Applications: Coupled Differential Equations

Each tank gains salt only in the brine that flows into it and loses salt only in the brine that flows out of it. Brine leaving a tank carries that tank's concentration, the salt divided by the 40 liters, so each flow moves its rate times a concentration. The two equations are coupled: the salt in B depends on how much arrives from A, and the salt in A depends on what comes back from B. Solving the second equation for x and substituting into the first leaves one second-order equation in y alone, which the auxiliary equation solves.

1. First check that the volumes stay fixed. Tank A takes in 6 + 2 = 8 liters per minute and sends out 8; tank B takes in 8 and sends out 2 + 6 = 8. So each tank always holds 40 liters, and the concentrations are x40 and y40 kilograms per liter.
2. (a) Salt enters A only in the 2 liters per minute from B, carrying 2 × y40 = 0.05y kilograms per minute, and leaves in the 8 liters per minute sent to B, carrying 8 × x40 = 0.2x. So dxdm = −0.2x + 0.05y. Salt enters B at 0.2x and leaves in 2 + 6 = 8 liters per minute, carrying 0.2y, so dydm = 0.2x − 0.2y.
3. Eliminate x. The second equation gives x = 5dydm + y, so dxdm = 5d²ydm² + dydm. Substituting both into the first equation gives 5d²ydm² + dydm = −dydm − 0.2y + 0.05y, and dividing by 5 after collecting terms gives d²ydm² + 0.4dydm + 0.03y = 0.
4. The auxiliary equation λ² + 0.4λ + 0.03 = 0 factorizes as (λ + 0.1)(λ + 0.3) = 0, so y = Ae^−0.1m + Be^−0.3m. At the start y = 0, so B = −A, and dydm = 0.2 × 20 − 0 = 4, so −0.1A + 0.3A = 4 and A = 20. Hence y = 20e^−0.1m − 20e^−0.3m.
5. (b) The salt in B is greatest where dydm = −2e^−0.1m + 6e^−0.3m = 0, that is where e^0.2m = 3, so m = 5ln 3 ≈ 5.49 minutes. Then e^−0.1m = 1√3 and e^−0.3m = 13√3, so y = 20√3 − 203√3 = 403√3 ≈ 7.70 kilograms.
6. Check: x = 5dydm + y = 10e^−0.1m + 10e^−0.3m, which is 20 at the start, as it should be. At m = 5ln 3 it is 10√3 + 103√3 = 403√3, the same as y: tank B holds the most salt exactly when the two tanks are equally salty, because then salt flows into B as fast as it flows out.

answer(a) dxdm = −0.2x + 0.05y and dydm = 0.2x − 0.2y; (b) tank B holds the most salt, 403√3 ≈ 7.70 kilograms, after 5ln 3 ≈ 5.49 minutes

techniqueCoupled Differential Equations · Forming Differential Equations · The Auxiliary Equation

examsH2

The blood loses drug in two ways, to the tissues and to the kidneys, and gains some back from the tissues; the tissues only exchange drug with the blood. That gives a linear system with a matrix of constant rates. Each eigenvalue of the matrix is a rate of decay, and its eigenvector is a split between blood and tissues that keeps its proportions while it decays. The dose is written as a combination of the two eigenvectors, and each part then decays at its own rate.

1. The blood loses 0.2x + 0.3x = 0.5x milligrams per hour and gains 0.2y, and the tissues gain 0.2x and lose 0.2y. So dxdh = −0.5x + 0.2y and dydh = 0.2x − 0.2y, with matrix −0.50.20.2−0.2.
2. The trace is −0.7 and the determinant is 0.1 − 0.04 = 0.06, so the characteristic polynomial is λ² + 0.7λ + 0.06 = (λ + 0.1)(λ + 0.6), and the eigenvalues are −0.1 and −0.6. For −0.1: −0.4x + 0.2y = 0, so y = 2x and an eigenvector is 12. For −0.6: 0.1x + 0.2y = 0, so an eigenvector is 2−1.
3. (a) The general solution is xy = c₁12e^−0.1h + c₂2−1e^−0.6h. At h = 0: c₁ + 2c₂ = 100 and 2c₁ − c₂ = 0, so c₂ = 2c₁, 5c₁ = 100, c₁ = 20 and c₂ = 40. Hence x = 20e^−0.1h + 80e^−0.6h and y = 40e^−0.1h − 40e^−0.6h.
4. (b) The tissues hold more when 40e^−0.1h − 40e^−0.6h > 20e^−0.1h + 80e^−0.6h, that is when 20e^−0.1h > 120e^−0.6h. Multiplying both sides by e^0.6h20 gives e^0.5h > 6, so h > 2ln 6 ≈ 3.58 hours.
5. As time goes on, the e^−0.6h terms shrink much faster than the e^−0.1h terms, so yx approaches 4020 = 2. The drug settles into the split of the eigenvector 12, twice as much in the tissues as in the blood, while the whole amount decays at the slower rate.
6. Check: at h = 2ln 6, e^−0.1h = 6^−0.2 ≈ 0.6988 and e^−0.6h = 6^−1.2 ≈ 0.1165, so x ≈ 13.98 + 9.32 and y ≈ 27.95 − 4.66, and both come to about 23.29 milligrams, equal as they should be. Both formulas also give amounts that are never negative, since y = 40e^−0.1h(1 − e^−0.5h).

answer(a) x = 20e^−0.1h + 80e^−0.6h and y = 40e^−0.1h − 40e^−0.6h milligrams; (b) after 2ln 6 ≈ 3.58 hours, and the ratio of the amount in the tissues to the amount in the blood approaches 2

techniqueSolving a Coupled System with Eigenvalues · Finding an Eigenvector · The Characteristic Polynomial

Each room's temperature changes at a rate made of its gains and its losses: the heater in A, the flow of heat between the rooms, and the loss to the outside. The settled temperatures are where both rates are zero. Measured from those settled temperatures, what is left is a linear system with no constant terms. Its matrix is symmetric, because the rooms exchange heat in the same way, and its eigenvectors are the average of the two rooms and their difference. Each of those settles on its own, at its own rate.

1. Room A gains 3 from the heater and 0.2(y − x) from B, and loses 0.1(x − 5) to the outside, so dxdh = 3 + 0.2(y − x) − 0.1(x − 5) = −0.3x + 0.2y + 3.5. Room B has no heater, so dydh = 0.2(x − y) − 0.1(y − 5) = 0.2x − 0.3y + 0.5.
2. (a) The rooms settle where both rates are zero: 0.3x − 0.2y = 3.5 and 0.2x − 0.3y = −0.5. Subtracting the second equation from the first gives 0.1x + 0.1y = 4, so x + y = 40, and adding them gives 0.5x − 0.5y = 3, so x − y = 6. Hence the rooms settle at x = 23°C and y = 17°C.
3. Measure from the settled temperatures, with p = x − 23 and q = y − 17. Then dpdh = −0.3p + 0.2q and dqdh = 0.2p − 0.3q, with matrix −0.30.20.2−0.3. Its eigenvalues are −0.1, with eigenvector 11, and −0.5, with eigenvector 1−1: the average of the two rooms settles at the rate 0.1 and their difference at the rate 0.5.
4. At the start p = 5 − 23 = −18 and q = 5 − 17 = −12, which is −1511 − 31−1. So x = 23 − 15e^−0.1h − 3e^−0.5h and y = 17 − 15e^−0.1h + 3e^−0.5h, and the gap between the rooms is x − y = 6 − 6e^−0.5h.
5. (b) The gap is 3°C when 6e^−0.5h = 3, so e^0.5h = 2 and h = 2ln 2 ≈ 1.39 hours. Then e^−0.1h = 2^−0.2 ≈ 0.8706 and e^−0.5h = 0.5, so x ≈ 23 − 13.06 − 1.5 = 8.44°C and y ≈ 17 − 13.06 + 1.5 = 5.44°C.
6. Check: at h = 0 the formulas give 5°C in both rooms, with dxdh = 1.5 + 1.5 = 3 and dydh = 1.5 − 1.5 = 0, the heater warming A alone. At h = 1.39 the gap is already half its final 6°C, while the average, 20 − 15e^−0.1h ≈ 6.94°C, has risen only about an eighth of the way from 5°C to its final 20°C, because the average settles five times more slowly than the gap.

answer(a) dxdh = −0.3x + 0.2y + 3.5 and dydh = 0.2x − 0.3y + 0.5; the rooms settle at 23°C and 17°C; (b) after 2ln 2 ≈ 1.39 hours, when room A is at about 8.44°C and room B at about 5.44°C

techniqueCoupled Differential Equations · Solving a Coupled System with Eigenvalues · Phase Portraits and Equilibrium Types

The balance point is where both rates are zero. Near it, the deviations from the balance obey a linear system with no constant terms, and the eigenvalues of its matrix decide what happens. Two negative eigenvalues pull every start back to the balance. Eigenvalues of opposite signs make a saddle: a deviation along the eigenvector of the negative eigenvalue decays, a deviation along the other grows, and any other start has some of the growing part.

1. At the balance point both rates are zero: −p + 2q = 14 and 3p − 2q = 6. Adding the two equations gives 2p = 20, so p = 10, and then 2q = 14 + 10 = 24, so q = 12.
2. Measure from the balance, with u = p − 10 and v = q − 12. Then dudt = 2(v + 12) − (u + 10) − 14 = −u + 2v and dvdt = 3(u + 10) − 2(v + 12) − 6 = 3u − 2v, with matrix −123−2.
3. (a) The characteristic polynomial is λ² + 3λ + (2 − 6) = λ² + 3λ − 4 = (λ − 1)(λ + 4), so the eigenvalues are 1 and −4. They have opposite signs, so the balance point is a saddle: it is unstable, and spending that starts near it does not in general return to it.
4. For the eigenvalue 1: −2u + 2v = 0, so an eigenvector is 11, along which both spendings grow together like e^t. For −4: 3u + 2v = 0, so an eigenvector is 2−3. Only a start whose deviation is a multiple of this one decays to the balance.
5. (b) With q = 15, v = 3, so the deviation must be −1 × 2−3 = −23, which gives u = −2 and p = 8 billion dollars a year. Check: at (8, 15), dpdt = 30 − 8 − 14 = 8 and dqdt = 24 − 30 − 6 = −12, which points along 2−3, straight toward (10, 12).
6. If P spends more, say 9, the deviation is −13 = 0.611 − 0.82−3. The first part grows like e^t with a positive coefficient, so both spendings rise without limit: the race escalates. Below 8 that coefficient is negative, and both spendings fall until the model stops applying at zero spending.

answer(a) the balance point is p = 10 and q = 12 billion dollars a year; the eigenvalues are 1 and −4, so it is a saddle, and spending that starts near it does not in general return to it; (b) P must spend 8 billion dollars a year; if P spends more, both spendings grow without limit and the race escalates

techniquePhase Portraits and Equilibrium Types · Solving a Coupled System with Eigenvalues · Finding an Eigenvector

Where both species are present and neither number changes, both brackets are zero. Close to that balance, write each number as its balance value plus a small deviation. Each rate is a population times its bracket, and the bracket is zero at the balance, so to first order only the change in the bracket counts, multiplied by the balance population; a product of two small deviations is dropped. That leaves a linear system whose eigenvalues classify the balance and whose eigenvectors show the directions in which the numbers move.

1. Both brackets are zero where x + 1.5y = 7 and 2x + y = 10. The second gives y = 10 − 2x, and then x + 15 − 3x = 7, so x = 4 and y = 2: 400 of X and 200 of Y.
2. Write x = 4 + u and y = 2 + v. The first bracket becomes −u − 1.5v and the second becomes −2u − v. Keeping only the terms of first order, dudt ≈ 0.05 × 4(−u − 1.5v) = −0.2u − 0.3v and dvdt ≈ 0.05 × 2(−2u − v) = −0.2u − 0.1v.
3. (a) The matrix −0.2−0.3−0.2−0.1 has characteristic polynomial λ² + 0.3λ + (0.02 − 0.06) = λ² + 0.3λ − 0.04 = (λ − 0.1)(λ + 0.4). The eigenvalues 0.1 and −0.4 have opposite signs, so the balance is a saddle: it is unstable, and the two species cannot stay near it together.
4. For 0.1: −0.3u − 0.3v = 0, so an eigenvector is 1−1, one species gaining what the other loses. For −0.4: 0.2u − 0.3v = 0, so an eigenvector is 32. The count gives u = 0.3 and v = −0.1, which is 0.181−1 + 0.0432.
5. (b) The growing part, 0.18e^0.1t1−1, has X rising and Y falling, so the numbers are moving in favor of X. After 10 weeks, u = 0.18e¹ + 0.12e⁻⁴ ≈ 0.4893 + 0.0022 = 0.4915 and v = −0.18e¹ + 0.08e⁻⁴ ≈ −0.4893 + 0.0015 = −0.4878, so there are about 449 of X and 151 of Y.
6. Check: stepping the full equations forward numerically for 10 weeks from x = 4.3 and y = 1.9 gives about 447 of X and 156 of Y, close to the linear approximation. Carried on further, the full model ends with 700 of X and no Y at all.

answer(a) the balance is 400 of X and 200 of Y; near it the eigenvalues are 0.1 and −0.4, so it is a saddle and unstable; (b) the numbers are moving in favor of X, with about 449 of X and 151 of Y after 10 weeks

techniquePhase Portraits and Equilibrium Types · Solving a Coupled System with Eigenvalues

The matrix of a linear model near a balance says how every small deviation changes. When its eigenvalues are complex, a ± bi, the solutions are e to the power at times sines and cosines of bt: the real part a says whether the swings grow or die away, and the imaginary part b is the rate of turning, in radians per year. The direction of turning comes from the rates of change at any one point.

1. The matrix is −0.2−2.50.20. Its trace is −0.2 and its determinant is 0 − (−2.5)(0.2) = 0.5, so the characteristic polynomial is λ² + 0.2λ + 0.5.
2. (a) By the quadratic formula, λ = −0.2 ± √0.04 − 22 = −0.2 ± √−1.962 = −0.1 ± 0.7i. The real part, −0.1, is negative, so the swings die away, and the imaginary part makes the numbers turn round the balance: it is a stable spiral.
3. For the direction, take the point u = 1, v = 0: a thousand extra hares and the usual number of lynx. There dudt = −0.2 and dvdt = 0.2, so the lynx begin to increase as the hares begin to fall. With hares across and lynx up, the path turns counterclockwise, so each peak in the hares comes before the matching peak in the lynx.
4. (b) Every solution is e^−0.1t times a combination of cos 0.7t and sin 0.7t, and those repeat each time 0.7t increases by 2π. So one peak in the hares follows another after 2π0.7 ≈ 8.98 years.
5. Over those 8.98 years the factor e^−0.1t is multiplied by e^{−0.1 × 2π/0.7} = e^−2π/7 ≈ 0.41, so each peak's height above the balance is about 0.41 times the one before. Check: with λ = −0.1 + 0.7i, λ² = 0.01 − 0.14i − 0.49 = −0.48 − 0.14i and 0.2λ = −0.02 + 0.14i, so λ² + 0.2λ + 0.5 = −0.48 − 0.02 + 0.5 = 0.

answer(a) the eigenvalues are −0.1 ± 0.7i, so the balance is a stable spiral; it turns counterclockwise with hares across and lynx up, so each peak in the hares comes before the matching peak in the lynx; (b) 2π0.7 ≈ 8.98 years from one hare peak to the next, and each peak's height is e^−2π/7 ≈ 0.41 times the one before

techniquePhase Portraits and Equilibrium Types · The Characteristic Polynomial

Naming the angular velocity v = dx/ds turns one second-order equation into two first-order ones: the angle changes at the rate v, and the equation itself says how v changes. The eigenvalues of the matrix of that pair are the roots of the auxiliary equation. Real negative eigenvalues mean the motion dies away without oscillating, although a start with a large enough velocity can still cross zero once, so the particular solution must be checked. The two constants in the solution are fixed by the angle and the angular velocity at release.

1. Let v = dxds, the angular velocity in radians per second. Then dxds = v and, from the equation, dvds = −4x − 5v. The matrix of this pair is 01−4−5.
2. (a) Its characteristic polynomial is λ² + 5λ + 4 = (λ + 1)(λ + 4), so the eigenvalues are −1 and −4. Both are real and negative, so the motion dies away without oscillating: the equilibrium x = 0, v = 0 is a stable node.
3. For −1 the first row gives v = −x, so an eigenvector is 1−1; for −4 it gives v = −4x, so an eigenvector is 1−4. So x = c₁e^−s + c₂e^−4s and v = −c₁e^−s − 4c₂e^−4s. At release x = 1.2 and v = 0, so c₁ + c₂ = 1.2 and c₁ + 4c₂ = 0, which give c₂ = −0.4 and c₁ = 1.6.
4. So x = 1.6e^−s − 0.4e^−4s = 0.4e^−s(4 − e^−3s). For s > 0, e^−3s < 1, so the bracket stays above 3 and x stays positive: the angle shrinks toward 0 but never becomes negative, and the door does not swing past the closed position.
5. (b) The door swings shut at the speed −v = 1.6e^−s − 1.6e^−4s. This is greatest where its derivative, −1.6e^−s + 6.4e^−4s, is zero, that is where e^3s = 4, so s = ln 43 ≈ 0.462 seconds. Then e^−s = 4^−1/3 and e^−4s = 14 × 4^−1/3, so the greatest speed is 1.6 × 34 × 4^−1/3 = 1.2 × 4^−1/3 ≈ 0.756 radians per second.
6. Check: at that moment dvds = −4x − 5v should be zero. There x = 1.6 × 4^−1/3 − 0.1 × 4^−1/3 = 1.5 × 4^−1/3 ≈ 0.945 and v ≈ −0.756, and −4 × 0.945 + 5 × 0.756 = −3.78 + 3.78 = 0. A speed of 0.756 radians per second is about 43 degrees per second, a steady close for a door.

answer(a) dxds = v and dvds = −4x − 5v; the eigenvalues are −1 and −4, both real and negative, and x = 1.6e^−s − 0.4e^−4s stays positive, so the door does not swing past the closed position; (b) the greatest closing speed is 1.2 × 4^−1/3 ≈ 0.756 radians per second, reached after ln 43 ≈ 0.462 seconds

techniqueA Second-Order Equation as a Coupled System · Phase Portraits and Equilibrium Types · Damped Oscillations

Naming the current I = dq/dm turns the second-order circuit equation into two first-order ones: the charge changes at the rate I, and the circuit equation says how I changes. With L = 0.25 and C = 4 in these units, the pair has a matrix whose eigenvalues are purely imaginary. A real part of zero means the swings neither grow nor die away, so the equilibrium is a center. The initial charge and current then fix the solution.

1. Name the current: dqdm = I. Then d²qdm² = dIdm, and the circuit equation gives 0.25dIdm = −q4, so dIdm = −q. The pair has matrix 01−10.
2. (a) The characteristic polynomial is λ² + 1, so the eigenvalues are ± i. Their real part is zero, so the swings neither grow nor die away: the equilibrium q = 0, I = 0 is a center, and the charge swings back and forth between the two plates forever, once every 2π ≈ 6.28 milliseconds.
3. With eigenvalues ± i the solutions are combinations of cos m and sin m. At m = 0, q = 20 and I = dqdm = 0, so q = 20cos m and I = −20sin m.
4. (b) The current is greatest in size when sin m = ± 1, so the greatest size of the current is 20 milliamperes. It first reaches that size at m = π2 ≈ 1.57 milliseconds, when the capacitor is empty, q = 20cosπ2 = 0. The minus sign in I = −20sin m says that it flows in the direction that discharges the capacitor.
5. Check by energy: at the start the capacitor stores q²2C = (20 × 10⁻⁶)²2 × 4 × 10⁻⁶ = 5 × 10⁻⁵ joules, and at the greatest current the coil stores 12LI² = 12 × 0.25 × 0.02² = 5 × 10⁻⁵ joules, the same. In a real circuit a little resistance gives the eigenvalues a small negative real part, and the swings slowly die away.

answer(a) dqdm = I and dIdm = −q; the eigenvalues are ± i, a center, so the charge swings between 20 and −20 microcoulombs forever, once every 2π ≈ 6.28 milliseconds; (b) 20 milliamperes, first after π2 ≈ 1.57 milliseconds

techniqueA Second-Order Equation as a Coupled System · Phase Portraits and Equilibrium Types · Simple Harmonic Motion

The molybdenum only decays, so its equation involves x alone and is solved first. The technetium is made at the rate at which the molybdenum decays and is lost at its own rate of decay, so its equation contains x. Once x is known, that becomes a first-order linear equation in y, solved by trying a multiple of the same exponential and adding the solution with right side zero. The matrix of the pair is triangular, so its eigenvalues are its diagonal entries, −a and −b, and the solution is built from those two exponentials.

1. Each isotope decays at its decay constant times the number of its atoms. The molybdenum only decays, so dxdh = −ax. The technetium gains one atom for each molybdenum atom that decays and loses its own atoms as they decay, so dydh = ax − by.
2. The first equation involves x alone, and with x = N at the start its solution is x = Ne^−ah.
3. Substituting gives dydh + by = aNe^−ah. Try y = Ke^−ah: then −aK + bK = aN, so K = aNb − a. Since b − a = ln 2(16 − 166) = 10ln 266, this is K = N10. Adding Ce^−bh, the solution with right side zero, and using y = 0 at the start gives C = −N10.
4. (a) So x = Ne^−ah and y = N10(e^−ah − e^−bh). This is zero at the start and positive afterward, because b > a makes e^−bh the smaller of the two exponentials.
5. (b) The technetium is greatest where dydh = N10(−ae^−ah + be^−bh) = 0, that is where e^{(b − a)h} = ba = 11. So h = ln 11b − a = 6.6ln 11ln 2 ≈ 22.8 hours.
6. At that moment dydh = ax − by = 0, so ax = by: the two activities are equal. Check: e^−ah = 2^−22.83/66 ≈ 0.7866 and e^−bh = 2^−22.83/6 ≈ 0.0715, so y ≈ N10(0.7866 − 0.0715) ≈ 0.0715N. Then ax ≈ 0.01050 × 0.7866N ≈ 0.00826N and by ≈ 0.1155 × 0.0715N ≈ 0.00826N decays per hour, equal. Hospitals draw the technetium off about once a day, close to this time.

answer(a) dxdh = −ax and dydh = ax − by, with x = Ne^−ah and y = N10(e^−ah − e^−bh); (b) after 6.6ln 11ln 2 ≈ 22.8 hours, when the two activities are equal

techniqueCoupled Differential Equations · Solving a Coupled System with Eigenvalues · The Integrating Factor

Each sugar is lost at its own rate and gained at the other's, so the two rates of change are equal and opposite and the total never changes. The eigenvalues of the matrix are the rates at which the parts of the mix along the eigenvectors change. An eigenvalue of zero belongs to a part that does not change at all, so every mix along that eigenvector is at rest, and a negative eigenvalue belongs to a part that dies away. Splitting the starting mix along the two eigenvectors gives the whole solution.

1. Glucose is lost at 0.25x and gained at 0.25y, and fructose the other way round, so dxdh = −0.25x + 0.25y and dydh = 0.25x − 0.25y. Adding the two equations gives ddh(x + y) = 0, so the total stays at x + y = 1.
2. (a) The matrix −0.250.250.25−0.25 has trace −0.5 and determinant 0.0625 − 0.0625 = 0, so its characteristic polynomial is λ² + 0.5λ = λ(λ + 0.5), and the eigenvalues are 0 and −0.5. For 0 an eigenvector is 11, and for −0.5 it is 1−1.
3. The eigenvalue 0 means that any mix with x = y is at rest: there is a whole line of equilibria, not a single point. The eigenvalue −0.5 means that the difference between the glucose and the fructose dies away. So the syrup settles at the point of that line with total 1: half glucose and half fructose.
4. The start, pure glucose, is 10 = 0.511 + 0.51−1. The first part stays and the second decays like e^−0.5h, so x = 0.5 + 0.5e^−0.5h and y = 0.5 − 0.5e^−0.5h.
5. (b) y = 0.42 when 0.5e^−0.5h = 0.08, so e^−0.5h = 0.16 and h = 2ln10.16 = 2ln 6.25 = 4ln 2.5 ≈ 3.67 hours. Check: e^{−0.5 × 3.665} ≈ 0.160, so y ≈ 0.5 − 0.080 = 0.420 and x ≈ 0.580, which add to 1.

answer(a) dxdh = −0.25x + 0.25y and dydh = 0.25x − 0.25y; the eigenvalues are 0, whose eigenvector gives a whole line of mixes at rest with equal glucose and fructose, and −0.5, for the difference between the two sugars, which dies away, so the syrup settles at half glucose and half fructose, 0.5 of each; (b) 4ln 2.5 ≈ 3.67 hours

techniqueSolving a Coupled System with Eigenvalues · Phase Portraits and Equilibrium Types · Finding an Eigenvector