# 15.2. First-Order Linear Differential Equations. First-Order Linear Differential Equations Bernoulli Equations Applications

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1 00 CHAPTER 5 Differential Equations SECTION 5. First-Orer Linear Differential Equations First-Orer Linear Differential Equations Bernoulli Equations Applications First-Orer Linear Differential Equations In this section, ou will see how integrating factors help to solve a ver important class of first-orer ifferential equations first-orer linear ifferential equations. Definition of First-Orer Linear Differential Equation A first-orer linear ifferential equation is an equation of the form Px Qx where P an Q are continuous functions of x. This first-orer linear ifferential equation is sai to be in stanar form. To solve a first-orer linear ifferential equation, ou can use an integrating factor ux, which converts the left sie into the erivative of the prouct ux. That is, ou nee a factor ux such that ux ux uxpx ux uxpx ux ux uxpx ux Px ux ux ln ux Px C ANNA JOHNSON PELL WHEELER ( ) Anna Johnson Pell Wheeler was aware a master s egree from the Universit of Iowa for her thesis The Extension of Galois Theor to Linear Differential Equations in 904. Influence b Davi Hilbert, she worke on integral equations while stuing infinite linear spaces. ux Ce Px. Because ou on t nee the most general integrating factor, let C. Multipling the original equation b ux e Px prouces Px Qx e Px Pxe Px QxePx epx QxePx. The general solution is given b epx Qxe Px C.

2 SECTION 5. First-Orer Linear Differential Equations 0 THEOREM 5.3 Solution of a First-Orer Linear Differential Equation An integrating factor for the first-orer linear ifferential equation Px Qx is ux e Px. The solution of the ifferential equation is e Px Qxe Px C. STUDY TIP Rather than memorizing this formula, just remember that multiplication b the integrating factor epx converts the left sie of the ifferential equation into the erivative of the prouct e Px. EXAMPLE Solving a First-Orer Linear Differential Equation Fin the general solution of Solution x x. The stanar form of the given equation is Px Qx x x. Thus, Px x, an ou have Px ln x x Stanar form e Px e ln x x. Integrating factor Figure 5.5 C = 0 C = C = 4 C = 3 C = C = x C = Therefore, multipling both sies of the stanar form b x iels x x 3 x x x x x x ln x C x ln x C. General solution Several solution curves for C,, 0,,, 3, an 4 are shown in Figure 5.5.

3 0 CHAPTER 5 Differential Equations EXAMPLE Solving a First-Orer Linear Differential Equation Fin the general solution of tan t, < t <. C = C = Figure 5.6 π π C = C = C = 0 t Solution The equation is alrea in the stanar form Thus, Pt tan t, an Pt tan t ln cos t which implies that the integrating factor is e Pt e ln cos t Integrating factor A quick check shows that cos t is also an integrating factor. Thus, multipling b cos t prouces tan t cos t. cos t cos t cos t cos t cos t sin t C tan t C sec t. Several solution curves are shown in Figure 5.6. General solution Pt Qt. Bernoulli Equations A well-known nonlinear equation that reuces to a linear one with an appropriate substitution is the Bernoulli equation, name after James Bernoulli ( ). Px Qx n Bernoulli equation This equation is linear if n 0, an has separable variables if n. Thus, in the following evelopment, assume that n 0 an n. Begin b multipling b n an n to obtain n Px n Qx n n npx n nqx n npx n nqx which is a linear equation in the variable n. Letting z n prouces the linear equation z npxz nqx. Finall, b Theorem 5.3, the general solution of the Bernoulli equation is n e npx nqxe npx C.

4 SECTION 5. First-Orer Linear Differential Equations 03 EXAMPLE 3 Solving a Bernoulli Equation Fin the general solution of Solution For this Bernoulli equation, let n 3, an use the substitution z 4 Let z n 3. z 4 3. Differentiate. Multipling the original equation b 43 prouces 3 x xe x 4 3 4x 4 4xe x z 4xz 4xe x. Original equation Multipl both sies b 4 3. Linear equation: This equation is linear in z. Using Px 4x prouces Px 4x x which implies that factor prouces e x x xe x 3. z Pxz Qx is an integrating factor. Multipling the linear equation b this z 4xz 4xe x ze x 4xze x 4xe x 4xe zex ze x x 4xe x Linear equation Exact equation Write left sie as total ifferential. Integrate both sies. ze x e x C z e x Ce x. Divie both sies b e x. Finall, substituting z 4, the general solution is 4 e x Ce x. General solution So far ou have stuie several tpes of first-orer ifferential equations. Of these, the separable variables case is usuall the simplest, an solution b an integrating factor is usuall a last resort. Summar of First-Orer Differential Equations Metho Form of Equation. Separable variables:. Homogeneous: Mx N 0 Mx, Nx, 0, where M an N are nth-egree homogeneous 3. Exact: Mx, Nx, 0, where M Nx 4. Integrating factor: ux, Mx, ux, Nx, 0 is exact 5. Linear: Px Qx 6. Bernoulli equation: Px Qx n

5 04 CHAPTER 5 Differential Equations Applications One tpe of problem that can be escribe in terms of a ifferential equation involves chemical mixtures, as illustrate in the next example. 4 gal/min Figure gal/min EXAMPLE 4 A Mixture Problem A tank contains 50 gallons of a solution compose of 90% water an 0% alcohol. A secon solution containing 50% water an 50% alcohol is ae to the tank at the rate of 4 gallons per minute. As the secon solution is being ae, the tank is being raine at the rate of 5 gallons per minute, as shown in Figure 5.7. Assuming the solution in the tank is stirre constantl, how much alcohol is in the tank after 0 minutes? Solution Let be the number of gallons of alcohol in the tank at an time t. You know that 5 when t 0. Because the number of gallons of solution in the tank at an time is 50 t, an the tank loses 5 gallons of solution per minute, it must lose 5 50 t gallons of alcohol per minute. Furthermore, because the tank is gaining gallons of alcohol per minute, the rate of change of alcohol in the tank is given b 5 50 t To solve this linear equation, let Pt 550 t an obtain Pt 5 50 t 5 ln 50 t. Because t < 50, ou can rop the absolute value signs an conclue that e Pt e 5 ln50t Thus, the general solution is 50 t 5 50 t 5 50 t C 4 Because 5 when t 0, ou have 5 50 C505 which means that the particular solution is 50 t Finall, when t 0, the amount of alcohol in the tank is 50 t t 5. C50 t t t C gal which represents a solution containing 33.6% alcohol.

6 SECTION 5. First-Orer Linear Differential Equations 05 In most falling-bo problems iscusse so far in the text, we have neglecte air resistance. The next example inclues this factor. In the example, the air resistance on the falling object is assume to be proportional to its velocit v. If g is the gravitational constant, the ownwar force F on a falling object of mass m is given b the ifference mg kv. But b Newton s Secon Law of Motion, ou know that F ma mv, which iels the following ifferential equation. m v mg kv v k m v g EXAMPLE 5 A Falling Object with Air Resistance An object of mass m is roppe from a hovering helicopter. Fin its velocit as a function of time t, assuming that the air resistance is proportional to the velocit of the object. Solution The velocit v satisfies the equation v kv m g where g is the gravitational constant an k is the constant of proportionalit. Letting b km, ou can separate variables to obtain v g bv v g bv b ln g bv t C ln g bv bt bc g bv Ce bt. Because the object was roppe, v 0 when t 0; thus g C, an it follows that bv g ge bt v g gebt b mg k ektm. E NOTE Notice in Example 5 that the velocit approaches a limit of mgk as a result of the air resistance. For falling-bo problems in which air resistance is neglecte, the velocit increases without boun. Figure 5.8 L S R I A simple electrical circuit consists of electric current I (in amperes), a resistance R (in ohms), an inuctance L (in henrs), an a constant electromotive force E (in volts), as shown in Figure 5.8. Accoring to Kirchhoff s Secon Law, if the switch S is close when t 0, the applie electromotive force (voltage) is equal to the sum of the voltage rops in the rest of the circuit. This in turn means that the current I satisfies the ifferential equation L I RI E.

7 06 CHAPTER 5 Differential Equations EXAMPLE 6 An Electric Circuit Problem Fin the current I as a function of time t (in secons), given that I satisfies the ifferential equation LI RI sin t, where R an L are nonzero constants. Solution In stanar form, the given linear equation is I R L I sin t. L Let Pt RL, so that e Pt e RLt, an, b Theorem 5.3, Ie RLt L e RLt sin t Thus, the general solution is I 4L R erlt R sin t L cos t C. I e RLt 4L R erlt R sin t L cos t C 4L R R sin t L cos t CeRLt. E X E R C I S E S F O R S E C T I O N 5. True or False? In Exercises an, etermine whether the statement is true or false. If it is false, explain wh or give an example that shows it is false. x x. is a first-orer linear ifferential equation.. is a first-orer linear ifferential equation. x e x In Exercises 3 an 4, (a) sketch an approximate solution of the ifferential equation satisfing the initial conition b han on the irection fiel, (b) fin the particular solution that satisfies the initial conition, an (c) use a graphing utilit to graph the particular solution. Compare the graph with the han-rawn graph of part (a). Differential Equation Initial Conition In Exercises 5, solve the first-orer linear ifferential equation x 3x 4 x 3x cos x sin x 0 0. sin x 0. x x. x x 5 e 5x 3. ex 4. sin x x 0, 0, Figure for 3 Figure for x In Exercises 3 8, fin the particular solution of the ifferential equation that satisfies the bounar conition. Differential Equation cos x x 3 e x e x 0 tan x sec x cos x sec x sec x x 0 Bounar Conition

8 SECTION 5. First-Orer Linear Differential Equations 07 In Exercises 9 4, solve the Bernoulli ifferential equation x x In Exercises 5 8, (a) use a graphing utilit to graph the irection fiel for the ifferential equation, (b) fin the particular solutions of the ifferential equation passing through the specifie points, an (c) use a graphing utilit to graph the particular solutions on the irection fiel x x 3 x 3 3 Differential Equation x x x x3 cot x x x x Electrical Circuits In Exercises 9 3, use the ifferential equation for electrical circuits given b L I RI E. In this equation, I is the current, R is the resistance, L is the inuctance, an E is the electromotive force (voltage). 9. Solve the ifferential equation given a constant voltage E Use the result of Exercise 9 to fin the equation for the current if I0 0, E 0 0 volts, R 550 ohms, an L 4 henrs. When oes the current reach 90% of its limiting value? 3. Solve the ifferential equation given a perioic electromotive force E 0 sin t. 3. Verif that the solution of Exercise 3 can be written in the form I ce RLt Points 0, 3, 0, E 0 R L sint x x, 4,, 8 0, 7, 0,,, 3, x x e x where, the phase angle, is given b arctanlr. (Note that the exponential term approaches 0 as t. This implies that the current approaches a perioic function.) 33. Population Growth When preicting population growth, emographers must consier birth an eath rates as well as the net change cause b the ifference between the rates of immigration an emigration. Let P be the population at time t an let N be the net increase per unit time resulting from the ifference between immigration an emigration. Thus, the rate of growth of the population is given b P kp N, N is constant. Solve this ifferential equation to fin P as a function of time if at time t 0 the size of the population is P Investment Growth A large corporation starts at time t 0 to continuousl invest part of its receipts at a rate of P ollars per ear in a fun for future corporate expansion. Assume that the fun earns r percent interest per ear compoune continuousl. Thus, the rate of growth of the amount A in the fun is given b A ra P where A 0 when t 0. Solve this ifferential equation for A as a function of t. Investment Growth Exercise 34. In Exercises 35 an 36, use the result of 35. Fin A for the following. (a) P \$00,000, r 6%, an t 5 ears (b) P \$50,000, r 5%, an t 0 ears 36. Fin t if the corporation nees \$800,000 an it can invest \$75,000 per ear in a fun earning 8% interest compoune continuousl. 37. Intravenous Feeing Glucose is ae intravenousl to the bloostream at the rate of q units per minute, an the bo removes glucose from the bloostream at a rate proportional to the amount present. Assume Qt is the amount of glucose in the bloostream at time t. (a) Determine the ifferential equation escribing the rate of change with respect to time of glucose in the bloostream. (b) Solve the ifferential equation from part (a), letting Q Q 0 when t 0. (c) Fin the limit of Qt as t. 38. Learning Curve The management at a certain factor has foun that the maximum number of units a worker can prouce in a a is 30. The rate of increase in the number of units N prouce with respect to time t in as b a new emploee is proportional to 30 N. (a) Determine the ifferential equation escribing the rate of change of performance with respect to time. (b) Solve the ifferential equation from part (a). (c) Fin the particular solution for a new emploee who prouce ten units on the first a at the factor an 9 units on the twentieth a.

9 08 CHAPTER 5 Differential Equations Mixture In Exercises 39 44, consier a tank that at time t 0 contains v 0 gallons of a solution of which, b weight, q 0 pouns is soluble concentrate. Another solution containing q pouns of the concentrate per gallon is running into the tank at the rate of r gallons per minute. The solution in the tank is kept well stirre an is withrawn at the rate of r gallons per minute. 39. If Q is the amount of concentrate in the solution at an time t, show that Q r Q v 0 r r t q r. 40. If Q is the amount of concentrate in the solution at an time t, write the ifferential equation for the rate of change of Q with respect to t if r r r. 4. A 00-gallon tank is full of a solution containing 5 pouns of concentrate. Starting at time t 0, istille water is amitte to the tank at a rate of 0 gallons per minute, an the well-stirre solution is withrawn at the same rate. (a) Fin the amount of concentrate Q in the solution as a function of t. (b) Fin the time at which the amount of concentrate in the tank reaches 5 pouns. (c) Fin the quantit of the concentrate in the solution as t. 4. Repeat Exercise 4, assuming that the solution entering the tank contains 0.05 poun of concentrate per gallon. 43. A 00-gallon tank is half full of istille water. At time t 0, a solution containing 0.5 poun of concentrate per gallon enters the tank at the rate of 5 gallons per minute, an the well-stirre mixture is withrawn at the rate of 3 gallons per minute. (a) At what time will the tank be full? (b) At the time the tank is full, how man pouns of concentrate will it contain? 44. Repeat Exercise 43, assuming that the solution entering the tank contains poun of concentrate per gallon. In Exercises 45 48, match the ifferential equation with its solution. Differential Equation x (a) Ce x 46. (b) Ce x 47. (c) x C 48. () Ce x 0 x 0 x x Solution 5. x 0 5. e x e x cos x cos x x ex 55. x cos x x x x x x e x x x x 0 6. x 4 x x x 3x x e x 0 x S E C T I O N PROJECT Weight Loss A person s weight epens on both the amount of calories consume an the energ use. Moreover, the amount of energ use epens on a person s weight the average amount of energ use b a person is 7.5 calories per poun per a. Thus, the more weight a person loses, the less energ the person uses (assuming that the person maintains a constant level of activit). An equation that can be use to moel weight loss is w C w where w is the person s weight (in pouns), t is the time in as, an C is the constant ail calorie consumption. (a) Fin the general solution of the ifferential equation. (b) Consier a person who weighs 80 pouns an begins a iet of 500 calories per a. How long will it take the person to lose 0 pouns? How long will it take the person to lose 35 pouns? (c) Use a graphing utilit to graph the solution. What is the limiting weight of the person? () Repeat parts (b) an (c) for a person who weighs 00 pouns when the iet is starte. In Exercises 49 64, solve the first-orer ifferential equation b an appropriate metho ex e x x

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