dy dx and so we can rewrite the equation as If we now integrate both sides of this equation, we get xy x 2 C Integrating both sides, we would have

Transcription

1 LINEAR DIFFERENTIAL EQUATIONS A first-der linear differential equation is one that can be put into the fm 1 d Py Q 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 eample of a linear equation is y y because, f 0, it can be written in the fm y 1 y Notice that this differential equation is not separable because it s impossible to fact the epression f y as a function of times a function of y. But we can still solve the equation by noticing, by the Product Rule, that y y y and so we can rewrite the equation as y If we now integrate both sides of this equation, we get y C y C If we had been given the differential equation in the fm of Equation, we would have had to take the preliminary step of multiplying each side of the equation by. It turns out that every first-der linear differential equation can be solved in a similar fashion by multiplying both sides of Equation 1 by a suitable function I called an integrating fact. We try to find I so that the left side of Equation 1, when multiplied by I, becomes the derivative of the product Iy: 3 Iy Py Iy If we can find such a function I, then Equation 1 becomes Iy IQ Integrating both sides, we would have Iy y IQ d C so the solution would be 4 y 1 I y IQ d C To find such an I, we epand Equation 3 and cancel terms: Iy IPy Iy Iy Iy IP I 1

2 LINEAR DIFFERENTIAL EQUATIONS This is a separable differential equation f I, which we solve as follows: y di I y P d ln I y P d P d I Ae where A e C. We are looking f a particular integrating fact, not the most general one, so we take A 1 and use 5 P d I e Thus, a fmula f the general solution to Equation 1 is provided by Equation 4, where I is given by Equation 5. Instead of memizing this fmula, however, we just remember the fm of the integrating fact. To solve the linear differential equation ypy Q, multiply both sides by P d the integrating fact I e and integrate both sides. EXAMPLE 1 Solve the differential equation. d 3 y 6 SOLUTION The given equation is linear since it has the fm of Equation 1 with P 3 and Q 6. An integrating fact is I e 3 d e 3 Figure 1 shows the graphs of several members of the family of solutions in Eample 1. Notice that they all approach as l. 6 C= C=1 C=0 C=_1 _ Multiplying both sides of the differential equation by Integrating both sides, we have e 3 e 3 d 3 e 3y 6 e 3 d d e 3y 6 e 3, we get FIGURE 1 C= 3 e 3 y y 6 e 3 d e 3 C y Ce 3 EXAMPLE Find the solution of the initial-value problem yy 1 0 y1 SOLUTION We must first divide both sides by the coefficient of y to put the differential equation into standard fm: 6 y 1 y 1 0 The integrating fact is I e 1 d e ln

3 LINEAR DIFFERENTIAL EQUATIONS 3 Multiplication of Equation 6 by gives yy 1 y 1 The solution of the initial-value problem in Eample is shown in Figure. 5 (1, ) 0 4 _5 FIGURE Then and so Since y1, we have y y 1 d ln C y ln C ln 1 C 1 Therefe, the solution to the initial-value problem is y ln C EXAMPLE 3 Solve y y 1. SOLUTION The given equation is in the standard fm f a linear equation. Multiplying by the integrating fact e d e we get e ye y e Even though the solutions of the differential equation in Eample 3 are epressed in terms of an integral, they can still be graphed by a computer algebra system (Figure 3). FIGURE 3.5 C= _.5.5 _.5 C=_ (e y) e Therefe e y y e d C Recall from Section 6.4 that e d can t be epressed in terms of elementary functions. Nonetheless, it s a perfectly good function and we can leave the answer as y e y e d Ce Another way of writing the solution is y e y e t Ce (Any number can be chosen f the lower limit of integration.) 0 APPLICATION TO ELECTRIC CIRCUITS E R switch L Let s consider the simple electric circuit shown in Figure 4: An electromotive fce (usually a battery generat) produces a voltage of Et volts (V) and a current of It amperes (A) at time t. The circuit also contains a resist with a resistance of R ohms ( ) and an induct with an inductance of L henries (H). Ohm s Law gives the drop in voltage due to the resist as RI. The voltage drop due to the induct is LdI. One of Kirchhoff s laws says that the sum of the voltage drops is equal to the supplied voltage Et. Thus, we have FIGURE 4 7 L di RI Et which is a first-der linear differential equation. The solution gives the current I at time t.

4 4 LINEAR DIFFERENTIAL EQUATIONS EXAMPLE 4 Suppose that in the simple circuit of Figure 4 the resistance is 1 and the inductance is 4 H. If a battery gives a constant voltage of 60 V and the switch is closed when t 0 so the current starts with I0 0, find (a) It, (b) the current after 1 s, and (c) the limiting value of the current. The differential equation in Eample 4 is both linear and separable, so an alternative method is to solve it as a separable equation. If we replace the battery by a generat, however, we get an equation that is linear but not separable (Eample 5). SOLUTION (a) If we put L 4, R 1, and Et 60 in Equation 7, we obtain the initial-value problem 4 di di 1I 60 3I 15 I0 0 I0 0 Multiplying by the integrating fact e 3 e 3t, we get 3t di e 3e 3t I 15e 3t d e 3t I 15e 3t e 3t I y 15e 3t 5e 3t C It 5 Ce 3t Figure 5 shows how the current in Eample 4 approaches its limiting value. 6 y=5 Since I0 0, we have 5 C 0, so C 5 and It 51 e 3t (b) After 1 second the current is I1 51 e A 0.5 FIGURE 5 (c) lim It lim 51 t l t l e3t 5 5 lim t l e 3t EXAMPLE 5 Suppose that the resistance and inductance remain as in Eample 4 but, instead of the battery, we use a generat that produces a variable voltage of Et 60 sin 30t volts. Find It. SOLUTION This time the differential equation becomes Figure 6 shows the graph of the current when the battery is replaced by a generat. 0.5 _ FIGURE 6 4 di The same integrating fact 1I 60 sin 30t e 3t gives d e 3t 3t di I e 3e 3t I 15e 3t sin 30t Using Fmula 98 in the Table of Integrals, we have e 3t I y 15e 3t sin 30t 15 e 3t 3sin 30t 30 cos 30t C 909 I sin 30t 10 cos 30t Ce 3t di 3I 15 sin 30t

5 LINEAR DIFFERENTIAL EQUATIONS 5 Since I0 0, we get C 0 so It sin 30t 10 cos 30t e 3t EXERCISES 1 4 Determine whether the differential equation is linear. 1. y e y y. y sin 3 y 3. y ln y 0 4. y cos y tan 5 14 Solve the differential equation. 5. y y e 6. y 5y 7. y y 8. yy cos 9. y y s y y sin y tan, d t du u 1 t, 14. A Click here f answers. d y t ln t dr r tet 15 0 Solve the initial-value problem. 15. y y, y0 16. t, t 0, y t 3 dv 17., tv 3t e t 18. y y 6, 0, v0 5 t 0 y1 0 y y y sin, y 0 0., d y 1 y1 0, 0 ; 1 Solve the differential equation and use a graphing calculat computer to graph several members of the family of solutions. How does the solution curve change as C varies? 1. y y cos, 0. y cos y cos 3. A Bernoulli differential equation (named after James Bernoulli) is of the fm n Py Qy d S Click here f solutions. Observe that, if n 0 1, the Bernoulli equation is linear. F other values of n, show that the substitution u y 1n transfms the Bernoulli equation into the linear equation 4 6 Use the method of Eercise 3 to solve the differential equation. 6. y y y 3 du d y y y 3 y y y 7. In the circuit shown in Figure 4, a battery supplies a constant voltage of 40 V, the inductance is H, the resistance is 10, and I0 0. (a) Find It. (b) Find the current after 0.1 s. 8. In the circuit shown in Figure 4, a generat supplies a voltage of Et 40 sin 60t volts, the inductance is 1 H, the resistance is 0, and I0 1 A. (a) Find It. (b) Find the current after 0.1 s. ; (c) Use a graphing device to draw the graph of the current function. 9. The figure shows a circuit containing an electromotive fce, a capacit with a capacitance of C farads (F), and a resist with a resistance of R ohms ( ). The voltage drop across the E capacit is QC, where Q is the charge (in coulombs), so in this case Kirchhoff s Law gives But I dq, so we have 1 npu 1 nq R dq C RI Q C Et 1 C Q Et Suppose the resistance is 5, the capacitance is 0.05 F, a battery gives a constant voltage of 60 V, and the initial charge is Q0 0 C. Find the charge and the current at time t. R

6 6 LINEAR DIFFERENTIAL EQUATIONS 30. In the circuit of Eercise 9, R, C 0.01 F, Q0 0, of 3 Lmin. If yt is the amount of salt (in kilograms) after and Et 10 sin 60t. Find the charge and the current at time t. t minutes, show that y satisfies the differential equation 31. Psychologists interested in learning they stu learning curves. A learning curve is the graph of a function Pt, the perfmance of someone learning a skill as a function of the training time t. The derivative dp represents the rate at which perfmance improves. (a) When do you think P increases most rapidly? What happens to dp as t increases? Eplain. (b) If M is the maimum level of perfmance of which the learner is capable, eplain why the differential equation dp km P k a positive constant is a reasonable model f learning. (c) Solve the differential equation in part (b) as a linear differential equation and use your solution to graph the learning curve. 3. Two new wkers were hired f an assembly line. Jim processed 5 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 Eercise 31 and assuming that P0 0, estimate the maimum number of units per hour that each wker is capable of processing. 33. In Section 7.6 we looked at miing problems in which the volume of fluid remained constant and saw that such problems give rise to separable equations. (See Eample 5 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 kgl is added at a rate of 5 Lmin. The solution is kept mied and is drained from the tank at a rate 3y 100 t Solve this equation and find the concentration after 0 minutes. 34. A tank with a capacity of 400 L is full of a miture of water and chline with a concentration of 0.05 g of chline per liter. In der to reduce the concentration of chline, fresh water is pumped into the tank at a rate of 4 Ls. The miture is kept stirred and is pumped out at a rate of 10 Ls. Find the amount of chline 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 proptional to the speed of the object. If st is the distance dropped after t seconds, then the speed is v st and the acceleration is a vt. If t is the acceleration due to gravity, then the downward fce on the object is mt cv, where c is a positive constant, and Newton s Second Law gives m dv mt cv (a) Solve this as a linear equation to show that v mt c 1 ectm (b) What is the limiting velocity? (c) Find the distance the object has fallen after t seconds. 36. If we igne 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 epression f the velocity of a falling object in Eercise 35(a) to find dvdm and show that heavier objects do fall faster than lighter ones.

7 LINEAR DIFFERENTIAL EQUATIONS 7 ANSWERS S Click here f solutions. 1. No 3. Yes 5. y 3 e Ce 7. y ln C 9. y 3 s C y 1 Ce 1 e e d u t t Ct y 1 3e 17. v t 3 e t 5e t 19. y cos 1. y sin cos C 6 C= C=1 C= C=_1 C= 4 5. y C (a) It 4 4e 5t (b) 4 4e A 9. Qt 31 e 4t, It 1e 4t 31. (a) At the beginning; stays positive, but decreases (c) Pt M Ce kt P(t) M P(0) y t 40, t 3 ; 0.75 kgl 35. (b) mtc (c) mtct mce ctm m tc t

8 8 LINEAR DIFFERENTIAL EQUATIONS SOLUTIONS 1. y 0 + e y = y is not linear since it cannot be put into the standard linear fm (1), y 0 + P () y = Q(). 3. y 0 +ln y =0 y 0 y = ln y 0 +( ) y = ln, which is in the standard linear fm (1), so this equation is linear. 5. Comparing the given equation, y 0 +y =e, with the general fm, y 0 + P ()y = Q(),weseethatP () = and the integrating fact is I() =e P ()d = e d = e. Multiplying the differential equation by I() gives e y 0 +e y =e 3 e y 0 =e 3 e y = e 3 d e y = 3 e3 + C y = 3 e + Ce. 7. y 0 y = [divide by ] y 0 + y = ( ). I() =e P () d = e ( /) d = e ln = e ln = e ln(1/) =1/. Multiplying the differential equation ( ) byi() gives 1 y0 3 y = 1 y = (ln + C )= ln + C. 0 1 y = 1 1 y =ln + C 9. Since P () is the derivative of the coefficient of y 0 [P () =1and the coefficient is ], we can write the differential equation y 0 + y = in the easily integrable fm (y) 0 = y = 3 3/ + C y = 3 + C/. 11. I() =e d = e. Multiplying the differential equation y 0 +y = by I() gives 0 e y 0 +e y = e e y = e. Thus y = e e d + C = e 1 e 1 e d + C = 1 + Ce e 1 e d. 13. (1 + t) du du + u =1+t, t>0 [divide by 1+t] + 1 u =1 ( ), which has the fm 1+t u 0 + P (t) u = Q(t). The integrating fact is I(t) =e P (t) = e [1/(1+t)] = e ln(1+t) =1+t. Multiplying ( ) byi(t) gives us our iginal equation back. We rewrite it as [(1 + t) u] 0 =1+t. Thus, (1 + t) u = (1 + t) = t + 1 t + C u = t + 1 t + C 1+t u = t +t +C. (t +1) 15. y 0 = + y y 0 +( 1)y =. I() =e ( 1) d = e. Multiplying by e gives e y 0 e y = e 17. (e y) 0 = e e y = e d = e e + C [integration by parts with u =, dv = e d] y = 1+Ce. y(0) = 1+C = C =3,soy = 1+3e. dv tv =3t e t, v (0) = 5. I(t) =e ( t) = e t. Multiply the differential equation by I(t) to get t dv e te t v =3t 0 e t v =3t 5=v(0) = 0 1+C 1=C,sov = t 3 e t +5e t. e t v = 3t = t 3 + C v = t 3 e t + Ce t. 19. y 0 = y + sin y 0 1 y = sin. I() =e ( 1/) d = e ln = e ln 1 = 1. Multiplying by 1 gives 1 y0 1 y =sin 1 y 0 =sin 1 y = cos + C y = cos + C. y(π) =0 π ( 1) + Cπ =0 C = 1,soy = cos.

9 LINEAR DIFFERENTIAL EQUATIONS 9 1. y y =cos ( 6= 0), so I() =e (1/)d = e ln = (f >0). Multiplying the differential equation by I() gives y 0 + y = cos (y) 0 = cos. Thus, y = 1 cos d+ C = 1 [ sin +cos + C] =sin + cos + C The solutions are asymptotic to the y-ais (ecept f C = 1). In fact, f C> 1, y as 0 +,whereas f C< 1, y as 0 +. As gets larger, the solutions approimate y =sin me closely. The graphs f larger C lieabovethosefsmallerc. The distance between the graphs lessens as increases. 3. Setting u = y 1 n, du d =(1 n) y n d d = equation becomes un/(1 n) 1 n yn du 1 n d = un/(1 n) 1 n du. Then the Bernoulli differential d du d + P ()u1/(1 n) = Q()u n/(1 n) du +(1 n)p ()u = Q()(1 n). d 5. y 0 + y = y3.heren =3, P () =, Q() = 1 and setting u = y, u satisfies u 0 4u =. Then I() =e ( 4/)d = 4 and u = 4 d + C 6 Thus, y = ± C 4 + 1/. 5 = C = C (a) di di +10I =40 +5I =0. Then the integrating fact is e 5 = e 5t. Multiplying the differential equation by the integrating fact gives e 5t di +5Ie5t =0e 5t e 5t I 0 =0e 5t I(t) =e 5t 0e 5t + C =4+Ce 5t.But0=I(0) = 4 + C,soI(t) =4 4e 5t. (b) I(0.1) = 4 4e A 9. 5 dq +0Q =60with Q(0) = 0 C. Then the integrating fact is e 4 = e 4t, and multiplying the differential equation by the integrating fact gives e 4t dq +4e4t Q =1e 4t e 4t Q 0 =1e 4t Q(t) =e 4t 1e 4t + C =3+Ce 4t.But0=Q(0) = 3 + C so Q(t) =3 1 e 4t is the charge at time t and I = dq/ =1e 4t is the current at time t. 31. (a) P increases most rapidly at the beginning, since there are usually many simple, easily-learned sub-skills associated with learning a skill. As t increases, we would epect dp/ to remain positive, but decrease. This is because as time progresses, the only points left to learn are the me difficult ones. (b) dp = k(m P ) is always positive, so the level of perfmance P is increasing. As P gets close to M, dp/ gets close to 0; that is, the perfmance levels off, as eplained in part (a).

10 10 LINEAR DIFFERENTIAL EQUATIONS (c) dp + kp = km,soi(t) =e k = e kt. Multiplying the differential equation by I(t) gives e kt dp + kpekt = kme kt e kt P 0 = kme kt P (t) =e kt kme kt + C = M + Ce kt, k>0. Furtherme, it is reasonable to assume that 0 P (0) M,so M C y(0) = 0 kg. Salt is added at a rate of 0.4 kg 5 L = kg. Since solution is drained from the tank at a L min min rate of 3 L/min, but salt solution is added at a rate of 5 L/min, the tank, which starts out with 100 L of water, contains (100 + t) L of liquid after t min. Thus, the salt concentration at time t is leaves the tank at a rate of y(t) t and leaves the tank, we get = it is linear. I(t) =ep t kg 3 L = L min 3y t 3y t. Rewriting this equation as + y(t) t kg. Salt therefe L kg. Combining the rates at which salt enters min t y =,weseethat =ep 3 ln(100 + t) = (100 + t) 3/. Multiplying the differential equation by I(t) gives (100 + t) 3/ +3(100+t)1/ y =(100+t) 3/ 0 (100 + t) 3/ y =(100+t) 3/ (100 + t) 3/ y = ( t)5/ + C y = (100 + t)+c( t) 3/.Now0=y(0) = (100) + C / =40+ 1 C 1000 C = 40,000, soy = (100 + t) 40,000( t) 3/ kg. From this solution (no pun intended), we calculate the salt concentration at time t to be C(t) = y(t) t = 40,000 (100 + t) + 5/ 5 C(0) = 40, / kg L and y(0) = 5 (140) 40,000(140) 3/ kg. kg. In particular, L 35. (a) dv + c m v = g and I(t) =e (c/m) = e (c/m)t, and multiplying the differential equation by I(t) gives e (c/m)t dv + vce(c/m)t 0 = ge (c/m)t e v (c/m)t = ge (c/m)t. Hence, m v(t) =e (c/m)t ge (c/m)t + K = mg/c + Ke (c/m)t. But the object is dropped from rest, so v(0) = 0 and K = mg/c. Thus, the velocity at time t is v(t) =(mg/c) 1 e (c/m)t. (b) lim v(t) =mg/c t (c) s(t) = v(t) =(mg/c) t +(m/c)e (c/m)t + c 1 where c 1 = s(0) m g/c. s(0) is the initial position, so s(0) = 0 and s(t) =(mg/c) t +(m/c)e (c/m)t m g/c.