Linear Differential Equations A first-order linear differential equation is one that can be put into the form dy ϩ P͑x͒y Q͑x͒ dx where P and Q are continuous functions on a given interval This type of equation occurs frequently in various sciences, as we will see An example of a linear equation is xyЈ ϩ y 2x because, for x 0, it can be written in the form yЈ ϩ y2 x Notice that this differential equation is not separable because it’s impossible to factor the expression for yЈ as a function of x times a function of y But we can still solve the equation by noticing, by the Product Rule, that xyЈ ϩ y ͑xy͒Ј and so we can rewrite the equation as ͑xy͒Ј 2x If we now integrate both sides of this equation, we get xy x ϩ C or C x yxϩ If we had been given the differential equation in the form of Equation 2, we would have had to take the preliminary step of multiplying each side of the equation by x It turns out that every first-order linear differential equation can be solved in a similar fashion by multiplying both sides of Equation by a suitable function I͑x͒ called an integrating factor We try to find I so that the left side of Equation 1, when multiplied by I͑x͒, becomes the derivative of the product I͑x͒y: I͑x͒͑yЈ ϩ P͑x͒y͒ ͑I͑x͒y͒Ј If we can find such a function I , then Equation becomes ͑I͑x͒y͒Ј I͑x͒Q͑x͒ Integrating both sides, we would have I͑x͒y y I͑x͒Q͑x͒ dx ϩ C so the solution would be y͑x͒ I͑x͒ ͫy ͬ I͑x͒Q͑x͒ dx ϩ C To find such an I, we expand Equation and cancel terms: I͑x͒yЈ ϩ I͑x͒P͑x͒y ͑I͑x͒y͒Ј IЈ͑x͒y ϩ I͑x͒yЈ I͑x͒P͑x͒ IЈ͑x͒ ■ LINEAR DIFFERENTIAL EQUATIONS This is a separable differential equation for I , which we solve as follows: y dI y P͑x͒ dx I ԽԽ ln I y P͑x͒ dx I Ae x P͑x͒ dx where A Ϯe C We are looking for a particular integrating factor, not the most general one, so we take A and use I͑x͒ e x P͑x͒ dx Thus, a formula for the general solution to Equation is provided by Equation 4, where I is given by Equation Instead of memorizing this formula, however, we just remember the form of the integrating factor To solve the linear differential equation yЈ ϩ P͑x͒y Q͑x͒, multiply both sides by the integrating factor I͑x͒ e x P͑x͒ dx and integrate both sides EXAMPLE Solve the differential equation dy ϩ 3x y 6x dx SOLUTION The given equation is linear since it has the form of Equation with P͑x͒ 3x and Q͑x͒ 6x An integrating factor is I͑x͒ e x 3x dx ex 3 Multiplying both sides of the differential equation by e x , we get Figure shows the graphs of several members of the family of solutions in Example Notice that they all approach as x l ϱ ■ ■ ex dy 3 ϩ 3x 2e x y 6x 2e x dx C=2 d x3 ͑e y͒ 6x 2e x dx or C=1 C=0 Integrating both sides, we have C=_1 _1.5 1.8 3 e x y y 6x 2e x dx 2e x ϩ C C=_2 _3 y ϩ CeϪx FIGURE EXAMPLE Find the solution of the initial-value problem x yЈ ϩ xy xϾ0 y͑1͒ SOLUTION We must first divide both sides by the coefficient of yЈ to put the differential equation into standard form: yЈ ϩ 1 y x x xϾ0 The integrating factor is I͑x͒ e x ͑1͞x͒ dx e ln x x LINEAR DIFFERENTIAL EQUATIONS ■ Multiplication of Equation by x gives x xyЈ ϩ y ■ ■ The solution of the initial-value problem in Example is shown in Figure ln x ϩ C x y and so x dx ln x ϩ C x xy y Then ͑xy͒Ј or Since y͑1͒ 2, we have (1, 2) 2 ln ϩ C C Therefore, the solution to the initial-value problem is _5 ln x ϩ x y FIGURE EXAMPLE Solve yЈ ϩ 2xy SOLUTION The given equation is in the standard form for a linear equation Multiplying by the integrating factor e x 2x dx e x we get 2 e x yЈ ϩ 2xe x y e x (e x y)Ј e x or Even though the solutions of the differential equation in Example are expressed in terms of an integral, they can still be graphed by a computer algebra system (Figure 3) ■ ■ 2 2 e x y y e x dx ϩ C Therefore Recall from Section 5.8 that x e x dx can’t be expressed in terms of elementary functions Nonetheless, it’s a perfectly good function and we can leave the answer as 2.5 y eϪx C=2 _2.5 2.5 ye dx ϩ CeϪx Another way of writing the solution is C=_2 y eϪx y _2.5 FIGURE x2 x e t dt ϩ CeϪx (Any number can be chosen for the lower limit of integration.) Application to Electric Circuits R E L switch In Section 7.2 we considered the simple electric circuit shown in Figure 4: An electromotive force (usually a battery or generator) produces a voltage of E͑t͒ volts (V) and a current of I͑t͒ amperes (A) at time t The circuit also contains a resistor with a resistance of R ohms (⍀) and an inductor with an inductance of L henries (H) Ohm’s Law gives the drop in voltage due to the resistor as RI The voltage drop due to the inductor is L͑dI͞dt͒ One of Kirchhoff’s laws says that the sum of the voltage drops is equal to the supplied voltage E͑t͒ Thus, we have FIGURE L dI ϩ RI E͑t͒ dt which is a first-order linear differential equation The solution gives the current I at time t ■ LINEAR DIFFERENTIAL EQUATIONS EXAMPLE Suppose that in the simple circuit of Figure the resistance is 12 ⍀ and the inductance is H If a battery gives a constant voltage of 60 V and the switch is closed when t so the current starts with I͑0͒ 0, find (a) I͑t͒, (b) the current after s, and (c) the limiting value of the current SOLUTION The differential equation in Example is both linear and separable, so an alternative method is to solve it as a separable equation (Example in Section 7.3) If we replace the battery by a generator, however, we get an equation that is linear but not separable (Example 5) ■ ■ (a) If we put L 4, R 12, and E͑t͒ 60 in Equation 7, we obtain the initial-value problem or dI ϩ 12I 60 dt I͑0͒ dI ϩ 3I 15 dt I͑0͒ Multiplying by the integrating factor e x dt e 3t, we get e 3t dI ϩ 3e 3tI 15e 3t dt d 3t ͑e I͒ 15e 3t dt e 3tI y 15e 3t dt 5e 3t ϩ C I͑t͒ ϩ CeϪ3t ■ ■ Figure shows how the current in Example approaches its limiting value Since I͑0͒ 0, we have ϩ C 0, so C Ϫ5 and I͑t͒ 5͑1 Ϫ eϪ3t ͒ (b) After second the current is y=5 I͑1͒ 5͑1 Ϫ eϪ3 ͒ Ϸ 4.75 A lim I͑t͒ lim 5͑1 Ϫ eϪ3t ͒ (c) tlϱ Ϫ lim eϪ3t 2.5 tlϱ tlϱ 5Ϫ05 FIGURE EXAMPLE Suppose that the resistance and inductance remain as in Example but, instead of the battery, we use a generator that produces a variable voltage of E͑t͒ 60 sin 30t volts Find I͑t͒ SOLUTION This time the differential equation becomes Figure shows the graph of the current when the battery is replaced by a generator ■ ■ dI ϩ 12I 60 sin 30t dt or dI ϩ 3I 15 sin 30t dt The same integrating factor e 3t gives dI d 3t ͑e I͒ e 3t ϩ 3e 3tI 15e 3t sin 30t dt dt 2.5 Using Formula 98 in the Table of Integrals, we have e 3tI y 15e 3t sin 30t dt 15 _2 FIGURE e 3t ͑3 sin 30t Ϫ 30 cos 30t͒ ϩ C 909 I 101 ͑sin 30t Ϫ 10 cos 30t͒ ϩ CeϪ3t LINEAR DIFFERENTIAL EQUATIONS ■ Since I͑0͒ 0, we get 50 Ϫ 101 ϩC0 I͑t͒ 101 ͑sin 30t Ϫ 10 cos 30t͒ ϩ 101 eϪ3t so 50 Exercises A Click here for answers 1–4 S Observe that, if n or 1, the Bernoulli equation is linear For other values of n, show that the substitution u y 1Ϫn transforms the Bernoulli equation into the linear equation Click here for solutions Determine whether the differential equation is linear yЈ ϩ e x y x y xyЈ ϩ ln x Ϫ x y ■ ■ 5–14 ■ ■ du ϩ ͑1 Ϫ n͒P͑x͒u ͑1 Ϫ n͒Q͑x͒ dx y ϩ sin x x 3yЈ yЈ ϩ cos y tan x ■ ■ ■ ■ ■ ■ ■ 24–26 ■ Solve the differential equation yЈ ϩ 2y 2e 24 xyЈ ϩ y Ϫxy yЈ x ϩ 5y x xyЈ Ϫ 2y x ■ 10 ϩ xy xyЈ dy x sin 2x ϩ y tan x, dx ■ ■ Ϫ͞2 Ͻ x Ͻ ͞2 15–20 ■ 17 ■ ■ ■ ■ ■ ■ ■ ■ Solve the initial-value problem 15 yЈ x ϩ y, 16 t ; ■ ■ ■ ■ ■ ■ ■ ■ ■ of E͑t͒ 40 sin 60t volts, the inductance is H, the resistance is 20 ⍀, and I͑0͒ A (a) Find I͑t͒ (b) Find the current after 0.1 s (c) Use a graphing device to draw the graph of the current function a capacitor with a capacitance of C farads (F), and a resistor with a resistance of R ohms (⍀) The voltage drop across the t Ͼ 0, y͑1͒ C dv Ϫ 2tv 3t 2e t , v͑0͒ dt 18 2xyЈ ϩ y 6x, x Ͼ 0, 19 xyЈ y ϩ x sin x, dy y Ϫ x, dx xϩ1 20 x ■ ■ ■ ■ ■ E y͑4͒ 20 R y͑͒ y͑1͒ 0, ■ ■ xϾ0 ■ ■ ■ ■ ■ ■ capacitor is Q͞C, where Q is the charge (in coulombs), so in this case Kirchhoff’s Law gives ; 21–22 Solve the differential equation and use a graphing calculator or computer to graph several members of the family of solutions How does the solution curve change as C varies? 21 xyЈ ϩ y x cos x, ■ ■ 29 The figure shows a circuit containing an electromotive force, y͑0͒ dy ϩ 2y t 3, dt ■ 28 In the circuit shown in Figure 4, a generator supplies a voltage dr ϩ r te t dt ■ ■ voltage of 40 V, the inductance is H, the resistance is 10 ⍀, and I͑0͒ (a) Find I͑t͒ (b) Find the current after 0.1 s du ϩ u ϩ t, t Ͼ 13 ͑1 ϩ t͒ dt 14 t ln t y3 y x x 27 In the circuit shown in Figure 4, a battery supplies a constant dy ϩ 2xy x 11 dx 12 25 yЈ ϩ 26 yЈ ϩ y xy x yЈ ϩ 2xy cos x xyЈ ϩ y sx Use the method of Exercise 23 to solve the differential equation ■ ■ ■ ■ ■ xϾ0 ■ RI ϩ But I dQ͞dt (see Example in Section 3.3), so we have 22 yЈ ϩ ͑cos x͒y cos x ■ ■ ■ ■ ■ 23 A Bernoulli differential equation (named after James Bernoulli) is of the form dy ϩ P͑x͒y Q͑x͒y n dx ■ Q E͑t͒ C ■ R dQ ϩ Q E͑t͒ dt C Suppose the resistance is ⍀, the capacitance is 0.05 F, a battery gives a constant voltage of 60 V, and the initial charge is Q͑0͒ C Find the charge and the current at time t ■ ■ LINEAR DIFFERENTIAL EQUATIONS 30 In the circuit of Exercise 29, R ⍀, C 0.01 F, Q͑0͒ 0, and E͑t͒ 10 sin 60t Find the charge and the current at time t 31 Let P͑t͒ be the performance level of someone learning a skill as a function of the training time t The graph of P is called a learning curve In Exercise 13 in Section 7.1 we proposed the differential equation dP k͓M Ϫ P͑t͔͒ dt as a reasonable model for learning, where k is a positive constant Solve it as a linear differential equation and use your solution to graph the learning curve 32 Two new workers were hired for an assembly line Jim processed 25 units during the first hour and 45 units during the second hour Mark processed 35 units during the first hour and 50 units the second hour Using the model of Exercise 31 and assuming that P͑0͒ 0, estimate the maximum number of units per hour that each worker is capable of processing 33 In Section 7.3 we looked at mixing problems in which the volume of fluid remained constant and saw that such problems give rise to separable equations (See Example in that section.) If the rates of flow into and out of the system are different, then the volume is not constant and the resulting differential equation is linear but not separable A tank contains 100 L of water A solution with a salt concentration of 0.4 kg͞L is added at a rate of L͞min The solution is kept mixed and is drained from the tank at a rate of L͞min If y͑t͒ is the amount of salt (in kilograms) after t minutes, show that y satisfies the differential equation 3y dy 2Ϫ dt 100 ϩ 2t Solve this equation and find the concentration after 20 minutes 34 A tank with a capacity of 400 L is full of a mixture of water and chlorine with a concentration of 0.05 g of chlorine per liter In order to reduce the concentration of chlorine, fresh water is pumped into the tank at a rate of L͞s The mixture is kept stirred and is pumped out at a rate of 10 L͞s Find the amount of chlorine in the tank as a function of time 35 An object with mass m is dropped from rest and we assume that the air resistance is proportional to the speed of the object If s͑t͒ is the distance dropped after t seconds, then the speed is v sЈ͑t͒ and the acceleration is a vЈ͑t͒ If t is the acceleration due to gravity, then the downward force on the object is mt Ϫ cv, where c is a positive constant, and Newton’s Second Law gives dv m mt Ϫ cv dt (a) Solve this as a linear equation to show that v mt ͑1 Ϫ eϪct͞m ͒ c (b) What is the limiting velocity? (c) Find the distance the object has fallen after t seconds 36 If we ignore air resistance, we can conclude that heavier objects fall no faster than lighter objects But if we take air resistance into account, our conclusion changes Use the expression for the velocity of a falling object in Exercise 35(a) to find dv͞dm and show that heavier objects fall faster than lighter ones LINEAR DIFFERENTIAL EQUATIONS ■ Answers No y e x ϩ CeϪ2x Yes Խ Խ y x ln x ϩ Cx 11 13 15 19 21 25 27 29 31 Click here for solutions S Ϫx y sx ϩ C͞x Ϫx 2 y Ϯ͓Cx ϩ 2͑͞5x͔͒Ϫ1͞2 (a) I͑t͒ Ϫ 4eϪ5t (b) Ϫ 4eϪ1͞2 Ϸ 1.57 A Ϫ4t Q͑t͒ 3͑1 Ϫ e ͒, I͑t͒ 12eϪ4t P(t) P͑t͒ M ϩ CeϪkt M y x ϩ Ce Ϫ e x e dx u ͑t ϩ 2t ϩ 2C͓͒͞2͑t ϩ 1͔͒ 2 y Ϫx Ϫ ϩ 3e x 17 v t 3e t ϩ 5e t y Ϫx cos x Ϫ x y sin x ϩ ͑cos x͒͞x ϩ C͞x x P(0) 33 y ͑100 ϩ 2t͒ Ϫ 40,000͑100 ϩ 2t͒Ϫ3͞2; 0.2275 kg͞L C=2 C=1 C=0.2 35 (b) mt͞c 10 C=_1 C=_2 _4 (c) ͑mt͞c͓͒t ϩ ͑m͞c͒eϪct͞m ͔ Ϫ m 2t͞c t ■ LINEAR DIFFERENTIAL EQUATIONS Solutions: Linear Differential Equations y + ex y = x2 y is not linear since it cannot be put into the standard linear form (1), y + P (x) y = Q(x) xy0 + ln x − x2 y = ⇒ xy − x2 y = − ln x ⇒ y0 + (−x) y = − ln x , which is in the standard linear x form (1), so this equation is linear Comparing the given equation, y0 + 2y = 2ex , with the general form, y0 + P (x)y = Q(x), we see that P (x) = R R and the integrating factor is I(x) = e P (x)dx = e dx = e2x Multiplying the differential equation by I(x) gives ¡ ¢0 R e2x y + 2e2x y = 2e3x ⇒ e2x y = 2e3x ⇒ e2x y = 2e3x dx ⇒ e2x y = 23 e3x + C ⇒ y = 23 ex + Ce−2x xy − 2y = x I(x) = e R [divide by x] ⇒ P (x) dx =e R (−2/x) dx equation (∗) by I(x) gives µ y + − x ¶ y = x (∗) −2 = e−2 ln|x| = eln|x| 1 y − 3y= x2 x x ⇒ y = x2 (ln |x| + C ) = x2 ln |x| + Cx2 µ = eln(1/x ) = 1/x2 Multiplying the differential ¶0 1 y = y = ln |x| + C ⇒ ⇒ x2 x x2 Since P (x) is the derivative of the coefficient of y0 [P (x) = and the coefficient is x], we can write the √ √ differential equation xy + y = x in the easily integrable form (xy)0 = x ⇒ xy = 23 x3/2 + C ⇒ √ y = 23 x + C/x R = ex Multiplying the differential equation y + 2xy = x2 by I(x) gives ³ ´0 2 2 ex y + 2xex y = x2 ex ⇒ ex y = x2 ex Thus 11 I(x) = e y = e−x 13 (1 + t) 2x dx hR i h R 2 x2 ex dx + C = e−x 12 xex − x2 e du + u = + t, t > [divide by + t] dt ⇒ u0 + P (t) u = Q(t) The integrating factor is I(t) = e R i 2 R dx + C = 12 x + Ce−x − e−x x2 e dx du + u = (∗), which has the form dt 1+t P (t) dt =e R [1/(1+t)] dt = eln(1+t) = + t Multiplying (∗) by I(t) gives us our original equation back We rewrite it as [(1 + t) u]0 = + t Thus, (1 + t) u = 15 y = x + y ⇒ R (1 + t) dt = t + 12 t2 + C t + 12 t2 + C t2 + 2t + 2C or u = 1+t (t + 1) R y0 + (−1)y = x I(x) = e (−1) dx = e−x Multiplying by e−x gives e−x y − e−x y = xe−x R (e−x y)0 = xe−x ⇒ e−x y = xe−x dx = −xe−x − e−x + C [integration by parts with u = x, ⇒ dv = e−x dx] 17 ⇒ u= ⇒ y = −x − + Cex y(0) = ⇒ −1 + C = ⇒ C = 3, so y = −x − + 3ex R 2 dv − 2tv = 3t2 et , v (0) = I(t) = e (−2t)dt = e−t Multiply the differential equation by I(t) to get dt ³ ´0 R dv 2 2 e−t − 2te−t v = 3t2 ⇒ e−t v = 3t2 ⇒ e−t v = 3t2 dt = t3 + C ⇒ v = t3 et + Cet dt 2 = v(0) = · + C · = C, so v = t3 et + 5et LINEAR DIFFERENTIAL EQUATIONS ■ R −1 1 y = x sin x I(x) = e (−1/x) dx = e− ln x = eln x = x x ả0 1 1 y = sin x ⇒ y = − cos x + C Multiplying by gives y0 − y = sin x ⇒ x x x x x 19 xy0 = y + x2 sin x ⇒ y0 − ⇒ y = −x cos x + Cx y(π) = ⇒ −π · (−1) + Cπ = ⇒ C = −1, so y = −x cos x − x R y = cos x (x 6= 0), so I(x) = e (1/x)dx = eln|x| = x (for x x > 0) Multiplying the differential equation by I(x) gives 21 y + xy0 + y = x cos x ⇒ (xy)0 = x cos x Thus, ·Z ¸ 1 y= x cos x dx + C = [x sin x + cos x + C] x x = sin x + C cos x + x x The solutions are asymptotic to the y-axis (except for C = −1) In fact, for C > −1, y → ∞ as x → 0+ , whereas for C < −1, y → −∞ as x → 0+ As x gets larger, the solutions approximate y = sin x more closely The graphs for larger C lie above those for smaller C The distance between the graphs lessens as x increases 23 Setting u = y 1−n , equation becomes dy du dy yn du un/(1−n) du = (1 − n) y −n or = = Then the Bernoulli differential dx dx dx − n dx − n dx un/(1−n) du du + P (x)u1/(1−n) = Q(x)un/(1−n) or + (1 − n)P (x)u = Q(x)(1 − n) − n dx dx y3 2 4u y = Here n = 3, P (x) = , Q(x) = and setting u = y −2 , u satisfies u0 − = − x x x x x x µZ µ ¶ ¶ R 2 Then I(x) = e (−4/x)dx = x−4 and u = x4 − dx + C = x4 + C = Cx4 + x 5x5 5x 25 y + ả1/2 Thus, y = ± Cx4 + 5x 27 (a) R dI dI + 10I = 40 or + 5I = 20 Then the integrating factor is e dt dt equation by the integrating factor gives e5t I(t) = e−5t £R dI + 5Ie5t = 20e5t dt dt ¡ ⇒ = e5t Multiplying the differential e5t I Â0 = 20e5t Ô 20e5t dt + C = + Ce−5t But = I(0) = + C, so I(t) = − 4e−5t (b) I(0.1) = − 4e−0.5 ≈ 1.57 A 29 R dQ + 20Q = 60 with Q(0) = C Then the integrating factor is e dt equation by the integrating factor gives e4t Q(t) = e−4t £R dQ + 4e4t Q = 12e4t dt dt ⇒ ¡ = e4t , and multiplying the differential e4t Q ¢0 = 12e4t Â Ô Ă 12e4t dt + C = + Ce−4t But = Q(0) = + C so Q(t) = − e−4t is the charge at time t and I = dQ/dt = 12e−4t is the current at time t 10 ■ LINEAR DIFFERENTIAL EQUATIONS 31 R dP + kP = kM , so I(t) = e k dt = ekt Multiplying the differential dt dP equation by I(t) gives ekt + kP ekt = kM ekt ⇒ dt ¡ kt ¢0 e P = kM ekt ⇒ ¡R ¢ P (t) = e−kt kM ekt dt + C = M + Ce−kt , k > Furthermore, it is reasonable to assume that ≤ P (0) ≤ M , so −M ≤ C ≤ µ 33 y(0) = kg Salt is added at a rate of 0.4 kg L ảà L ả =2 kg Since solution is drained from the tank at a rate of L/min, but salt solution is added at a rate of L/min, the tank, which starts out with 100 L of water, contains (100 + 2t) L of liquid after t Thus, the salt concentration at time t is µ y(t) kg 100 + 2t L ¶µ ¶ y(t) kg Salt therefore 100 + 2t L 3y kg Combining the rates at which salt enters 100 + 2t ả dy 3y dy and leaves the tank, we get =2− Rewriting this equation as + y = 2, we see that dt 100 + 2t dt 100 + 2t àZ ả ¢ ¡ dt it is linear I(t) = exp = exp 32 ln(100 + 2t) = (100 + 2t)3/2 Multiplying the differential 100 + 2t leaves the tank at a rate of equation by I(t) gives (100 + 2t)3/2 h (100 + 2t)3/2 y i0 L = dy + 3(100 + 2t)1/2 y = 2(100 + 2t)3/2 dt = 2(100 + 2t)3/2 ⇒ ⇒ (100 + 2t)3/2 y = 25 (100 + 2t)5/2 + C ⇒ y = 25 (100 + 2t) + C(100 + 2t)−3/2 Now = y(0) = 25 (100) + C · 100−3/2 = 40 + 1000 C ⇒ i h C = −40,000, so y = 25 (100 + 2t) − 40,000(100 + 2t)−3/2 kg From this solution (no pun intended), we y(t) calculate the salt concentration at time t to be C(t) = = 100 + 2t " −40,000 5/2 (100 + 2t) + # kg In particular, L −40,000 kg + ≈ 0.2275 and y(20) = 25 (140) − 40,000(140)−3/2 ≈ 31.85 kg 1405/2 L R c dv + v = g and I(t) = e (c/m)dt = e(c/m)t , and multiplying the differential equation by I(t) gives 35 (a) dt m h i0 dv vce(c/m)t e(c/m)t + = ge(c/m)t ⇒ e(c/m)t v = ge(c/m)t Hence, dt i hRm v(t) = e−(c/m)t ge(c/m)t dt + K = mg/c + Ke−(c/m)t But the object is dropped from rest, so v(0) = i h and K = −mg/c Thus, the velocity at time t is v(t) = (mg/c) − e−(c/m)t C(20) = (b) lim v(t) = mg/c t→∞ h i v(t) dt = (mg/c) t + (m/c)e−(c/m)t + c1 where c1 = s(0) − m2 g/c2 s(0) is the initial position, i h so s(0) = and s(t) = (mg/c) t + (m/c)e−(c/m)t − m2 g/c2 (c) s(t) = R