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Recently Asked Differential equations Questions

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Q1:3. The temperature of an object in a bath at a time t is governed by the differential equation: dT dt -k(T-25t)=0, the change in the temperature is an increase of 3°F and the temperature of the object is 12°F. a. Find k. b. Determine the temperature of the object at time t.See Answer
Q2:1. You will be constructing a slope field(a qualitative analysis) for the following scenario. Suppose that the population of weasels in a particular park is modeled by the differential equation (in thousands of weasels): dP .0153P(P-32) dt a. Find and draw any equilibrium solutions. b. Draw the isoclines for P=10 and P-40. Make sure you indicate the slopes for the isoclines. c. Indicate the areas where the solution curves are increasing and decreasing. d. Find the changes in concavity. Then determine the concavity of the functions e. Describe what will happen to the population if the population starts at 20,000 weasels. Draw a solution curve to demonstrate your findings.See Answer
Q3:6. Consider the following system of equations: 3x1-x3 + 2x4x5 = 1 x13x2+x33x4 + 2x5 = −1 - 4x1-2x2-3x3 + 6x4 - x5 = 5 x4 + 4x5 = -2 a. Construct the matrix equation and the augmented matrix for the above system. b. The reduced row echelon form for the system is: [1 0 0 0 0 100 -16/7 33/7 -2/7 -2/7 0 0 1 0 15/7-41/7 Lo 0 0 1 4 -2 Write the solution to the system in parametric and vector form. Then describe the solution geometrically.See Answer
Q4:5. Use Gaussian elimination to row reduce the following matrix to row echelon form BY HAND (label as the ref) and reduced row echelon form (label as the rref), then give the rank(A). Make sure that you include your row operations: 630 -1 2 -5 303 A=0See Answer
Q5: Name Problem Set - Population Dynamics Perhaps the most successful conjunction of math and biology has been in the study of ecology and evolution. This problem set focuses on one aspect of population biology: population size. We will try to get an understanding, via mathematics, of how populations change over time. THE EXPONENTIAL EQUATION Let's think of a population of individuals that reproduces asexually via fission. It could be yeast or bacteria, it doesn't really matter what it is, just what it does. We start with 1 individual at time (t) = 0. 1. If after every hour, each individual divides into two individuals, list how many individuals there will be at each time interval: time, t (h) 0 1 2 3 4 5 6 #indiv (N) But tables like this would get very unwieldy after even one day of growth. So, let's try to simplify the relationship with a function that tells us how many individuals there are at any interval. 2. Complete the following function by giving 2 an exponent (HINT: it will be a variable): N₁= 2- Throughout this problem set, N will refer to population size and N, is population at time t. We're not going to worry about what the units of time are, however. Now we will try to make this a more general equation that can be applied to populations with different growth rates. It is possible to represent 2 as any combination of numbers, e.g., : 2=1+1,=√ =√√2² 148 So 74 we can begin to make the function in question 2 more general by substituting a new combination of values for 2. Because it is an exponential relationship, we'll try using e and to make it equal to 2, we will raise it to the power of 0.693. For this population: N₁ = 2¹ = e0.693t We're still not at the general equation, however, because we haven't really changed the equation, just its form. It's still just 2¹. Now we said that this population doubles every hour. But not every population does. So, which part of this equation is responsible for making it double every hour? 0.693. e is a constant, so if we want the population's rate of change to be something different from doubling every hour, we should change the exponent. To make the exponent a variable, let's replace 0.693 (which applies just to this population) with r: rt N₁ = e't So now depending on the population, we can have r = whatever is appropriate: 0.693 in these yeast or 0.214 in some flies or 0.014 in elephants, for example. But there's one more thing! Not all populations begin at the same size. If you have a sexually reproducing population, you definitely need more than one individual, and populations can start with many more than one. So, where should we add the value for the population at time 0? Remember, that in this example No = 1, therefore, our final, general equation becomes: N₁ = Noert This equation demonstrates the size of a population as a function of time and is called the exponential equation. (NOTE: This is the "integrated form" of the equation. We will also look at the differential form later.) No is a constant it is a constant for any one population, but it will differ among populations. It is the variable time in whatever units are appropriate. We'll just consider it unitless. r is perhaps the most important component of the equation, and it determines the rate at which a population grows over time. ris called the intrinsic rate of natural increase, and it is a constant for each population that represents essentially how quickly a population can grow when there are no limitations. POPULATION DEMOGRAPHICS - integrated form Population size will change based on the number of (3.) minus the number of (4.) (Ignore immigration and emigration.) r is essentially the first answer minus the second answer per each individual under conditions that are optimal for population growth. 5. Give an example of a biological factor that would result in a difference in r between two populations. All else being equal, there would be a difference in r between two populations, if one population had a higher/larger than the other. Now this gives us the growth of a population over time. But what we are really interested in is taking any given point in time and predicting how quickly the population will grow at that point in time. So let's make a differential equation relative to time. No 6. Differentiate the exponential equation with respect to time. The result is below, just show how to get there. (HINT: before differentiating, start by taking the natural log of both sides. If you get to the end and it N doesn't look like the final answer, then it might help to remember that: ert 7. First, let's plot the integrated form of the exponential equation on the axes below. Plot three different functions, each with N(0) = 1 but with three different r's: 0.1, 0.2 and 0.5. You do not need to show your work, but of course you must label the functions: 1000 dN(t) dt means. It Now what does this tell us about population size and how it changes? Keep in mind what means the change in population size over a given time. Most of the time we think about dN (t) as dt⇒0. That is, at any instant, how quickly the population is growing. dt 900 800 700 dN, dt 600 €500 400 300 200 100 0 0 =rN₁ - differentiated form 10 = 20 30 40 t 50 60 70 80 8. OK, so what do these three curves tell you about r? As r increases, how do the curves change? dN₁ dt 9. Another approach to problem 6 above is to ask what happens to the differential equation that you derived in question 6.) as r increases? (Think about 10. What does the exponential equation (in its integrated/non-differential form) predict about population size after a long time? Solve for the limit as t→∞. Show your work. 11. Does your answer to 10 make biological sense? Explain. THE LOGISTIC EQUATION So let's agree that the exponential equation is unrealistic in at least one way: there are no populations that are infinitely large. Real populations can generally be grouped into two major groups: (1) those that are unstable and fluctuate wildly and (2) those that are relatively stable. The ones that fluctuate wildly tend to have booms and busts. They increase rapidly and then quickly die back. Locusts that swarm and destroy crops are a good example. The populations that are more stable are the ones we're going to discuss now. The idea is that in a given habitat, each population has a carrying capacity. This is sort of a ceiling of stable population size. That is, there is a certain amount of room and food and water and other resources in a habitat and these combine to determine the number of individuals that can live there sustainably. That population size that is the maximum sustainable population size is the carrying capacity, symbolized as K. dN₁ t = dt So, how do make the model better fit reality and incorporate K into our equation? We want the population size not to change when it is equal to K. In other words, when N = K, 0. Because we are talking about changes, we should start with the differential form of the equation. of K by adding to the differential version of the exponential equation: K-N₁ So, we can incorporate the effect dN₁ =rN₁ dt K Set N = K and see what you get for This equation is called the logistic equation and when the integrated form is graphed it produces the logistic growth curve. Note that this form here is the differentiated form. dN₁ t dt The version above is the more common way to see this equation, but if you want to see how many individuals there are in a population at any time, you would need the integrated form: Ket - integrated form N₁ Note that we use No here. It's a constant just like in the exponential equation. A) r= 0.1, B) r = 0.1, C) r = 0.2, 12. On the graph to the right, plot the integrated form of the logistic No= 1, No= 250, No=1, e + equation (using whatever means you wish) for the three given populations. Label them, of course (3 pts.): K = 500 K = 500 K = 500 K-No No - differentiated form N(t) 550 500 450 400 350 300 250 200 150 100 50 0 0 10 20 30 40 50 60 70 80 90 100 tSee Answer
Q6:Question 7 Solve the initial value problem: +6 6 dy + 18y = cost — sint, y(0) = 0, y' (0) = 2 - 23 325 cost +973 sin t) e-3t. cost - 32 sint 23 325 (-3 cos 3t - 59 sin 3t) e−³ + 23 325 23 325 Cos 3t+ 23 325 59 sin 3t) e-³t+ 23 325 O(-32 cost +593 sint) e-³t - + 23 325 11 cost + 2 sint cost 11 325 sin t cost-sintSee Answer
Q7:Question 4 In a mass-spring system where mass m = 1 kg, spring constant k = 2, and damping c = 0, the position of the mass x (t) can be modelled by the differential equation below: x" + 4x = F(t) F(t) is the external force applied to the mass and F(t) = {1tf0<t<1 if-1<t≤0 (a) Compute and plot the Fourier series of F(t) to verify your answer. (b) Hence find and sketch the steady period solution of x (t). (14 marks) (14 marks)See Answer
Q8:The self-study problem from the last lecture showed that the basic SIR model can be simplified to: du dt dv . = -Rouv (Rou-1)v = V where u is the proportion of hosts that are susceptible • v is the proportion of hosts that are infected (and infectious) w is the proportion of hosts that are recovered • Ro=BN is the basic reproduction number 3 is the ("pairwise") transmission rate is the recovery rate Why do we not need the third equation? Basic reproduction number, Ro The basic reproduction number ¹, Ro, is the most important quantity in mathematical epidemiology. The basic reproduction number tells you the average number of secondary infections produced by one infected individual in an otherwise susceptible population. Simply put, it is the ratio of the rate at which new infections are produced by each infected host when the entire population is susceptible (3N), to the rate at which infections are lost (7). ¹Note that Ro is not a rate. It is dimensionless. Sometimes it is referred to as the basic reproduction ratio, but should never be referred to as the basic reproduction rate!/nreproduction ratio, but should never be referred to as the basic reproduction rate! Algebraic analysis Find and determine the stability of any equilibria. What do you notice? Algebraic analysis Let's tackle this another way. Suppose initially that almost everyone is susceptible (u(0) ≈ 1). When can an epidemic take off (the epidemic threshold)? 5/nGeometric analysis What would a phase plane look like for this system? Note that u + v ≤ 1, so we have to solve the system on a triangle. Q: What are the nullclines and direction field? Geometric analysis Let's sketch two phase planes to understand the dynamics of the model. 7/nNote that So the host population size (density) must be greater than N, for an epidemic to occur. Q: How do the transmission rate, population size, and recovery rate affect whether an epidemic will occur or not? Final size of an epidemic How many people will get infected during an epidemic? Will everyone become infected? For this SIR model, any point on v = 0 is a disease-free equilibrium (DFE) and the epidemic eventually burns out, leaving the population in one of these states. The final size of an epidemic is the total number of individuals who become infected over the course of the epidemic. If all individuals are initially susceptible, it is given by the number of individuals eventually in the recovered class. 10/nFinal size of an epidemic Recall that Then Idea: If we solve this equation for w as a function of u with initial condition w = wo =0 when u = Uo = 1 then we find a relationship between w and u that is valid everywhere on the solution trajectory that starts arbitrarily close to the DFE uo = (1,0,0). Self-study problem 1) Integrated from (uo, Wo) = (1,0) to a general point (u₁, W₁) and rearrange to find equations for w₁ in terms of u₁ and for u₁ in terms of W₁. 2) Using the fact that the epidemic eventually burns out ( v→0 as t→∞), and your equation for u above to shown that the final size of the epidemic, w₁, satisfies 1- Wy=e-Rowy 3) Sketch 1- w and e-Row for Ro <1 and Ro> 1 to show that at most there is a single root (solution) to the equation above, giving the final size of an epidemic. 11See Answer
Q9:Problem 3. (6 pts) A 2 inch, schedule-40 steel pipe carries saturated steam at 60 psia through a laboratory that is 60 ft long. (a) Use the shell-balance approach to derive the differential equation that governs heat conduction through the insulated pipe. Derive two equations, one in terms of heat flux (q) and the other in terms of temperature (T). (3 pts) (b) The pipe is insulated with 1.5 inches of 85% magnesia that costs $0.75 per foot. How long must the steam line be in service to justify the insulation cost if the heating cost for the steam is $0.68 per 105 Btu? The outside surface convective heat-transfer coefficient may be taken as 5 Btu/hr ft² °F. (3 pts)See Answer
Q10:Problem 2. (6 pts) A furnace wall is to be designed to transmit a maximum heat flux of 200 Btu/hr ft² of wall area. The inside and outside wall temperatures are to be 2000 and 300°F, respectively. (a) Use the shell-balance approach to derive the differential equation that governs heat conduction through an insulated wall. Derive two equations, one in terms of heat flux (q) and the other in terms of temperature (T). (b) Determine the most economical arrangement of bricks measuring 9 by 4.5 by 3 inches if the bricks are made from one of two materials, one with a k of 0.44 Btu/hr ft °F and a maximum usable temperature of 1500°F and the other with a k of 0.94 Btu/hr ft °F and a maximum usable temperature of 2200°F. Bricks made of each material cost the same amount and may be laid in any manner. Problem 3. (6 pts)See Answer
Q11:Problem 1. (6 pts) A nuclear reactor has spherical fuel elements that produce heat according to the relationship à -à-[¹-(3)] where is the volumetric heat generation rate, kW/m³, and ro is the radius of the sphere. (a) Use the shell-balance approach and derive the differential equation that governs the temperature profile within the sphere. (3 pts) (b) Using this differential equation solve the expression for the temperature difference between the center and surface of the sphere (i.e., Tecenter-Tsurface). (3 pts)See Answer
Q12:7. Direct Solution of Resonance Condition my + ky Fcoswt ; " 2 F y" + w²y = coswt, forcing function at frequency of w. m y = y + y₂ = C₁sinwt + C₂coswt + Y #(30) Р Use Variation of Parameters to obtain Y. Formula Solution (See B/DP for derivation) Y //* = natural frequency = w#(27) == y₁ S y₂coswtdt Wy₁v₂ + S Y₂ y,coswtdt -#(47) 11/₂2 - sinut coswtcoswtdt wy,₂ W|y₁v₂| = |V₁ V₂ V₁¹ V₁₂ | = v₁y₁₂ — y₂y₁#(49) + coswt f. sinwtcoswtdt w|y -#(48) = (sinwt) (- wsinwt) - coswt (wcoswt) =- wwt + wt] = - w # (50) Note: Wy₁, y₂ #0 for y, y to be independent solutions (not proportional to each other.) (29)/nENGG 614 2023 Ninth Class 10-31-23 Revised of 10-30-23, 8:30pm 24/4/1 Y F/m But I sinwtcoswtdt → Y = sinut fwtdt W sinut W sinut 6 S But sin2wt=2sinotcoswt + udu W Ftsinuat 2mw + Ftsinut 2mw costat W 2 cos wt = 1+cos2wat 2 sin2wt 4ய 2w But 2)dt. 4 ய f sinwtcoswtdt # (51) utorint 2² 1+cos2wat 2 cosut (sinust) (sin2ust) 2 So -;fsinwtcoswtdt = m cont cosut W -f sinwtcoswtdt#(52) So y(t) = C₁y₁ +₂₂ - sinwtcoswtdt + wit 2w costsin wat 2w Ftsinust 2m -#(54) Equation 54 is identical to equation (38) as before for final value of y(t) when 0→w. But we still need to evaluate, C₁, C₂ for rest conditions, i.e. for y₁ = y₁ = 0 Ftsinat 2mw 772 #(53) -#(55) Page 17 of 22/nENGG 614 2023 Ninth Class 10-31-23 Revised of 10-30-23, 8:30pm Let y(0) = y(0) = 0 y'(t) = y(t) = C,sinat + C₂coswt + y(0) = 0 = C₂.0 + C₂1 + C₂ = 0#(57) wC coswt - C₂wsinwt + 0 = wC₁ - 0 + 2mw F 2mw y(t) = C₁ = 0#(59) Ftsinuat 2mw So the complete solution for system initially at rest is: Ftsinust 2mw -#(56) F.0.0 2mw -(0 + w.0.1) -(sinwt + wtcoswt) As before by undetermined coefficients and use of L'Hopital's Rule (see equation (38) page 13.) Note: If you generate arbitrary constants of integration when evaluating the integrals in the Variation of Parameters formula, you will get the complete general solution; i.e., C₁y₁ + C₂y₂ + Y, not just the Y. But we already know y, so no need to repeat it. 8. Damped Free Vibration, 80 my + 8y + ky = 0, y(0) = yo ↓ initial displacement y + (y + (y = 0#(4) y (0) = y,# (3) ↓ initial velocity Page 18 of 22/nDimensions of 4km 8 (MT²) (M) (+)** Case I: If 4km/8² <1, discriminant is positive and < 1, so both roots are negative. Case II: If 4km/8² = 1, discriminant is zero and roots 2m y(t)=r₁c₁e¹² +r₂C₂¹²² y₁ = C₁ + C₁₂; C₁ = y₁ - C₂ 1 y₁ = r₁₂C₁ + r₂C₂ 11 =r₁(y₁ - C₂) +r₂C₂ --√√ 2m =(√₂-T₁) C₂+T₁Y₁ +r 0 = Case III: If 4km/8² > 1, discriminant is negative and roots are complex conjugates Case I: roots, real & unequal, both negative = = dimensionless 1 4km 7₂-7₂ Particular Solution in terms of ICs on y and y': y₁(t₂-₁)-v₁+ry_ryo-yo #(63) -#(64)See Answer
Q13:Forced, undamped vibration, continued... Suppose the forcing function is sinusoidal = Fsinet The defining ODE for displacement, y(t), is: my + ky = Fsinet y(t) = C₁sinwt + C₂coswt + Y Р General Solution ENGG 614 2023 Ninth Class 10-31-23 Revised of 10-30-23, 8:30pm Substitute Y in the ODE and obtain: Let Y = Asinet; don't need Bcoset since both y and y", the only dependent variable terms, P are both sines. - A0²sinet + Aw²sinet = ( )sinwt# (40) A 1 = (-)( ²¹ =)#(41) y(t) = C₁sinwt + C₂coswt + ((sinot# (42) m (30 Page 15 of 22/n-A0'sinet + Aw sinet y(t) = C₁sinwt + C₂coswt + (=) ( ²1 )sinot#(42) m msinwt# (40) ^ = ( ) ( )* (41) Let y = y = 0, system at rest at t = 0 when F is applied For y(0) = 0: 0 = C₁0 + C₂·1+(.).0; C₂ = = 0 2 So the solution for y(t) is: For y (0) = 0: _y(t) = wc₁coswt — wc,sinwt + () 1¹ecoset 0 = wC₁ • 1 - W. 0.0 + C₁ = ( =)))*(¹4) wsin Ot-Osinut L'Hopital's Rule, 0 [The resonance condition] y(t) = ( ) ( mw W FO 2 m(w²-0²) y(t) = ( − 2)(~)( ¹²² )sinwt + (_)(_ ¹ )sinot m ய = (-) (¹) - Osinw tw sinet] y(t) = (( wtcos@t-sino t -20 #(45) ²), 0→w . 1See Answer
Q14:Homework #2: Check this result in the ODE & the two ICs. Example: Tedesco et al, page 115¹ ¹Tedesco, McDougal and Ross, Structural Dynamics, page 115 inches ENGG 614 2023 Ninth Class 10-31-23 Revised of 10-30-23, 8:30pm w = √k/m Time for one cycle =T= 2πT W = 2500 x 32.2 x 12 1000 = 0.202 sec 1 = 31.1 rad/sec Page 14 of 22 time in seconds in radians/nTime for one cycle = T = tsinut y(t) = (m) (sin 2 2500 x 32.2 x 12 w = √/k/m = 2500 zx12: = 31.1 rad/sec 1000 y = 3.7231.1t mw 2π W -= 0.202 sec F = 600 lb 600 lb x 12 in/ft 31.1 slugs x 31.1 rads/sec 2mw m= 1000 lb 32.2 ft/sec² = 7.44 inches/sec time in seconds = 31.1 slugs = 3.72 inches/sec in radiansSee Answer
Q15:(OV) H = D(E,H) (T,P) D(V,H) ST,P) - (ON), - (OV) S.N SV (57)--(55), S.N as as (2),,-- ().. ƏN P.N (D) - (D).. UN V.N 1. Derive the following Maxwell's equations using Stokes Theorem (G).-), (The book from where I got these equation uses U instead of E.)/n2. Using Jacobian Transform, derive the following equation, where all the symbols have their usual meanings: (F). - [¹ (r). - ] I found this problem from a paper. Hint: you can start with the following Jacobian equation. (SP) - (T,H) H (P,H) You may also need the following equations: (SP), -7(SP),+V (37), -T(3T), if you wish to go further, you may try to derive the following equation: Ꮴ . -VC₂ C₂ (SP)₂ + 7 (or), - v (or), P T V Hint: Calculate the determinants of numerator and denominator, and proceed from there. (E,H) (SF) - ST.P). H (V,H) (T,P)See Answer
Q16:1. 2. 3. 4. 5. Solve the given differential equation (2+ ye x) dx + (xe xy -2y) dy = 0. ху [5 Marks] Find the Fourier series for the function f(x)=2x, -<x< 1. Solve the linear differential equation: dy 2 dx +−y=x4 The End X Solve the following using method of undetermined coefficients: Use Simpson's rule with n=4 to estimate the value of the definite integral [10 marks] [5 marks] y' +y'- 12y = 2e ²x [5 Marks] [5 Marks] dxSee Answer
Q17:Homework #2: Show that the particular solutions, equations 26 and 27, satisfy the boundary conditions and the ODE (see equation 3) appropriate to each region.See Answer
Q18:Homework #1: If y, and y, are solutions of ay + by + cy = 0, prove that C₁y₁ + C₂y₂ is also a solution. So back to the estuary problem. We now can write the general solution to the estuary equation, but will need boundary conditions to get a particular solution. First thing to notice is that W, the waste load, does not appear in the ODE. The small element Ax, see on Figure 6-2 in the original analysis of the mass balance, was arbitrarily chosen outside the zone of waste discharge. So we really have two separate estuary zones to analyze; i.e., upstream and downstream of the plane of waste discharge. Then we have, switching from C for waste concentration to L for waste BOD concentration and writing the general solutions for equation 3 in the upstream and downstream regions: L₁ = C₁e¹² +₂e; L₁ = C₂e¹*+C₂e; In which ₁ T and upstream region call it region I U = = 2/1 (¹ + √/¹ 1 2E r = - 00 < x≤0 2 2E 0≤x < 00 1 + = 2/1 (1 - √ 1 + AKE) 4KE downstream region call it region II OK to write, Both solutions will hold at as will be seen via choice of BC's. (10) (11) (12) (13)See Answer
Q19:3. Two identical masses are connected to each other, and to a nearby wall, using a number of linear springs. The variables and y denote the displacements of the masses from the wall relative to their positions when the springs are relaxed, and m and k are constants. Which of the following constitute possible equations of motion for this system? (a) μä + 2kx = 3ky, (b) μë + 2kx = 3ky, pij +2ky = 4kx; pÿj +2ky = 3kx; (c) μä + 7kx = 7ky, μÿ +7ky = 5kx; (d) μä + 7kx = 5ky, pÿj +5ky = 5kx. If you reject the plausibility of any of these systems of equations, explain why.See Answer
Q20:1. The five masses shown are constrained to translate only horizontally. Each line segment of this color represents a linear spring with stiffness k and each line segment of this color represents a linear dashpot with damping coefficient b. The system is depicted in a reference configuration in which all the springs are relaxed. 441 142 H43 Let x, denote the displacement of the block with mass coupled differential equations governing the dynamics of 4 145 , to the right of its reference location. Find the five ₁ through 25.See Answer

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