🧠:<think>Okay, so I'm trying to help this user who's a skeptical science journalist investigating claims about an exoplanet called Exo-Z. They have two main questions here, both involving some calculations and analysis. Let me break this down step by step.First, the user wants to calculate the minimum and maximum distances of Exo-Z from its star. I remember that for an elliptical orbit, the periapsis is the closest point and apoapsis is the farthest. The formula for periapsis is a(1 - e) and apoapsis is a(1 + e), where 'a' is the semi-major axis and 'e' is the eccentricity. Given that a is 5 AU and e is 0.6, plugging those numbers in should give me the distances. Calculating that, periapsis would be 5*(1 - 0.6) = 2 AU, and apoapsis would be 5*(1 + 0.6) = 8 AU. That's quite a range. Now, considering the possibility of life, I know that extreme temperature variations could be a problem. If the planet is too close, it might get scorched, and too far could lead to freezing. So, even though it's within the habitable zone at 5 AU, the extremes might make it hard for life to survive.Moving on to the second part, the user wants to verify the orbital period using Kepler's Third Law. I recall the formula is P² = (4π²/G(M+m)) * a³. But since the planet's mass is negligible compared to the star, we can simplify it. If the star's mass is similar to the Sun, then Kepler's law simplifies to P² = a³, where P is in Earth years and a in AU. Plugging in a = 5 AU, P² = 125, so P is the square root of 125, which is approximately 11.18 years. But the agency claims it's 4 years. That's a big discrepancy. So, either the semi-major axis is wrong, the period is wrong, or the star's mass is different. Since the user mentioned the star is similar to the Sun, the period should be around 11 years, not 4. This suggests the agency's claim might not be accurate.I should also consider the implications. If the period is off, it could mean the data was misinterpreted or there's some other factor. Maybe the exoplanet isn't in a stable orbit, or perhaps there's another body influencing it. But without more information, it's hard to say. The journalist should probably ask for more data or independent verification.Overall, the calculations show that while the exoplanet's orbit is possible, the extreme distances and the incorrect orbital period raise questions about the agency's claims. The journalist should be cautious and seek additional evidence before reporting on the habitability of Exo-Z.</think>

🧠:<think>Okay, so I have this problem to solve about my novel machine learning algorithm. It's divided into two parts: first, deriving the gradient and Hessian of a cost function, and second, proving convergence of an iterative update rule. Let me take it step by step.Starting with the first part: Gradient and Hessian Analysis. The cost function is given by:[J(theta) = sum_{i=1}^{n} left( y_i - f(x_i, theta) right)^2 + lambda |theta|_2^2]Alright, so this looks like a standard least squares cost function with an L2 regularization term. The function ( f(x_i, theta) ) is non-linear, which means that the gradient and Hessian won't be as straightforward as in linear models.First, I need to find the gradient vector ( nabla_theta J(theta) ). The gradient is the vector of partial derivatives of J with respect to each component of ( theta ). Let me denote ( f_i = f(x_i, theta) ) for simplicity.So, expanding the cost function:[J(theta) = sum_{i=1}^{n} (y_i - f_i)^2 + lambda sum_{j=1}^{m} theta_j^2]Where ( m ) is the number of parameters in ( theta ).Taking the partial derivative with respect to ( theta_k ):For the first term, using the chain rule:[frac{partial}{partial theta_k} sum_{i=1}^{n} (y_i - f_i)^2 = sum_{i=1}^{n} 2(y_i - f_i) cdot frac{partial f_i}{partial theta_k}]And for the regularization term:[frac{partial}{partial theta_k} left( lambda sum_{j=1}^{m} theta_j^2 right) = 2 lambda theta_k]So, putting it together, the gradient is:[nabla_theta J(theta) = sum_{i=1}^{n} 2(y_i - f_i) nabla_theta f_i + 2 lambda theta]Wait, that seems right. Each term in the sum is the derivative of the squared error for each data point, multiplied by the gradient of ( f_i ), and then adding the derivative of the regularization term.Now, moving on to the Hessian matrix ( H_theta J(theta) ). The Hessian is the matrix of second partial derivatives. So, for each pair ( (theta_k, theta_l) ), we need to compute:[frac{partial^2 J}{partial theta_k partial theta_l} = sum_{i=1}^{n} 2 left[ frac{partial f_i}{partial theta_k} frac{partial f_i}{partial theta_l} + (y_i - f_i) frac{partial^2 f_i}{partial theta_k partial theta_l} right] + 2 lambda delta_{kl}]Where ( delta_{kl} ) is the Kronecker delta, which is 1 if ( k = l ) and 0 otherwise.Let me verify this. The second derivative of the squared error term involves the product of the first derivatives (which gives the outer product of the gradient) and the second derivative of ( f_i ) times the residual ( (y_i - f_i) ). The regularization term contributes ( 2 lambda ) on the diagonal.So, the Hessian can be written as:[H_theta J(theta) = sum_{i=1}^{n} 2 left[ nabla_theta f_i (nabla_theta f_i)^top + (y_i - f_i) nabla_theta^2 f_i right] + 2 lambda I]Where ( I ) is the identity matrix. This makes sense because the Hessian combines the curvature information from both the model and the regularization.Okay, that seems solid. I think I have the expressions for the gradient and Hessian.Moving on to the second part: Convergence Analysis. The update rule is given by:[theta_{k+1} = theta_k - eta H_theta^{-1} J(theta_k) nabla_theta J(theta_k)]Wait, hold on. That notation is a bit confusing. Is it ( H_theta^{-1} J(theta_k) ) multiplied by ( nabla_theta J(theta_k) )? Or is it ( H_theta^{-1} ) evaluated at ( J(theta_k) ) times the gradient?Wait, actually, the way it's written is ( H_theta^{-1} J(theta_k) nabla_theta J(theta_k) ). Hmm, that seems a bit off. Typically, the update rule in Newton's method is ( theta_{k+1} = theta_k - H^{-1} nabla J ). So perhaps the notation here is a bit unclear.Wait, maybe it's supposed to be ( H_theta J(theta_k)^{-1} nabla_theta J(theta_k) ). That would make more sense, because you're multiplying the inverse Hessian by the gradient. So perhaps the update rule is:[theta_{k+1} = theta_k - eta H_theta J(theta_k)^{-1} nabla_theta J(theta_k)]But the original problem statement says ( H_theta^{-1} J(theta_k) nabla_theta J(theta_k) ). Hmm, maybe it's a typo or notation issue. Alternatively, perhaps it's ( H_theta^{-1} (J(theta_k)) nabla_theta J(theta_k) ), but that still doesn't parse well.Wait, perhaps it's ( H_theta^{-1} ) evaluated at ( theta_k ), multiplied by ( J(theta_k) ) and then by ( nabla_theta J(theta_k) ). That seems convoluted. Alternatively, maybe it's ( H_theta^{-1} nabla_theta J(theta_k) ), scaled by ( J(theta_k) ). That would be unusual.Wait, perhaps the problem statement meant:[theta_{k+1} = theta_k - eta H_theta^{-1} (nabla_theta J(theta_k)) nabla_theta J(theta_k)]But that would be a rank-1 update, which is not typical. Alternatively, perhaps the update is:[theta_{k+1} = theta_k - eta H_theta^{-1} (theta_k) nabla_theta J(theta_k)]Which is more standard, where ( H_theta^{-1} ) is the inverse Hessian at ( theta_k ), multiplied by the gradient at ( theta_k ).Given the original notation, it's a bit ambiguous, but I think the intended update rule is:[theta_{k+1} = theta_k - eta H_theta^{-1} (theta_k) nabla_theta J(theta_k)]So, I'll proceed under that assumption, as it's the standard Newton-Raphson update with a learning rate.Now, to prove convergence under the assumptions that ( H_theta J(theta) ) is positive definite and ( J(theta) ) is strictly convex.First, since ( J(theta) ) is strictly convex and the Hessian is positive definite everywhere, the function has a unique global minimum ( theta^* ). The update rule is a scaled Newton step.In Newton's method, when the Hessian is positive definite and the function is twice continuously differentiable, the method converges quadratically to the minimum provided the initial guess is sufficiently close. However, the problem doesn't specify a learning rate ( eta ), but in the update rule, it's included.Wait, actually, in standard Newton's method, the step is ( theta_{k+1} = theta_k - H^{-1} nabla J ). Here, they have an additional learning rate ( eta ). So, it's a damped Newton method, which can help with convergence even when the initial guess is not close.Given that ( H ) is positive definite, ( H^{-1} ) is also positive definite, so the direction ( -H^{-1} nabla J ) is a descent direction.Since ( J ) is strictly convex, any descent direction will lead to a decrease in the function value. Moreover, with a proper choice of ( eta ), the sequence ( theta_k ) will converge to ( theta^* ).But to formalize this, I need to use some convergence theorems.One approach is to use the properties of descent methods. Since ( J ) is strictly convex and the Hessian is positive definite, the function is also strongly convex if the Hessian is bounded below by a positive definite matrix.But in our case, the Hessian is positive definite, but not necessarily bounded away from zero in a uniform way. However, since ( J ) is strictly convex, the level sets are compact, which might help.Alternatively, since the Hessian is positive definite, the function is convex, and the Newton direction is a descent direction. If we choose ( eta ) such that the step satisfies the Armijo condition, then the sequence will converge.But the problem states to assume ( H ) is positive definite and ( J ) is strictly convex, so perhaps we can use the fact that Newton's method with exact line search converges quadratically, but with a fixed learning rate, it might converge linearly.Alternatively, since ( J ) is twice differentiable, strictly convex, and the Hessian is positive definite, we can use the convergence results for Newton's method.Wait, but in the update rule, they have ( eta H^{-1} nabla J ). So, it's a scaled Newton step. If ( eta ) is chosen such that ( 0 < eta < 2 / lambda_{text{max}} ), where ( lambda_{text{max}} ) is the largest eigenvalue of ( H ), then the step is a descent direction and the function value decreases.But since ( H ) is positive definite, all its eigenvalues are positive, so such an ( eta ) exists.Moreover, because ( J ) is strictly convex, the sequence ( theta_k ) will converge to the unique minimum ( theta^* ).Alternatively, using the fact that the update is a contraction mapping if ( eta ) is chosen appropriately.Wait, perhaps I can use the fact that the function is strongly convex if the Hessian is bounded below by a positive definite matrix. But in our case, the Hessian is only positive definite, not necessarily bounded below in a uniform way.But given that ( J ) is strictly convex and the Hessian is positive definite, the function is also coercive if it's defined over the entire space, which it is, since it's a sum of squares plus an L2 norm.Thus, the function tends to infinity as ( |theta| ) tends to infinity, so the level sets are compact. Therefore, any descent sequence will converge to the global minimum.But to make it more precise, perhaps I can use the convergence theorem for Newton's method under these assumptions.In Boyd and Vandenberghe's Convex Optimization, they discuss Newton's method for unconstrained optimization. Specifically, when the function is twice differentiable, the Hessian is Lipschitz continuous, and the initial point is sufficiently close to the optimum, Newton's method converges quadratically.But in our case, we have a fixed learning rate, so it's a damped Newton method. The convergence can be linear under certain conditions.Alternatively, since the Hessian is positive definite, the function is convex, and the gradient is Lipschitz continuous (if ( f ) is smooth enough), then the sequence will converge.Wait, perhaps I can use the fact that the function is strongly convex, but actually, strong convexity requires that the Hessian is bounded below by a positive definite matrix uniformly, which we don't necessarily have here. We only have positive definiteness, not necessarily uniform.But given that ( J ) is strictly convex and the Hessian is positive definite, and assuming that ( f ) is smooth enough (twice continuously differentiable), then the function is convex and the Newton direction is a descent direction.If we choose ( eta ) such that the step satisfies the sufficient decrease condition, then the function value decreases sufficiently, and the sequence converges.Alternatively, if ( eta ) is chosen as a constant less than 2 divided by the maximum eigenvalue of the Hessian, then the convergence is guaranteed.But since the Hessian can vary with ( theta ), it's tricky. However, under the assumption that the Hessian is positive definite everywhere, and the function is strictly convex, the sequence will converge to the global minimum.I think the key points are:1. ( J(theta) ) is strictly convex with a unique global minimum.2. The Hessian is positive definite, so the Newton direction is a descent direction.3. With a proper learning rate ( eta ), the function value decreases at each step.4. The sequence ( theta_k ) is bounded and converges to ( theta^* ).So, putting it all together, under these assumptions, the iterative update rule will converge to the global minimum.I think that's the gist of it. I might need to write it more formally, but that's the reasoning.</think>

🧠:<think>Okay, so I have this differential equation modeling personal growth, and I need to find the general solution and then the specific solution with an initial condition. Let me try to unpack this step by step.First, the differential equation is given by:[ frac{dG(t)}{dt} = a cdot M(t) cdot G(t) + b cdot E(t) cdot (1 - G(t)) ]And they've given me specific forms for M(t) and E(t):- ( M(t) = e^{-kt} )- ( E(t) = sin(omega t) )So, substituting these into the equation, it becomes:[ frac{dG}{dt} = a e^{-kt} G(t) + b sin(omega t) (1 - G(t)) ]Hmm, that looks like a linear differential equation. Let me rewrite it in the standard linear form:[ frac{dG}{dt} + P(t) G(t) = Q(t) ]To do that, I need to collect the terms involving G(t) on the left side. Let me move the terms around:First, expand the right-hand side:[ frac{dG}{dt} = a e^{-kt} G(t) + b sin(omega t) - b sin(omega t) G(t) ]Now, bring all the G(t) terms to the left:[ frac{dG}{dt} - a e^{-kt} G(t) + b sin(omega t) G(t) = b sin(omega t) ]Factor out G(t):[ frac{dG}{dt} + left( -a e^{-kt} + b sin(omega t) right) G(t) = b sin(omega t) ]So, in standard linear form, this is:[ frac{dG}{dt} + P(t) G(t) = Q(t) ]Where:- ( P(t) = -a e^{-kt} + b sin(omega t) )- ( Q(t) = b sin(omega t) )Alright, so to solve this linear differential equation, I need an integrating factor, μ(t), which is given by:[ mu(t) = e^{int P(t) dt} ]Let me compute that integral:[ int P(t) dt = int left( -a e^{-kt} + b sin(omega t) right) dt ]Breaking this into two integrals:1. ( int -a e^{-kt} dt )2. ( int b sin(omega t) dt )Let's compute each separately.First integral:[ int -a e^{-kt} dt ]The integral of ( e^{-kt} ) is ( -frac{1}{k} e^{-kt} ), so multiplying by -a:[ -a cdot left( -frac{1}{k} e^{-kt} right) = frac{a}{k} e^{-kt} ]Second integral:[ int b sin(omega t) dt ]The integral of ( sin(omega t) ) is ( -frac{1}{omega} cos(omega t) ), so multiplying by b:[ b cdot left( -frac{1}{omega} cos(omega t) right) = -frac{b}{omega} cos(omega t) ]Putting both integrals together:[ int P(t) dt = frac{a}{k} e^{-kt} - frac{b}{omega} cos(omega t) + C ]But since we're computing the integrating factor, we can ignore the constant of integration, C. So,[ mu(t) = e^{frac{a}{k} e^{-kt} - frac{b}{omega} cos(omega t)} ]Hmm, that looks a bit complicated. Let me write it as:[ mu(t) = e^{frac{a}{k} e^{-kt}} cdot e^{-frac{b}{omega} cos(omega t)} ]So, the integrating factor is the product of two exponentials. I don't think this simplifies much further, so I'll keep it as is.Now, the solution to the linear differential equation is given by:[ G(t) = frac{1}{mu(t)} left( int mu(t) Q(t) dt + C right) ]Where C is the constant of integration. Let's plug in μ(t) and Q(t):First, μ(t) is:[ e^{frac{a}{k} e^{-kt} - frac{b}{omega} cos(omega t)} ]And Q(t) is:[ b sin(omega t) ]So, the integral becomes:[ int mu(t) Q(t) dt = int b sin(omega t) cdot e^{frac{a}{k} e^{-kt} - frac{b}{omega} cos(omega t)} dt ]This integral looks quite challenging. Let me see if I can simplify it or find a substitution.Looking at the exponent:[ frac{a}{k} e^{-kt} - frac{b}{omega} cos(omega t) ]Let me denote:Let me consider substitution for the exponent. Let me set:Let me denote ( u = -frac{b}{omega} cos(omega t) ). Then, du/dt = ( frac{b}{omega} cdot omega sin(omega t) ) = ( b sin(omega t) ). Hmm, interesting. So, du = ( b sin(omega t) dt ). That seems promising because the integral has ( b sin(omega t) dt ).But wait, the exponent also has ( frac{a}{k} e^{-kt} ). So, the exponent is the sum of two terms: one involving ( e^{-kt} ) and another involving ( cos(omega t) ). So, the integral is:[ int e^{frac{a}{k} e^{-kt} + u} cdot frac{du}{b} ]Wait, hold on. Let me write it again.Let me denote:Let me set ( u = -frac{b}{omega} cos(omega t) ). Then, du = ( b sin(omega t) dt ), so ( dt = frac{du}{b sin(omega t)} ). But in the integral, we have ( b sin(omega t) dt ), which is just du.So, the integral becomes:[ int e^{frac{a}{k} e^{-kt} + u} du ]Wait, but ( frac{a}{k} e^{-kt} ) is still a function of t, not u. So, unless I can express ( e^{-kt} ) in terms of u, which I don't think is straightforward.Alternatively, perhaps I can split the exponent:[ e^{frac{a}{k} e^{-kt} - frac{b}{omega} cos(omega t)} = e^{frac{a}{k} e^{-kt}} cdot e^{-frac{b}{omega} cos(omega t)} ]So, the integral becomes:[ int b sin(omega t) cdot e^{frac{a}{k} e^{-kt}} cdot e^{-frac{b}{omega} cos(omega t)} dt ]Which is:[ e^{frac{a}{k} e^{-kt}} cdot int b sin(omega t) cdot e^{-frac{b}{omega} cos(omega t)} dt ]Now, focusing on the integral:[ int b sin(omega t) cdot e^{-frac{b}{omega} cos(omega t)} dt ]Let me make a substitution here. Let me set:Let ( v = -frac{b}{omega} cos(omega t) )Then, dv/dt = ( frac{b}{omega} cdot omega sin(omega t) ) = ( b sin(omega t) )So, dv = ( b sin(omega t) dt )Therefore, the integral becomes:[ int e^{v} dv = e^{v} + C = e^{-frac{b}{omega} cos(omega t)} + C ]So, putting it all back together:The integral ( int mu(t) Q(t) dt ) is:[ e^{frac{a}{k} e^{-kt}} cdot left( e^{-frac{b}{omega} cos(omega t)} + C right) ]Wait, hold on. Let me retrace:We had:[ int b sin(omega t) cdot e^{frac{a}{k} e^{-kt}} cdot e^{-frac{b}{omega} cos(omega t)} dt ]Which I rewrote as:[ e^{frac{a}{k} e^{-kt}} cdot int b sin(omega t) cdot e^{-frac{b}{omega} cos(omega t)} dt ]Then, substituting ( v = -frac{b}{omega} cos(omega t) ), so that ( dv = b sin(omega t) dt ), the integral becomes:[ e^{frac{a}{k} e^{-kt}} cdot int e^{v} dv = e^{frac{a}{k} e^{-kt}} cdot e^{v} + C ]Substituting back for v:[ e^{frac{a}{k} e^{-kt}} cdot e^{-frac{b}{omega} cos(omega t)} + C ]Therefore, the integral is:[ e^{frac{a}{k} e^{-kt} - frac{b}{omega} cos(omega t)} + C ]So, going back to the solution formula:[ G(t) = frac{1}{mu(t)} left( int mu(t) Q(t) dt + C right) ]We have:[ mu(t) = e^{frac{a}{k} e^{-kt} - frac{b}{omega} cos(omega t)} ]And:[ int mu(t) Q(t) dt = e^{frac{a}{k} e^{-kt} - frac{b}{omega} cos(omega t)} + C ]Therefore, plugging into G(t):[ G(t) = frac{1}{e^{frac{a}{k} e^{-kt} - frac{b}{omega} cos(omega t)}} left( e^{frac{a}{k} e^{-kt} - frac{b}{omega} cos(omega t)} + C right) ]Simplify this expression:The exponential in the denominator cancels with the exponential in the numerator:[ G(t) = 1 + C e^{-left( frac{a}{k} e^{-kt} - frac{b}{omega} cos(omega t) right)} ]Wait, let me verify:[ frac{1}{mu(t)} cdot mu(t) = 1 ]And:[ frac{1}{mu(t)} cdot C = C e^{-left( frac{a}{k} e^{-kt} - frac{b}{omega} cos(omega t) right)} ]Yes, that's correct.Therefore, the general solution is:[ G(t) = 1 + C e^{-frac{a}{k} e^{-kt} + frac{b}{omega} cos(omega t)} ]Wait, let me make sure about the signs:The exponent in μ(t) is ( frac{a}{k} e^{-kt} - frac{b}{omega} cos(omega t) ), so when we take the reciprocal, it becomes negative:[ e^{-left( frac{a}{k} e^{-kt} - frac{b}{omega} cos(omega t) right)} = e^{-frac{a}{k} e^{-kt} + frac{b}{omega} cos(omega t)} ]Yes, that's correct.So, the general solution is:[ G(t) = 1 + C e^{-frac{a}{k} e^{-kt} + frac{b}{omega} cos(omega t)} ]Alright, that seems like the general solution. Now, moving on to part 2, where we have the initial condition ( G(0) = G_0 ). Let's find the specific solution.First, let's compute G(0). Plugging t=0 into the general solution:[ G(0) = 1 + C e^{-frac{a}{k} e^{0} + frac{b}{omega} cos(0)} ]Simplify the exponents:- ( e^{0} = 1 )- ( cos(0) = 1 )So,[ G(0) = 1 + C e^{-frac{a}{k} + frac{b}{omega}} ]But we know that ( G(0) = G_0 ), so:[ G_0 = 1 + C e^{-frac{a}{k} + frac{b}{omega}} ]Solving for C:[ C e^{-frac{a}{k} + frac{b}{omega}} = G_0 - 1 ][ C = (G_0 - 1) e^{frac{a}{k} - frac{b}{omega}} ]Therefore, plugging back into the general solution:[ G(t) = 1 + (G_0 - 1) e^{frac{a}{k} - frac{b}{omega}} cdot e^{-frac{a}{k} e^{-kt} + frac{b}{omega} cos(omega t)} ]We can combine the exponents:[ G(t) = 1 + (G_0 - 1) e^{frac{a}{k} - frac{b}{omega} - frac{a}{k} e^{-kt} + frac{b}{omega} cos(omega t)} ]Let me factor out the constants:Notice that ( frac{a}{k} - frac{a}{k} e^{-kt} = frac{a}{k} (1 - e^{-kt}) )Similarly, ( -frac{b}{omega} + frac{b}{omega} cos(omega t) = frac{b}{omega} ( cos(omega t) - 1 ) )So, the exponent becomes:[ frac{a}{k} (1 - e^{-kt}) + frac{b}{omega} ( cos(omega t) - 1 ) ]Therefore, the solution can be written as:[ G(t) = 1 + (G_0 - 1) e^{frac{a}{k} (1 - e^{-kt}) + frac{b}{omega} ( cos(omega t) - 1 )} ]This seems to be the specific solution satisfying the initial condition.Let me just double-check the steps to make sure I didn't make any mistakes.1. Started with the differential equation, substituted M(t) and E(t).2. Rewrote it in standard linear form, identified P(t) and Q(t).3. Computed the integrating factor μ(t) by integrating P(t), which resulted in an exponential of two terms.4. Set up the integral for the solution, noticed that part of the integral could be simplified with substitution, leading to the integral simplifying nicely.5. Plugged back into the solution formula, simplified, and found the general solution.6. Applied the initial condition to solve for the constant C, substituted back, and simplified the exponent.Everything seems to check out. I don't see any algebraic errors, and the substitution steps make sense. The integral was a bit tricky, but breaking it down and using substitution for the part involving sin and cos worked out.So, I think this is the correct solution.Final Answer1. The general solution is:[ boxed{G(t) = 1 + C e^{-frac{a}{k} e^{-kt} + frac{b}{omega} cos(omega t)}} ]2. The specific solution with the initial condition ( G(0) = G_0 ) is:[ boxed{G(t) = 1 + (G_0 - 1) e^{frac{a}{k} (1 - e^{-kt}) + frac{b}{omega} (cos(omega t) - 1)}} ]</think>

🧠:<think>Alright, so I have two sub-problems to solve here, both related to Nikola's interests in Serbian geography and culture through mathematics. Let me tackle them one by one.Starting with Sub-problem 1: It's a differential equation modeling the flow rate of the Danube river. The equation given is:[ frac{dF(t)}{dt} + 2F(t) = 5sin(t) ]And the initial condition is ( F(0) = 3 ). Okay, so this is a linear first-order ordinary differential equation (ODE). I remember that the standard approach to solving such equations is using an integrating factor.First, let me write the equation in standard form. It's already in the form:[ frac{dF}{dt} + P(t)F = Q(t) ]where ( P(t) = 2 ) and ( Q(t) = 5sin(t) ). Since ( P(t) ) is a constant, the integrating factor ( mu(t) ) is given by:[ mu(t) = e^{int P(t) dt} = e^{int 2 dt} = e^{2t} ]Okay, so I multiply both sides of the differential equation by ( e^{2t} ):[ e^{2t} frac{dF}{dt} + 2e^{2t}F = 5e^{2t}sin(t) ]The left-hand side should now be the derivative of ( F(t)e^{2t} ). Let me check:[ frac{d}{dt}[F(t)e^{2t}] = F'(t)e^{2t} + 2F(t)e^{2t} ]Yes, that's exactly the left-hand side. So, integrating both sides with respect to ( t ):[ int frac{d}{dt}[F(t)e^{2t}] dt = int 5e^{2t}sin(t) dt ]So, the left side simplifies to ( F(t)e^{2t} ). Now, I need to compute the integral on the right side. Hmm, integrating ( 5e^{2t}sin(t) ) dt. That seems like it will require integration by parts or maybe using a formula for integrating products of exponentials and trigonometric functions.I recall that for integrals of the form ( int e^{at}sin(bt) dt ), the integral can be found using a standard technique. Let me set ( a = 2 ) and ( b = 1 ) here.The formula is:[ int e^{at}sin(bt) dt = frac{e^{at}}{a^2 + b^2}(asin(bt) - bcos(bt)) + C ]So, applying this formula with ( a = 2 ) and ( b = 1 ):[ int e^{2t}sin(t) dt = frac{e^{2t}}{2^2 + 1^2}(2sin(t) - 1cos(t)) + C ][ = frac{e^{2t}}{5}(2sin(t) - cos(t)) + C ]Therefore, the integral on the right side is:[ 5 times frac{e^{2t}}{5}(2sin(t) - cos(t)) + C ][ = e^{2t}(2sin(t) - cos(t)) + C ]So, putting it all together, we have:[ F(t)e^{2t} = e^{2t}(2sin(t) - cos(t)) + C ]Now, divide both sides by ( e^{2t} ):[ F(t) = 2sin(t) - cos(t) + Ce^{-2t} ]Now, apply the initial condition ( F(0) = 3 ). Let's plug in ( t = 0 ):[ F(0) = 2sin(0) - cos(0) + Ce^{-0} ][ 3 = 0 - 1 + C ][ 3 = -1 + C ][ C = 4 ]So, the solution is:[ F(t) = 2sin(t) - cos(t) + 4e^{-2t} ]Let me double-check my work. I used the integrating factor correctly, multiplied through, and recognized the integral formula. The integration by parts formula gave me the right integral, and substituting back seems correct. The initial condition was applied properly, leading to ( C = 4 ). So, I think this is correct.Moving on to Sub-problem 2: Nikola models the river's path as a complex function ( z(t) = x(t) + iy(t) ), where ( x(t) = e^{t/2}cos(t) ) and ( y(t) = e^{t/2}sin(t) ). We need to find the total length of the river's path from ( t = 0 ) to ( t = 2pi ).I remember that the length of a parametric curve ( z(t) = x(t) + iy(t) ) from ( t = a ) to ( t = b ) is given by:[ L = int_{a}^{b} sqrt{left(frac{dx}{dt}right)^2 + left(frac{dy}{dt}right)^2} dt ]So, I need to compute ( frac{dx}{dt} ) and ( frac{dy}{dt} ), square them, add, take the square root, and integrate from 0 to ( 2pi ).First, let's compute ( frac{dx}{dt} ) and ( frac{dy}{dt} ).Given ( x(t) = e^{t/2}cos(t) ), so:Using the product rule:[ frac{dx}{dt} = frac{d}{dt}[e^{t/2}] cos(t) + e^{t/2} frac{d}{dt}[cos(t)] ][ = left(frac{1}{2}e^{t/2}right)cos(t) - e^{t/2}sin(t) ][ = frac{1}{2}e^{t/2}cos(t) - e^{t/2}sin(t) ][ = e^{t/2}left(frac{1}{2}cos(t) - sin(t)right) ]Similarly, for ( y(t) = e^{t/2}sin(t) ):[ frac{dy}{dt} = frac{d}{dt}[e^{t/2}] sin(t) + e^{t/2} frac{d}{dt}[sin(t)] ][ = left(frac{1}{2}e^{t/2}right)sin(t) + e^{t/2}cos(t) ][ = frac{1}{2}e^{t/2}sin(t) + e^{t/2}cos(t) ][ = e^{t/2}left(frac{1}{2}sin(t) + cos(t)right) ]Okay, so now we have:[ frac{dx}{dt} = e^{t/2}left(frac{1}{2}cos(t) - sin(t)right) ][ frac{dy}{dt} = e^{t/2}left(frac{1}{2}sin(t) + cos(t)right) ]Now, let's compute ( left(frac{dx}{dt}right)^2 + left(frac{dy}{dt}right)^2 ).First, square both derivatives:[ left(frac{dx}{dt}right)^2 = e^{t} left(frac{1}{2}cos(t) - sin(t)right)^2 ][ left(frac{dy}{dt}right)^2 = e^{t} left(frac{1}{2}sin(t) + cos(t)right)^2 ]So, adding them together:[ left(frac{dx}{dt}right)^2 + left(frac{dy}{dt}right)^2 = e^{t} left[ left(frac{1}{2}cos(t) - sin(t)right)^2 + left(frac{1}{2}sin(t) + cos(t)right)^2 right] ]Let me compute the expression inside the brackets:Let me denote ( A = frac{1}{2}cos(t) - sin(t) ) and ( B = frac{1}{2}sin(t) + cos(t) ). Then, we need to compute ( A^2 + B^2 ).Compute ( A^2 ):[ A^2 = left(frac{1}{2}cos(t) - sin(t)right)^2 ][ = left(frac{1}{2}cos(t)right)^2 - 2 times frac{1}{2}cos(t) times sin(t) + (sin(t))^2 ][ = frac{1}{4}cos^2(t) - cos(t)sin(t) + sin^2(t) ]Compute ( B^2 ):[ B^2 = left(frac{1}{2}sin(t) + cos(t)right)^2 ][ = left(frac{1}{2}sin(t)right)^2 + 2 times frac{1}{2}sin(t) times cos(t) + (cos(t))^2 ][ = frac{1}{4}sin^2(t) + sin(t)cos(t) + cos^2(t) ]Now, add ( A^2 + B^2 ):[ A^2 + B^2 = left( frac{1}{4}cos^2(t) - cos(t)sin(t) + sin^2(t) right) + left( frac{1}{4}sin^2(t) + sin(t)cos(t) + cos^2(t) right) ]Let's combine like terms:- ( frac{1}{4}cos^2(t) + cos^2(t) = frac{5}{4}cos^2(t) )- ( -cos(t)sin(t) + sin(t)cos(t) = 0 )- ( sin^2(t) + frac{1}{4}sin^2(t) = frac{5}{4}sin^2(t) )So, altogether:[ A^2 + B^2 = frac{5}{4}cos^2(t) + frac{5}{4}sin^2(t) ][ = frac{5}{4}(cos^2(t) + sin^2(t)) ][ = frac{5}{4}(1) ][ = frac{5}{4} ]Wow, that simplified nicely! So, the expression inside the brackets is ( frac{5}{4} ). Therefore, the integrand becomes:[ sqrt{e^{t} times frac{5}{4}} = sqrt{frac{5}{4}e^{t}} = frac{sqrt{5}}{2}e^{t/2} ]So, the integral for the length ( L ) is:[ L = int_{0}^{2pi} frac{sqrt{5}}{2}e^{t/2} dt ]This is a straightforward integral. Let's factor out the constants:[ L = frac{sqrt{5}}{2} int_{0}^{2pi} e^{t/2} dt ]Let me compute the integral ( int e^{t/2} dt ). The antiderivative of ( e^{kt} ) is ( frac{1}{k}e^{kt} ), so here ( k = frac{1}{2} ):[ int e^{t/2} dt = 2e^{t/2} + C ]Therefore, evaluating from 0 to ( 2pi ):[ int_{0}^{2pi} e^{t/2} dt = 2e^{(2pi)/2} - 2e^{0} ][ = 2e^{pi} - 2(1) ][ = 2(e^{pi} - 1) ]So, plugging back into the expression for ( L ):[ L = frac{sqrt{5}}{2} times 2(e^{pi} - 1) ][ = sqrt{5}(e^{pi} - 1) ]Therefore, the total length of the river's path in Serbia is ( sqrt{5}(e^{pi} - 1) ).Let me just verify my steps. I computed the derivatives correctly using the product rule, squared them, added, and simplified. The cross terms canceled out, leaving me with a constant multiple of ( e^{t} ). Then, taking the square root simplified it further. The integral was straightforward, and the substitution gave me the result. Seems solid.So, summarizing:Sub-problem 1 solution: ( F(t) = 2sin(t) - cos(t) + 4e^{-2t} )Sub-problem 2 solution: The total length is ( sqrt{5}(e^{pi} - 1) )Final AnswerSub-problem 1: boxed{2sin(t) - cos(t) + 4e^{-2t}}Sub-problem 2: boxed{sqrt{5}(e^{pi} - 1)}</think>