t ) becomes very large, the exponential terms will approach zero, leaving the solution to oscillate with the particular solution terms. However, the oscillations are persistent and do not decay over time, meaning the deviation from normal blood flow will continue to oscillate.But Dr. Aria wants the blood flow to return to normal as quickly as possible. The persistent oscillations might not be ideal because they indicate that the deviation doesn't settle down to zero; instead, it keeps oscillating. This could mean that the new technique might not completely stabilize the blood flow, leading to ongoing fluctuations.However, the amplitude of these oscillations is relatively small. Let me compute the amplitude of the particular solution.The particular solution is:[ frac{1}{10} cos(t) + frac{3}{10} sin(t) ]This can be written as ( R cos(t - phi) ), where ( R ) is the amplitude.Compute ( R ):[ R = sqrt{left(frac{1}{10}right)^2 + left(frac{3}{10}right)^2} = sqrt{frac{1}{100} + frac{9}{100}} = sqrt{frac{10}{100}} = sqrt{frac{1}{10}} = frac{1}{sqrt{10}} approx 0.316 ]So, the amplitude is approximately 0.316, which is a relatively small deviation. The exponential terms decay to zero, so over time, the deviation will oscillate around zero with a small amplitude.Therefore, while the deviation doesn't return to exactly zero, it does approach a small oscillation around normal blood flow. This might be acceptable if the amplitude is within a safe range, but if Dr. Aria wants the blood flow to stabilize completely, she might need to adjust the parameters of her technique to eliminate the oscillatory component.Alternatively, perhaps the homogeneous solution's decay is fast enough that the transient response dies out quickly, leaving the oscillations as a steady-state response. In that case, the recovery might be considered quick if the exponential terms decay rapidly, even if there's a small persistent oscillation.Looking at the exponents, the terms are ( e^{-t} ) and ( e^{-2t} ). The ( e^{-2t} ) term decays faster than ( e^{-t} ). So, the transient response will diminish relatively quickly, and the system will enter the steady-state oscillation sooner rather than later.In conclusion, the solution shows that after the initial transient phase, the deviation from normal blood flow will oscillate with a small amplitude. This suggests that the new technique might lead to a quicker return to a stable, albeit slightly oscillating, blood flow compared to a system without such a particular solution. However, if complete stabilization is desired, further adjustments might be necessary.Final Answer1. The general solution is:[ boxed{y(t) = C_1 e^{-t} + C_2 e^{-2t} + frac{1}{10} cos(t) + frac{3}{10} sin(t)} ]2. The particular solution with the given initial conditions is:[ boxed{y(t) = -frac{1}{2} e^{-t} + frac{2}{5} e^{-2t} + frac{1}{10} cos(t) + frac{3}{10} sin(t)} ]The solution indicates that the deviation from normal blood flow will decay over time due to the exponential terms and eventually oscillate with a small amplitude, suggesting a relatively quick return to near-normal blood flow.</think>

🧠:<think>Okay, so I have this problem about a lucha libre wrestler who wants to design a special octagonal ring for a match. There are two parts: first, calculating the area of the regular octagon, and second, finding the radius of a circular logo that's tangent to each side of the octagon. Hmm, let me think about how to approach each part.Starting with the first problem: calculating the area of a regular octagon with each side being 10 meters. I remember that regular polygons have formulas for their areas based on the number of sides and the length of each side. For a regular polygon with 'n' sides, each of length 's', the area can be calculated using the formula:Area = (n * s^2) / (4 * tan(π/n))So, for an octagon, n is 8. Plugging in the values, we get:Area = (8 * 10^2) / (4 * tan(π/8))Let me compute that step by step.First, 10 squared is 100. So, 8 * 100 is 800.Then, the denominator is 4 * tan(π/8). I need to calculate tan(π/8). I know that π is approximately 3.1416, so π/8 is about 0.3927 radians.Calculating tan(0.3927). Hmm, I might need to use a calculator for that. Let me recall that tan(π/8) is equal to sqrt(2) - 1, which is approximately 0.4142. Wait, is that right? Let me verify.Yes, tan(π/8) can be calculated using the identity tan(θ/2) = (1 - cos θ)/sin θ. If θ is π/4, then tan(π/8) = (1 - cos(π/4))/sin(π/4). Cos(π/4) is sqrt(2)/2, so 1 - sqrt(2)/2 is approximately 1 - 0.7071 = 0.2929. Sin(π/4) is also sqrt(2)/2, approximately 0.7071. So, 0.2929 / 0.7071 is approximately 0.4142, which is sqrt(2) - 1. So, yes, tan(π/8) is sqrt(2) - 1, approximately 0.4142.So, the denominator is 4 * 0.4142, which is approximately 1.6568.Therefore, the area is 800 / 1.6568. Let me compute that.800 divided by 1.6568. Let me do this division step by step.First, 1.6568 goes into 800 how many times? Let me compute 800 / 1.6568.Alternatively, since 1.6568 is approximately 1.6568, I can write this as 800 / 1.6568 ≈ 800 / 1.6568.Let me compute this using a calculator method.1.6568 * 482 ≈ 1.6568 * 480 = 791.424, and 1.6568 * 2 = 3.3136, so total is approximately 794.7376. That's less than 800.1.6568 * 483 ≈ 794.7376 + 1.6568 ≈ 796.39441.6568 * 484 ≈ 796.3944 + 1.6568 ≈ 798.05121.6568 * 485 ≈ 798.0512 + 1.6568 ≈ 799.7081.6568 * 486 ≈ 799.708 + 1.6568 ≈ 801.3648So, 1.6568 * 485 ≈ 799.708, which is just under 800. The difference is 800 - 799.708 = 0.292.So, 0.292 / 1.6568 ≈ 0.176.Therefore, 485 + 0.176 ≈ 485.176.So, approximately 485.176 square meters.Wait, but let me check this with another method because I might have made an error in the multiplication.Alternatively, I can use the formula for the area of a regular octagon, which is also given by 2(1 + sqrt(2)) * s^2, where s is the side length.Yes, that's another formula I remember. Let me use that.So, Area = 2 * (1 + sqrt(2)) * (10)^2Compute that.First, sqrt(2) is approximately 1.4142.So, 1 + sqrt(2) ≈ 2.4142.Multiply by 2: 2 * 2.4142 ≈ 4.8284.Then, multiply by s^2, which is 100: 4.8284 * 100 = 482.84.Wait, that's different from the previous result. Hmm, so which one is correct?I think I might have messed up the first method because I used tan(π/8) as approximately 0.4142, but let me verify the formula.The general formula for the area of a regular polygon is (1/2) * perimeter * apothem.Alternatively, another formula is (n * s^2) / (4 * tan(π/n)).So, for n=8, s=10, that's (8 * 100) / (4 * tan(π/8)) = 800 / (4 * tan(π/8)) = 200 / tan(π/8).Since tan(π/8) is sqrt(2) - 1, which is approximately 0.4142, so 200 / 0.4142 ≈ 482.84.Wait, that's the same as the second formula. So, perhaps I made a mistake in my first calculation when I thought the area was approximately 485.176. It should actually be 482.84.So, the correct area is approximately 482.84 square meters.But let me compute it more precisely.Given that tan(π/8) is exactly sqrt(2) - 1, so tan(π/8) = sqrt(2) - 1 ≈ 1.4142 - 1 = 0.4142.So, 200 / (sqrt(2) - 1). To rationalize the denominator, multiply numerator and denominator by (sqrt(2) + 1):200 * (sqrt(2) + 1) / [(sqrt(2) - 1)(sqrt(2) + 1)] = 200*(sqrt(2) + 1)/(2 - 1) = 200*(sqrt(2) + 1)/1 = 200*(sqrt(2) + 1).Compute that:sqrt(2) ≈ 1.41421356So, sqrt(2) + 1 ≈ 2.41421356Multiply by 200: 2.41421356 * 200 = 482.842712So, approximately 482.843 square meters.Therefore, the area is approximately 482.843 square meters.So, that's the first part.Now, moving on to the second problem: calculating the radius of a circular logo that's tangent to each side of the octagon. So, the circle is inscribed in the octagon, meaning it's tangent to all eight sides.In a regular polygon, the radius of the inscribed circle (also called the apothem) can be calculated using the formula:Apothem = (s) / (2 * tan(π/n))Where s is the side length, and n is the number of sides.So, for our octagon, n=8, s=10.Therefore, Apothem = 10 / (2 * tan(π/8)) = 5 / tan(π/8)We already know that tan(π/8) is sqrt(2) - 1, so:Apothem = 5 / (sqrt(2) - 1)Again, rationalizing the denominator:Multiply numerator and denominator by (sqrt(2) + 1):Apothem = 5*(sqrt(2) + 1) / [(sqrt(2) - 1)(sqrt(2) + 1)] = 5*(sqrt(2) + 1)/(2 - 1) = 5*(sqrt(2) + 1)Compute that:sqrt(2) ≈ 1.41421356So, sqrt(2) + 1 ≈ 2.41421356Multiply by 5: 2.41421356 * 5 ≈ 12.0710678So, approximately 12.071 meters.Therefore, the radius of the circle is approximately 12.071 meters.Wait, but let me make sure I didn't confuse the apothem with something else. The apothem is indeed the radius of the inscribed circle, which is tangent to all sides. So, yes, that should be correct.Alternatively, I can think of the octagon as being composed of 8 isosceles triangles, each with a base of 10 meters and two sides equal to the radius of the circumscribed circle. The apothem would be the height of each of these triangles.But in this case, since we're dealing with the inscribed circle, the apothem is the radius we're looking for.So, yes, 12.071 meters is the correct radius.To summarize:1. The area of the octagonal ring is approximately 482.843 square meters.2. The radius of the inscribed circle is approximately 12.071 meters.I think that's it. Let me just double-check my calculations.For the area:Using the formula 2*(1 + sqrt(2))*s^2, with s=10:2*(1 + 1.41421356)*100 = 2*(2.41421356)*100 = 4.82842712*100 = 482.842712, which rounds to 482.843.For the apothem:5*(sqrt(2) + 1) = 5*(2.41421356) ≈ 12.0710678, which rounds to 12.071.Yes, that seems consistent.Final Answer1. The area of the octagonal ring is boxed{482.843} square meters.2. The radius of the circular logo is boxed{12.071} meters.</think>

🧠:<think>First, I need to determine how many photographs Alex needs to use for retraining to increase the accuracy from 85% to 95%. The improvement needed is 10 percentage points. Since the improvement is linearly proportional to the amount of additional training data, I'll calculate the number of photographs required to achieve this improvement.Next, I'll calculate the total storage required for all 5,000 photographs. Each photograph is 25 MB, so multiplying the number of photographs by the file size gives the total storage in megabytes. I'll then convert this to gigabytes since the cloud storage cost is per GB.Finally, I'll compute the monthly storage cost by multiplying the total storage in gigabytes by the cost per GB.</think>

🧠:<think>Okay, so I have this problem about a language learning app that uses differential equations to model a student's vocabulary knowledge. The first part is to find the general solution for V(t), and the second part is to compute the Knowledge Retention Index (KRI) and analyze its dependence on certain constants. Let me try to work through this step by step.Starting with part 1. The differential equation given is:dV/dt = k1*(L(t) - V(t)) + k2*A(t)We are told that L(t) = L0*e^{-λt} and A(t) = A0*e^{-μt}. So, substituting these into the equation, we get:dV/dt = k1*(L0*e^{-λt} - V(t)) + k2*A0*e^{-μt}Let me rewrite this equation:dV/dt + k1*V(t) = k1*L0*e^{-λt} + k2*A0*e^{-μt}This looks like a linear first-order differential equation. The standard form is:dV/dt + P(t)*V(t) = Q(t)In this case, P(t) is k1, which is a constant, and Q(t) is k1*L0*e^{-λt} + k2*A0*e^{-μt}.To solve this, I can use an integrating factor. The integrating factor μ(t) is given by:μ(t) = e^{∫P(t) dt} = e^{∫k1 dt} = e^{k1*t}Multiplying both sides of the differential equation by μ(t):e^{k1*t}*dV/dt + e^{k1*t}*k1*V(t) = e^{k1*t}*(k1*L0*e^{-λt} + k2*A0*e^{-μt})The left side is the derivative of [e^{k1*t}*V(t)] with respect to t. So, we can write:d/dt [e^{k1*t}*V(t)] = e^{k1*t}*(k1*L0*e^{-λt} + k2*A0*e^{-μt})Now, let's integrate both sides from 0 to t:∫₀ᵗ d/dt [e^{k1*s}*V(s)] ds = ∫₀ᵗ e^{k1*s}*(k1*L0*e^{-λs} + k2*A0*e^{-μs}) dsThe left side simplifies to e^{k1*t}*V(t) - e^{0}*V(0) = e^{k1*t}*V(t) - V(0)So, we have:e^{k1*t}*V(t) - V(0) = ∫₀ᵗ [k1*L0*e^{(k1 - λ)s} + k2*A0*e^{(k1 - μ)s}] dsNow, let's compute the integral on the right side. We can split it into two separate integrals:= k1*L0 ∫₀ᵗ e^{(k1 - λ)s} ds + k2*A0 ∫₀ᵗ e^{(k1 - μ)s} dsLet me compute each integral separately.First integral: ∫ e^{(k1 - λ)s} dsLet’s denote a = k1 - λ. Then, the integral becomes ∫ e^{a s} ds = (1/a) e^{a s} + CSimilarly, the second integral: ∫ e^{(k1 - μ)s} dsLet’s denote b = k1 - μ. Then, the integral becomes ∫ e^{b s} ds = (1/b) e^{b s} + CSo, plugging these back into the equation:e^{k1*t}*V(t) - V(0) = k1*L0*(1/(k1 - λ))*(e^{(k1 - λ)t} - 1) + k2*A0*(1/(k1 - μ))*(e^{(k1 - μ)t} - 1)Now, let's solve for V(t):V(t) = e^{-k1*t} [ V(0) + k1*L0/(k1 - λ)*(e^{(k1 - λ)t} - 1) + k2*A0/(k1 - μ)*(e^{(k1 - μ)t} - 1) ]Simplify each term:First term: V(0)*e^{-k1*t}Second term: k1*L0/(k1 - λ)*(e^{-k1*t}*(e^{(k1 - λ)t} - 1)) = k1*L0/(k1 - λ)*(e^{-λ t} - e^{-k1 t})Third term: k2*A0/(k1 - μ)*(e^{-k1*t}*(e^{(k1 - μ)t} - 1)) = k2*A0/(k1 - μ)*(e^{-μ t} - e^{-k1 t})So, combining all terms:V(t) = V(0)*e^{-k1*t} + (k1*L0)/(k1 - λ)*(e^{-λ t} - e^{-k1 t}) + (k2*A0)/(k1 - μ)*(e^{-μ t} - e^{-k1 t})This seems a bit complicated, but let me see if I can factor out e^{-k1 t}:V(t) = V(0)*e^{-k1*t} + (k1*L0)/(k1 - λ)*e^{-λ t} - (k1*L0)/(k1 - λ)*e^{-k1 t} + (k2*A0)/(k1 - μ)*e^{-μ t} - (k2*A0)/(k1 - μ)*e^{-k1 t}Combine the terms with e^{-k1 t}:= [V(0) - (k1*L0)/(k1 - λ) - (k2*A0)/(k1 - μ)]*e^{-k1 t} + (k1*L0)/(k1 - λ)*e^{-λ t} + (k2*A0)/(k1 - μ)*e^{-μ t}So, that's the general solution for V(t). It includes terms decaying exponentially with rates k1, λ, and μ.Wait, let me check if I made any mistakes in the integration. The integrating factor was correct, and the integral of e^{a s} is indeed (1/a)(e^{a t} - 1). So, the steps seem correct.Now, moving on to part 2. We need to compute the KRI, which is the integral from 0 to T of V(t)*A(t) dt.Given that A(t) = A0*e^{-μ t}, and V(t) is the solution we found above.So, KRI = ∫₀ᵀ V(t)*A(t) dt = ∫₀ᵀ [V(0)*e^{-k1 t} + (k1*L0)/(k1 - λ)*e^{-λ t} + (k2*A0)/(k1 - μ)*e^{-μ t} - (k1*L0)/(k1 - λ)*e^{-k1 t} - (k2*A0)/(k1 - μ)*e^{-k1 t}] * A0*e^{-μ t} dtWait, that seems messy. Let me write V(t) as:V(t) = C1*e^{-k1 t} + C2*e^{-λ t} + C3*e^{-μ t}Where:C1 = V(0) - (k1*L0)/(k1 - λ) - (k2*A0)/(k1 - μ)C2 = (k1*L0)/(k1 - λ)C3 = (k2*A0)/(k1 - μ)So, V(t) = C1*e^{-k1 t} + C2*e^{-λ t} + C3*e^{-μ t}Then, KRI = ∫₀ᵀ V(t)*A(t) dt = ∫₀ᵀ [C1*e^{-k1 t} + C2*e^{-λ t} + C3*e^{-μ t}] * A0*e^{-μ t} dtMultiplying through:= A0 ∫₀ᵀ [C1*e^{-(k1 + μ) t} + C2*e^{-(λ + μ) t} + C3*e^{-2μ t}] dtSo, this integral can be split into three separate integrals:= A0 [ C1 ∫₀ᵀ e^{-(k1 + μ) t} dt + C2 ∫₀ᵀ e^{-(λ + μ) t} dt + C3 ∫₀ᵀ e^{-2μ t} dt ]Compute each integral:First integral: ∫ e^{-(k1 + μ) t} dt = [ -1/(k1 + μ) e^{-(k1 + μ) t} ] from 0 to T = [ -1/(k1 + μ) (e^{-(k1 + μ) T} - 1) ] = (1 - e^{-(k1 + μ) T})/(k1 + μ)Second integral: ∫ e^{-(λ + μ) t} dt = (1 - e^{-(λ + μ) T})/(λ + μ)Third integral: ∫ e^{-2μ t} dt = (1 - e^{-2μ T})/(2μ)So, putting it all together:KRI = A0 [ C1*(1 - e^{-(k1 + μ) T})/(k1 + μ) + C2*(1 - e^{-(λ + μ) T})/(λ + μ) + C3*(1 - e^{-2μ T})/(2μ) ]Now, substitute back C1, C2, C3:C1 = V(0) - (k1 L0)/(k1 - λ) - (k2 A0)/(k1 - μ)C2 = (k1 L0)/(k1 - λ)C3 = (k2 A0)/(k1 - μ)So, substituting:KRI = A0 [ (V(0) - (k1 L0)/(k1 - λ) - (k2 A0)/(k1 - μ))*(1 - e^{-(k1 + μ) T})/(k1 + μ) + (k1 L0)/(k1 - λ)*(1 - e^{-(λ + μ) T})/(λ + μ) + (k2 A0)/(k1 - μ)*(1 - e^{-2μ T})/(2μ) ]This is the expression for KRI.Now, to analyze how changes in λ and μ affect KRI.Looking at the expression, KRI is a sum of three terms, each involving different combinations of λ and μ.1. The first term involves (1 - e^{-(k1 + μ) T})/(k1 + μ), which is a function of μ. As μ increases, the denominator increases, making the term smaller. Also, the exponent in the numerator becomes more negative, making e^{-(k1 + μ) T} smaller, so 1 - e^{-(k1 + μ) T} approaches 1. So, the first term is a decreasing function of μ.2. The second term involves (1 - e^{-(λ + μ) T})/(λ + μ). Similarly, as λ increases, the denominator increases, making the term smaller, and the exponent becomes more negative, making the numerator approach 1. So, this term is also a decreasing function of λ.3. The third term involves (1 - e^{-2μ T})/(2μ). As μ increases, the denominator increases, and the exponent becomes more negative, making the numerator approach 1. So, this term is also a decreasing function of μ.Therefore, KRI decreases as either λ or μ increases. This suggests that higher decay rates (λ and μ) lead to lower KRI, meaning less knowledge retention.Wait, but let me think again. The KRI is the integral of V(t)*A(t). If λ and μ are higher, that means L(t) and A(t) decay faster. So, the student's vocabulary growth is based on L(t) which is decaying, and their accuracy A(t) is also decaying. So, if these decay faster, the product V(t)*A(t) might be lower over time, leading to a lower KRI.But in the expression, each term is a function that decreases as λ or μ increases. So, yes, higher λ and μ lead to lower KRI.Alternatively, if λ and μ are smaller, meaning L(t) and A(t) decay more slowly, then the terms in KRI would be larger, leading to higher KRI.So, in summary, KRI decreases as λ or μ increases, and increases as λ or μ decreases.Wait, but let me check the signs. If λ increases, the second term in KRI becomes smaller because (1 - e^{-(λ + μ) T})/(λ + μ) decreases. Similarly, if μ increases, the first and third terms decrease. So, yes, overall KRI decreases with higher λ and μ.Therefore, the Knowledge Retention Index is inversely related to the decay rates λ and μ. Higher decay rates result in lower KRI, indicating less effective retention of knowledge over time.I think that's the analysis. Let me just recap:1. Solved the differential equation using integrating factor, got V(t) in terms of exponentials.2. Plugged V(t) into the KRI integral, expanded it into three terms, each involving different exponents.3. Analyzed how each term depends on λ and μ, concluded that higher λ and μ lead to lower KRI.I think that's solid. Maybe I should check if the integrating factor was correctly applied, but I think it was. The steps seem correct.Final Answer1. The general solution for ( V(t) ) is:[boxed{V(t) = left[V(0) - frac{k_1 L_0}{k_1 - lambda} - frac{k_2 A_0}{k_1 - mu}right] e^{-k_1 t} + frac{k_1 L_0}{k_1 - lambda} e^{-lambda t} + frac{k_2 A_0}{k_1 - mu} e^{-mu t}}]2. The expression for the Knowledge Retention Index (KRI) is:[boxed{text{KRI} = A_0 left[ frac{C_1 (1 - e^{-(k_1 + mu) T})}{k_1 + mu} + frac{C_2 (1 - e^{-(lambda + mu) T})}{lambda + mu} + frac{C_3 (1 - e^{-2mu T})}{2mu} right]}]where ( C_1 = V(0) - frac{k_1 L_0}{k_1 - lambda} - frac{k_2 A_0}{k_1 - mu} ), ( C_2 = frac{k_1 L_0}{k_1 - lambda} ), and ( C_3 = frac{k_2 A_0}{k_1 - mu} ). The KRI decreases as ( lambda ) and ( mu ) increase, indicating that higher decay rates reduce knowledge retention.</think>