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Exercises 1 More exercises are available in Elementary Differential Equations. If you have a problem to solve any of them, feel free to come to office hour. 1 Problem 1 Write down a differential equation of the form t → ∞. dy dt = ay + b whose solutions have the required behavior as 1. All solutions approach y = 3. 2. All other solution diverge from y = 2. 1.1 Solution b at We know that the general solution of dy dt = ay + b is y = − a + ce . If a < 0, the exponential term goes to b zero when t goes to infinity ⇒ y tends to a . If a > 0, the exponential term goes to infinity when t goes to infinity ⇒ y diverges from ab . 1. We want y to be something like y = 3 + cet with a < 0. If we choose a = −1 and c = 1, we get: dy = 3 − y. dt (1) 2. We want y to be something like y = 2 + ceat with a > 0. If we choose a = 1 and c = 1, we get: dy = −2 + y. dt 2 (2) Problem 2 A spherical raindrop evaporates at a rate proportional to its surface area. Write a differential equation for the volume of the raindrop as a function of time. 2.1 Solution We know that after a time ∆t, the raindrop will lose a volume ∆V , which is proportional to the surface of the raindrop 4πr2 (r is radius of the sphere). Thus, we can write: ∆V = −a · 4πr2 · (∆t) (3) with a > 0. The minus sign is there because the volume is decreasing. The volume of a sphere is given by: V = 4 3 πr . 3 (4) Therefore, we have: r= 3 V 4π 1 1/3 (5) We finally get (∆t and ∆V become infinitesimal): dV = −kV 2/3 . dt (6) with k > 0. 3 Problem 3 For small, slowly falling objects, the assumption that the drag force is proportional to the velocity is a good one. For larger, more rapidly falling objects, it is more accurate to assume that the drag force is proportional to the square of the velocity. 1. Write a differential equation for the velocity of a falling object of mass m if the drag force is proportional to the square of the velocity. 2. Determine the limiting velocity after a long time. 3.1 Solution 1. When the drag force was proportional to velocity (v), the equation was: m dv = mg − γv. dt (7) Now the drag force is proportional to the square of the velocity (v 2 ): m dv = mg − γv 2 . dt (8) You cannot use the general formula to find the solution because equation (8) is first order but nonlinear: you cannot write equation (8) as a linear combination of v and dv dt . 2. When the limiting velocity is reached, the system is said to have reached the equilibrium. When the equilibrium is reached the velocity is constant and thus, we have dv dt = 0. Therefore, we get: r mg (9) v(t) = γ 4 Problem 4 Solve each of the following initial value problems and plot the solutions for several values of y0 . Then, describe in a few words how the solution resemble, and differ from each, other. 1. dy dt = −y + 5, y(0) = y0 2. dy dt = −2y + 5, y(0) = y0 3. dy dt = −2y + 10, y(0) = y0 4.1 Solution To solve this problem, we just use the general solution. 1. y = 5 + (y0 − 5)e−t 2. y = 5 + (y0 − 5)e−2t 3. y = 5 2 + (y0 − 5)e−2t The equilibrium solution is 5 for (1) and (2). The equilibrium is for (2) and (3) than it is for (1). 2 5 2 for (3). The equilibrium is reached faster 5 Problem 5 Here is an alternative way to solve the equation: dy = ay − b. dt (10) 1. Solve the simpler equation: dy = ay. dt (11) Call the solution y1 (t), 2. Observe that the only difference between equation (10) and equation (11) is the constant −b in equation (10). Therefore it may seem reasonable to assume that the solutions of these two equations also differ only by a constant. Test thus assumption by trying to find a constant k such that y(t) = y1 (t)+k is a solution of equation (10). 5.1 Solution 1. We have successively: dy = ay, dt 1 dy = a, y dt d (ln y) = a, dt Z Z d (ln y) = dt a, dt dt (12) (13) (14) (15) ln y = at + c, (16) y = ec eat . (17) at Therefore, y1 = Ce . 2. Now, we want to find k such y1 (t) + k is a solution of equation (10). Thus, we want: d Ceat + k = a Ceat + k − b. dt (18) aCeat = aCeat + ak − b, (19) We have: k= The solution of equation (10) is y = 6 b a b . a (20) + Ceat , which is what we expected. Problem 6 A falling object satisfies the initial value problem: v dv = 9.8 − , dt 5 v(0) = 0. 1. Find the time that must elapse for the object to reach 98% of its limiting velocity. 2. How far does the object fall in the time found in part (1). 3 (21) (22) 6.1 Solution 1. The general solution of the equation is: t v(t) = 49(1 − e− 5 ). (23) We want: v(T ) = 0.98v(t∞ ), = 0.98 · 49. (24) Therefore, we get: T 1 − e− 5 = 0.98, 0.02 = e − T5 , T ln 0.02 = − , 5 T ≈ 19.56s 2. To find the distance travelled by the object in 19.56s, we need to integrate the speed: Z 19.56 t x= dt 49(1 − e− 5 ), (25) (26) (27) (28) (29) 0 19.56 x = 49 t + 5e−t 5 0 , (30) = 718.34 7 Problem 7 The half-life of a radioactive material is the time required for an amount of this material to decay to one-half its original value. Show that for any radioactive material that decays according to the equation dQ dt = −rQ, the half-life τ and the decay rate r satisfy the equation rτ = ln 2. 7.1 Solution We know that the solution of: dQ = −rQ, dt is Q(t) = Q0 e−rt . After one half-time, Q(τ ) = Q20 . Thus, we have: Q0 = Q0 e−rτ , 2 2 = erτ , rτ = ln 2. 8 (31) (32) (33) (34) Problem 8 Consider an electric circuit containing a capacitor, resistor, and battery. The charge Q(t) on the capacitor satisfies the equation: dQ Q + = V, (35) R dt C where R is the resistance, C is the capacitance, and V is the constant voltage supplied by the battery. 1. If Q(0) = 0, find Q(t) at any time t, sketch the graph of Q versus t. 2. Find the limiting value QL that Q(t) approaches after a long time. 3. Suppose that Q(t1 ) = QL and that at time t = t1 the battery is removed and the circuit closed again. Find Q(t) for t > t1 and sketch its graph. 4 8.1 Solution 1. Let’s rewrite equation (35): dQ Q V =− + , dt RC R with Q(0) = 0. We know that the solution is: t (36) Q(t) = V C + ke− RC . (37) 0 = V C + k. (38) t Q(t) = V C 1 − e− RC . (39) Using the initial condition, we find: Thus, we finally have: t 2. When t → ∞, e− RC → 0. Thus, the limiting value QL is V C. 3. We are solving an equation similar to equation (35) but the initial condition is different. Since there is no battery, equation (35) becomes: dQ Q + = 0, (40) R dt C with Q(t1 ) = V C. The general solution is: t Q(t) = ke− RC . (41) Using the initial condition, we find: t1 V C = ke− RC . (42) Thus, we finally get: Q(t) = V Ce− 9 t−t1 RC . Problem 9 Determine the order of the given differential equation and whether the equation is linear or nonlinear. 2 1. t2 ddt2y + t dy dt + 2y = sin t d3 y dt3 d2 y dt2 2. d4 y dt4 + 3. d2 y dt2 + sin(t + y) = sin t 9.1 + + dy dt +y =1 Solution 1. Second order because of d2 y dt2 , linear. 2. Fourth order because of d4 y dt4 , linear. 3. Second order because of 10 2 d y dt2 , nonlinear because of sin(t + y). Problem 10 Verify that each given function is a solution of the differential equation: 1. y00 − y = 0; y(t) = et . 2. ty0 − y = t2 ; y = 3t + t2 . 3. y0000 + 4y000 + 3y = t; y(t) = 3t . 5 (43) 10.1 Solution 1. dy = et , dt d2 y = et . dt2 2. (44) (45) et − et = 0. (46) dy = 3 + 2t. dt (47) t(3 + 2t) − (3t + t2 ) = t2 . (48) 3. 1 dy = , dt 3 d2 y = 0, dt2 3 d y = 0, dt3 d4 y = 0. dt4 3 11 t = t. 3 (49) (50) (51) (52) (53) Problem 11 Determine the order of the given partial differential equation and state whether the equation is linear or nonlinear. Partial derivatives are denoted by subscripts. 1. uxx + uyy + uzz = 0 2. uxx + uyy + uux + uuy + u = 0 3. uxxxx + 2uxxyy + uyyyy = 0 11.1 Solution 1. Second order, linear. 2. Second order, nonlinear because of uux and uuy . 3. Fourth order. 6