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# Engineering Mathematics MATH 3321

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This 32 page Class Notes was uploaded by Alvena McDermott on Saturday September 19, 2015. The Class Notes belongs to MATH 3321 at University of Houston taught by Staff in Fall. Since its upload, it has received 88 views. For similar materials see /class/208410/math-3321-university-of-houston in Mathmatics at University of Houston.

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23 Some Applications In this section we give some examples of applications of linear and separable differential equations 1 Orthogonal Trajectories The oneparameter family of curves ltw4VweNC mam w is a family of circles with center at the point 27 1 and radius If we differentiate this equation with respect to x we get 2z722y71y0 72 7y71 m This is the differential equation for the family of circles Note that if we choose a speci c point mo7 yo7 yo 7 1 on one of the circles7 then b gives the slope of the tangent line at 07110 Now consider the family of straight lines passing through the point 27 1 y71Kz72 C The differential equation for this family is y 7 verify this 1 Comparing equations b and d we see that right side of b is the negative reciprocal of the right side of Therefore we can conclude that if P mo yo is a point of intersection of one of the circles and one of the lines then the line and the circle are perpendicular orthogonal to each other at P The following figure shows the two families drawn in the same coordinate system A curve that intersects each member of a given family of curves at right angles orthog onally is called an orthogonal trajectory of the family Each line in c is an orthogonal trajectory of the family of circles a and conversely each circle in a is an orthogonal trajectory of the family of lines In general if Fmyc0 and GmyK0 are oneparameter families of curves such that each member of one family is an orthogonal trajectory of the other family then the two families are said to be orthogonal trajectories A procedure for nding a family of orthogonal trajectories Cz y K 0 for a given family of curves Fz y C 0 is as follows Step 1 Determine the differential equation for the given family Fz y C 0 Step 2 Replace y in that equation by ily the resulting equation is the differential equation for the family of orthogonal trajectories Step 3 Find the general solution of the new differential equation This is the family of orthogonal trajectories Example Find the orthogonal trajectories of the family of parabolas y sz SOLUTION You can verify that the differential equation for the family y Caz can be written as Replacing y by ily we get the equation 1 2 i i i if 7y which simpli es to y if y z 2y a separable equation Separating the variables7 we get 2343 7x or 2y dy imdm integrating with respect to m we have 1 2 y27 z20 or y20 This is a family of ellipses with center at the origin and major axis on the z axis I Exercises 231 Find the orthogonal trajectories for the family of curves 1 y 0mg 2 mCy4 3 y 0x2 2 4 y2 207 Find the orthogonal trajectories for the family of curves 0 The family of parabolas symmetric with respect to the y axis and vertex at the origin a The family of parabolas with vertical axis and vertex at the point 17 2 T The family of circles that pass through the origin and have their center on the z axis 00 The family of circles tangent to the m axis at 37 O Show that the given family is self orthogonal 9 y 4035 C 272 71 0274 2 m 10 3 2 Exponential Growth and Decay Radioactive Decay It has been observed and veri ed experimentally that the rate of decay of a radioactive material at time t is proportional to the amount of material present at time t Mathematically this says that if A At is the amount of radioactive material present at time 257 then A TA where r the constant of proportionality is negative To emphasize the fact that A is decreasing7 this equation is often written A ikA or ikA k gt 0 constant This is the form we shall use The constant of proportionality k is called the decay constant Note that this equation is both linear and separable and so we can use either method to solve it It is easy to show that the general solution is At 05 If A0 A0 is amount of material present at time t 07 then C A0 and At A0 5 Note that At 0 HalfLife An important property of a radioactive material is the length of time T it takes to decay to one half the initial amount This is the so called half life of the material Physicists and chemists characterize radioactive materials by their half lives To nd T we solve the equation A0 A0 ETkT for T A0 A0 eTkT eikT Conversely if we know the half life T of a radioactive material then the decay constant k is given by ln 2 k 7 T Example Cobalt 60 is a radioactive element that is used in medical radiology It has a half life of 53 years Suppose that an initial sample of cobalt 60 has a mass of 100 grams a Find the decay constant and determine an expression for the amount of the sample that will remain t years from now b How long will it take for 90 of the sample to decay SOLUTION a Since the half life T ln 2k we have ln2 ln2 kTEO131 With A0 100 the amount of material that will remain after t years is At 100 5013 b If 90 of the material decays then 10 which is 10 grams remains Therefore we solve the equation 100 5013 10 for t 1 0 1 701311 7 7 n 39 y e 701 0131t ln01 t 701317176 It will take approximately 176 years for 90 of the sample to decay I Population Growth Growth of an Investment It has been observed and veri ed experimentally that under ideal conditions a population eg bacteria fruit ies humans etc tends to increase at a rate proportional to the size of the population Therefore if P Pt is the size of a population at time t then dP E rP r gt 0 constant 1 In this case the constant of proportionality r is called the growth constant Similarly in a bank that compounds interest continuously the rate of increase of funds at time t is proportional to the amount of funds in the account at time t Thus equation 1 also represents the growth of a principal amount under continuous compounding Since the two cases are identical we ll focus on the population growth case 40 The general solution of equation 1 is Pt Cequot lf P0 P0 is the size of the population at time t 0 then Pt P0 5 is the size of the population at time 25 Note that Pt 00 In reality the rate of increase of a population does not continue to be proportional to the size of the population After some time has passed factors such as limitations on space or food supply introduc tion of diseases and so forth affect the growth rate the mathematical model is not valid indefinitely In contrast the model does hold inde nitely in the case of the growth of an investment under continuous compounding Doubling time The analog of the half life of a radioactive material is the so called doubling time the length of time T that it takes for a population to double in size Using the same analysis as above we have 2 A0 A0 ETT erT 2 TT ln 2 T M 7 In the banking investment and real estate communities there is a standard measure called the rule of 72 which states that the length of time approximately for a principal invested at r compounded continuously to double in value is 72r We know that the doubling time is T M z w n g B r r r r39 This is the origin of the rule of 7277 72 is used rather than 69 because it has more divisors I Example Scientists have observed that a small colony of penguins on a remote Antarctic island obeys the population growth law There were 2000 penguins initially and 3000 penguins 4 years later a How many penguins will there be after 10 years b How long will it take for the number of penguins to double SOLUTION Let Pt denote the number of penguins at time 25 Since P0 2000 we have Pt 2000 e 41 We use the fact that P4 3000 to determine the growth constant T 3000 2000 5 5 15 47 ln157 7 and so 1 15 n 7 0101 T 4 Therefore7 the number of penguins in the colony at any time t is Pt 2000 5010 a The number of penguins in the colony after 10 years is approximately P10 2000 503910110 2000 5101 e 5491 b To nd out how long it will take the number of penguins in the colony to double7 we need to solve 2000 5010 4000 for t ln 2 0101t 7 7 7 m e 7 27 0101t 7 ln 27 t 7 01017 686 years Note There is another way of expressing P that uses the exact value of r From the equation 3000 2000 5 we get r ln Thus t t t4 Pt 200051 111W 200051nl321 2000 l Exercises 232 1 A certain radioactive material is decaying at a rate proportional to the amount present If a sample of 50 grams of the material was present initially and after 2 hours the sample lost 10 of its mass7 find a An expression for the mass of the material remaining at any time t b The mass of the material after 4 hours c The half life of the material to What is the half life of a radioactive substance it takes 5 years for onethird of the material to decay OJ The size of a certain bacterial colony increases at a rate proportional to the size of the colony Suppose the colony occupied an area of 025 square centimeters initially7 and after 8 hours it occupied an area of 035 square centimeters 42 a Estimate the size of the colony 25 hours after the initial measurement b What is the expected size of the colony after 12 hours c Find the doubling time of the colony F A biologist observes that a certain bacterial colony triples every 4 hours and after 12 hours occupies 1 square centimeter a How much area did the colony occupy when rst observed b What is the doubling time for the colony 0 In 1980 the world population was approximately 45 billion and in the year 2000 it was approximately 6 billion Assume that the world population at each time 25 increases at a rate proportional to the population at time t Measure 25 in years after 1980 a Find the growth constant and give the world population at any time t b How long will it take for the world population to reach 9 billion double the 1980 population c The world population for 2002 was reported to be about 62 billion What population does the formula in a predict for the year 2002 It is estimated that the arable land on earth can support a maximum of 30 billion people Extrapolate from the data given in Exercise 5 to estimate the year when the a food supply becomes insufficient to support the world population 3 Newton s Law of CoolingHeating Newton s Law of Cooling states that the rate of change of the temperature u of an object is proportional to the difference between u and the constant temperature a of the surrounding medium eg7 air or water7 called the ambient temperature The mathematical formulation of this statement is du a mu 7 a m constant The constant of proportionality7 m7 in this model must be negative for if the object is warmer than the ambient temperature u 7 a gt 07 then its temperature will decrease dudt lt 07 which implies m lt 0 if the object is cooler than the ambient temperature u 7 a lt 07 then its temperature will increase dudt gt 07 which again implies m lt 0 To emphasize that the constant of proportionality is negative7 we write Newton s Law of Cooling as du a kw 7 a7 k gt 0 constant 1 43 This differential equation is both linear and separable so either method can be used to solve it As you can check the general solution is ut a 05 If the initial temperature of the object is 740 uo then u0a050a0 and 071070 Thus the temperature of the object at any time t is given by w 7 a 7 770 7 UV 2 The graphs of ut in the cases uo lt a and uo gt a are given below Note that limtnoo ut a in each case In the rst case u is increasing and its graph is concave down in the second case u is decreasing and its graph is concave up Example A metal bar with initial temperature 25 C is dropped into a container of boiling water 1000 C After 5 seconds the temperature of the bar is 350 C a What will the temperature of the bar be after 1 minute b How long will it take for the temperature of the bar to be Within 050 C of the boiling water SOLUTION Applying equation 2 the temperature of the bar at any time t is Tt 100 25 7100e t 100 7 75 5 The rst step is to determine the constant k Since T5 35 we have 35 100 7 75 a 75 a 65 75k ln6575 k 7 00286 Therefore Tt 100 7 75 570028675 44 a The temperature of the bar after 1 minute is7 approximately T60 100 7 75 50v0286lt60gt g 100 7 75 51717 g 8653quot b We want to calculate how long it will take for the temperature of the bar to reach 9950 Thus7 we solve the equation 995 100 7 75 500286t for t 995 100775 50028 775 50028 705 7008262 ln 0575 25 6066 seconds I Exercises 233 1 A thermometer is taken from a room where the temperature is 72 F to the outside where the temperature is 32quot F After 12 minute7 the thermometer reads 50 F a What will the thermometer read after it has been outside for 1 minute b How many minutes does the thermometer have to be outside for it to read 350 F 2 A metal ball at room temperature 200 C is dropped into a container of boiling water 1000 C given that the temperature of the ball increases 20 in 2 seconds7 find a The temperature of the ball after 6 seconds in the boiling water b How long it will take for the temperature of the ball to reach 900 C 3 Suppose that a corpse is discovered at 10 pm and its temperature is determined to be 85 F Two hours later7 its temperature is 740 F If the ambient temperature is 680 F7 estimate the time of death 4 Falling Objects With Air Resistance Consider an object with mass m in free fall near the surface of the earth The object experiences the downward force of gravity its weight as well as air resistance7 which may be modeled as a force that is proportional to velocity and acts in a direction opposite to the motion By Newton s second law of motion7 we have all k m i 7m 7 1 dt 9 7 where g is the gravitational acceleration constant and k gt 0 is the dmg coe cient that depends upon the density of the atmosphere and aerodynamic properties of the object Note if time is measured in seconds and distance in feet7 then 9 is approximately 32 feet 45 per second per second if time is measured in seconds and distance in meters7 then 9 is approximately 98 meters per second per second Rearrangement gives all iTU 97 where r km Once this equation is solved for the velocity v the height of the object is obtained by simple integration Exercises 234 1 a Solve the initial value problem in terms of r 97 and 120 b Show that vt a 7mgk as t a 00 This is called the terminal velocity of the object c lntegrate v to obtain the height y assuming an initial height y0 yo 2 An object with mass 10kg is dropped from a height of 200m Given that its drag coefficient is k 25 Nms7 after how many seconds does the object hit the ground 3 An object with mass 50 kg is dropped from a height of 200m lt hits the ground 10 seconds later Find the object s drag coefficient k 4 An object with mass 10 kg is projected upward from ground level with initial velocity 60 ms lt hits the ground 84 seconds later a Find the object s drag coefficient k b Find the maximum height c Find the velocity with which the object hits the ground 5 Mixing Problems Here we consider a tank7 or other type of container7 that contains a volume V of water in which some amount of impurity eg7 salt is dissolved Water containing the dissolved impurity at a known concentration ows or is pumped into the tank at a given volume flow mite7 and water ows out of the tank also at a given volume ow rate We assume that the water in the tank remains thoroughly mixed at all times Let At be the amount of impurity in the tank at time t The impurity concentration in the tank is then 1425V7 and At will satisfy a differential equation of the generic form 1 E in ow rate 7 out ow rate 46 These ow rates are products of the form concentration x volume ow rate So if we let Rm and Rout denote the volume ow rates and let kin denote the impurity concentration of the in ow7 then we have dA A E kinRin 7 V Rout When Rm Rout the volume V is constant If Rm 7 Rout but each is constant7 then V V0 l Rim 7 Rout t Exercises 235 1 A tank with a capacity of 2m3 2000 liters is initially full of pure water At time t 07 salt water with salt concentration 5 gramsliter begins to ow into the tank at a rate of 10 litersminute The well mixed solution in the tank is pumped out at the same rate a Set up7 and then solve7 the initial value problem for the amount of salt in the tank at time 25 minutes b Find the time when the salt concentration in the tank becomes 4 gramsliter to A 100 gallon tank is initially full of water At time t 07 a 20 hydrochloric acid solution begins to ow into the tank at a rate of 2 gallonsminute The well mixed solution in the tank is pumped out at the same rate a Set up7 and then solve7 the initial value problem for the amount of hydrochloric acid in the tank at time 25 minutes b Find the time when the hydrochloric acid concentration becomes 10 03 A room measuring 10 m x 5 m x 3 m initially contains air that is free of carbon monox ide At time t 07 air containing 3 carbon monoxide enters the room at a rate of 1 milminute7 and the well circulated air in the room leaves at the same rate a Set up7 and then solve7 the initial value problem for the amount of carbon monox ide in the room at time 25 minutes b Find the time when the carbon monoxide concentration in the room reaches 2 q A tank with a capacity of 1m3 1000 liters is initially half full of pure water At time t 07 4 salt solution begins to ow into the tank at a rate of 30 litersminute The well mixed solution in the tank is pumped out at a rate of 20 litersminute a Set up and then solve the initial value problem for the amount of salt in the tank between time t 0 and the time when the tank becomes full b Find the salt concentration of the solution in the tank during this process 01 A 100 gallon tank is initially full of pure water At time t 0 water containing salt at concentration 15 gramsgallon begins to ow into the tank at a rate of 1 gallonminute while the well mixed solution in the tank is pumped out at a rate of 2 gallonsminute a Set up and then solve the initial value problem for the amount of salt in the tank between time t 0 and the time when the tank becomes empty b Find the maximum amount of salt in the tank during this process 6 The Logistic Equation In the mid nineteenth century the Belgian mathematician RF Verhulst used the differential equation lagM 7 i lt1 where k and M are positive constants to study the population growth of various countries This equation is now known as the logistic equation and its solutions are called logistic functions Life scientists have used this equation to model the spread of an infectious disease through a population and social scientists have used it to study the ow of information In the case of an infectious disease if M denotes the number of people in the population and yt is the number of infected people at time t then the differential equation states that the rate of change of infected people is proportional to the product of the number of people who have the disease and the number of people who do not The constant M is called the carrying capacity of the enVironment Note that dydt gt 0 when 0 lt y lt M dydt 0 when y M and dydt lt 0 when y gt M The constant k is the intrinsic growth rate The differential equation 1 is separable It is also a Bernoulli equation We write the equation as 1 MM 24 mdyikdt 01 1 M 1 M dy 7 kt Cl partial fraction decomposition y M i y yik0 and integrate 1 1 Mlnlyli lnlMigl kt01 48 We can solve this equation for y as follows 1 y MIn Miy 7 kt01 In My MktM01Mkt02 02M01 7y y Mkt02 02 Mkt Mkt 02 7M7y e e 5 Ce C e Now in the context of this discussion y yt satis es 0 lt yt lt M Therefore yMi y gt 0 and we have 24 Mkt C M 7 y 6 Solving this equation for y we get C 4 y C eiMkt 39 Finally if y0 R R lt M then CM R R of which implies C 7 and t 7 Mk y R 7 R6 t The graph of this particular solution is shown below Note that y is an increasing function In the Exercises you are asked to show that the graph is concave up on 0 a and concave down on a 00 This means that the disease is spreading at an increasing rate up to time a after a the disease is still spreading but at a decreasing rate Note also that limtnoo yt M Example An in uenza virus is spreading through a small city with a population of 50000 people Assume that the virus spreads at a rate proportional to the product of the number of people who have been infected and the number of people who have not been infected lf 100 people were infected initially and 1000 were infected after 10 days find 49 a The number of people infected at any time t b How long it will take for half the population to be infected SOLUTION a Substituting the given data into equation b7 we have 10050 000 50 000 t 7 M l 100 49 900 5750900195 1 499 5750900 We can determine the constant k by applying the condition y10 1000 We have 50000 1000 149957500000k 4995750090019 49 75000009 ln49499 k g 00000046 Thus7 the number of people infected at time t is approximately D7 50000 y 1499e0v23t39 b To nd how long it will take for half the population to be infected7 we solve yt 257 000 for t 50 000 25 000 7 1 499570231 1 499 70231 7 5 499 t w E 27 days I Exercises 236 1 A rumor spreads through a small town with a population of 5000 at a rate propor tional to the product of the number of people who have heard the rumor and the number who have not heard it Suppose that 100 people initiated the rumor and that 500 people heard it after 3 days a How many people will have heard the rumor after 8 days b How long will it take for half the population to hear the rumor to Let y be the logistic function Show that dydt increases for y lt M2 and decreases for y gt M2 What can you conclude about dydt when y M2 50 3 Solve the logistic equation by means of the change of variables W U05 Mt vt 2v t Express the constant of integration in terms of the initial value y0 yo 4 Suppose that a population governed by a logistic model exists in an environment with carrying capacity of 800 If an initial population of 100 grows to 300 in 3 years7 nd the intrinsic growth rate k CHAPTER 1 Introduction to Differential Equations 11 Basic Terminology Most of the phenomena studied in the sciences and engineering involve processes that change with time For example7 it is well known that the rate of decay of a radioactive material at time t is proportional to the amount of material present at time t In mathematical terms this says that d dig ky k a negative constant 1 where y yt is the amount of material present at time t If an object7 suspended by a spring7 is oscillating up and down7 then Newton7s Second Law of Motion F ma combined with Hooke7s Law the restoring force of a spring is proportional to the displacement of the object results in the equation d2y 2 g k y 07 k a pos1t1ve constant 2 where y yt denotes the position of the object at time t The basic equation governing the diffusion of heat in a uniform rod of nite length L is given by Bu 7 3 at 812 where u uzt is the temperature of the rod at time t at position I on the rod Each of these equations is an example of what is known as a differential equation DIFFERENTIAL EQUATION A di ferential equation is an equation that contains an unknown function together with one or more of its derivatives Here are some additional examples of differential equations Example 1 2 I y i y a y y 1 b 12 7 235 2y 413 c g 27 0 Laplace7s equation d g 7 4 4 3671 TYPE As suggested by these examples7 a differential equation can be classi ed into one of two general categories determined by the type of unknown function appearing in the equation If the unknown function depends on a single independent variable then the equation is an ordinary di er ential equation if the unknown function depends on more than one independent variable then the equation is a partial di erential equationi According to this classi cation the differential equations 1 and 2 are ordinary differential equations and 3 is a partial differential equation In Exam ple 1 equations a b and d are ordinary differential equations and equation c is a partial differential equation Differential equations both ordinary and partial are also classi ed according to the highest ordered derivative of the unknown function ORDER The order of a differential equation is the order of the highest derivative of the unknown function appearing in the equation Equation 1 is a rst order equation and equations 2 and 3 are second order equations In Example 1 equation a is a rst order equation b and c are second order equations and equation d is a third order equation In general the higher the order the more complicated the equation In Chapter 2 we will consider some rst order equations and in Chapter 3 we will study certain kinds of second order equationsi Higher order equations and systems of equations will be considered in Chapter 6 The obvious question that we want to consider is that of solving a given differential equation SOLUTION A solution of a di erential equation is a function de ned on some interval I in the case of an ordinary differential equation or on some domain D in two or higher dimensional space in the case of a partial differential equation with the property that the equation reduces to an identity when the function is substituted into the equation Example 2 Given the secondorder ordinary differential equation 12 y 7 21 y 2y 413 Example 1 show that a 12 213 is a solution b 2x2 31 is not a solution SOLUTION a The rst step is to calculate the rst two derivatives of y y 12 213 y 21 612 y 2 121 Next we substitute y and its derivatives into the differential equation 122 121 7 2121 612 212 213 77 413 Simplifying the lefthand side we get 212 1213 7 412 712a3 212 413 i 413 and 413 7 413 The equation is satis ed y 12 213 is a solution b The rst two derivatives of 2 are 2 212 31 2 4x 3 ZN 4 Substituting into the differential equation we have 124 7 2141 3 2212 31 7 413i Simplifying the lefthand side we get 4x2781276x4126z 0413 The function 2 212 31 is not a solution of the differential equation I Example 3 Show that uz y cos I sinh y sin I cosh y is a solution of Laplace s equation 8211 8211 w wwi SOLUTION The rst step is to calculate the indicated partial derivatives 7 7 sin I sinh y cos I cosh y g 7 cos I sinh y 7 sin I cosh y 7 cos I cosh y sin I sinh y gig cos I sinh y sin I cosh yr Substituting into the differential equation we nd that 7 cos I sinh y 7 sin I cosh y cos I sinh y sin I cosh y 0 and the equation is satis ed uz y cos I sinh ysin z cosh y is a solution of Laplace s equationi Exercises 1 1 1i Classify the following differential equations With respect to type iiel7 ordinary or partial and order a b V y 2 zyy sin 1 y em tan 1 C 92 7 0 812 Bray 8y2 7 d2 3 cl zy2 If e f g y 751y ye 71 V BuBz MailBy dig dz2 dy d3 ZyE If i e 7 g 721 For each differential equation determine Whether or not the given functions are solutions 2 y 4y 0 sin 31 cos 21 2sin 21 dsy dy x 1 x 1 1 Big E76 yz7ls1nz e7 2z72cosz el 4 my y 0 y1r1n1I7 MI 12 5 z 1y my 7 y z 12 e 12 1 I2 1l d3 d2 d 6 E 7 ST 6 0 61621 02637 cl Cg constants7 262 3631 4 2 2 7 a a 0 11117yln12y27 u2zy 1373zy2l 812 By2 Sly 7y27z yze 172 21sinhz172l 8 82 9i 7 1626712 11117 67th cos I 11217 7th sin 27ml Find the set of all solutions of each of the following differential equations y 21lnz i y 32 i y 6zcos 21 dy i3l dz y dy xy70i Determine values of 7 if possible7 so that the given differential equation has a solution of the form y em yll74y0i iy2y 78y0i iyll76yl9y0i y 7 4y By 7 2y 0i iy 72y 5y0i Determine values of 7 if possible7 so that the given differential equation has a solution of the form y I7 d2y dy 2772 i 2 0 I dz2 zdzJr y i z2yzy 79y0i i 12y 7 31y 4y 0i 12 nParameter Family of Solutions General Solution Particular Solu tion Introduction You know from your experience in previous mathematics courses that the calculus of functions of several variables limits graphing quotm 39 39 39 quot and quot quot is more complicated than the calculus of functions of a single variable By extension therefore you would expect that the study of partial differential equations would be more complicated than the study of ordinary differential equations This is indeed the casel Since the intent of this material is to introduce some of the basic theory and methods for differential equations we shall con ne ourselves to ordinary differential equations from this point forward Hereafter the term di ereritial equation shall be interpreted to mean ordinary di erential equation Partial differential equations are studied in subsequent courses I We begin by considering the simple rstorder differential equation y 1 where f is some given function In this case we can nd y simply by integrating yfzdz FzC where F is an antiderivative of f and C is an arbitrary constant Not only did we nd a solution of the differential equation we found a whole family of solutions each member of which is determined by assigning a speci c value to the constant C In this context the arbitrary constant is called a parameter and the family of solutions is called a oneparameter family Remark ln calculus you learned that not only is each member of the family y FI C a solution of the differential equation but this family actually represents the set of all solutions of the equation that is there are no other solutions outside of this family I Example 1 The differential equation y 312 7 sin 21 has the oneparameter family of solutions 312 7 sin 21 dz IS 4 cos 21 C As noted above this family of solutions represents the set of all solutions of the equation I In a similar manner if we are given a second order equation of the form y f I then we can nd y by integrating twice with each integration step producing an arbitrary constant of integration Example 2 If y 61 4e27c then y 6x 4e dz 312 2e C1 y 312 2621 C1 dz 13 e21 C11 C2 C1 C2 arbitrary constants The set of functions y13621C11C2 is a twoparameter family of solutions of the differential equation y 61 4621 Again from calculus we can conclude that this family actually represents the set of all solutions of the differential equation there are no other solutions I nPARAMETER FAMILY OF SOLUTIONS The examples given above are very special cases In general to nd a set of solutions of an n th order differential equation we would expect intuitively to integrate n times with each integration step producing an arbitrary constant of integration As a result we expect an n th order differential equation to have an npammeter family of solutions SOLVING A DIFFERENTIAL EQUATION To solve an n th order differential equation means to nd an n parameter family of solutions It is important to understand that the two nls here are the same For example to solve a fourthorder differential equation we need to nd a fourparameter family of solutions Example 3 Show that y Ce is a oneparameter family of solutions of y Icy k a given constantl Equation 1 in Section 11 SOL U T ON y Ce 2 koala Substituting into the differential equation we get keel 7 k Gem keel we Thus y Ce is a oneparameter family of solutions You were shown in calculus that y Ce represents the set of all solutions of the equation Example 4 Show that y C112 C21 213 is a twoparameter family of solutions of 12y 7 21y 2y 413 SOLUTION We calculate the rst two derivatives of y and then substitute into the differential equation y C112C2I2x3 y 2C11 C2 612 y 2C1 121 7 12201 12x 7 21 2011 02 612 2 0112 021 213 39 4131 Simplifying the lefthand side and rearranging the terms we get 01212 7 412 212 02 721 21 1213 7 1213 413 i 413 010 020 413 413 413 413 Thus for any two constants C1 C2 the function y C112 C21 213 is a solution of the differential equation The set of functions y C112 C21 213 is a twoparameter family of solutions of the equation In Chapter 3 we will see that this twoparameter family represents the set of all solutions of the equation I GENERAL SOLUTIONSINGULAR SOLUTIONS For most of the equations that we will study in this course an nparameter family of solutions of a given nth order equation will represent the set of all solutions of the equation In such cases the term general solution is often used in place of nparameter family of solutions Because it less cumbersome we will use the term general solution77 rather than nparameter family of solutions77 recognizing that there is possible imprecision in the use of the term in some cases an nparameter family of solutions may not be the set of all solutions 1 Solutions of an nth order differential equation which are not included in an n parameter family of solutions are called singular solutionsi Example 5 Consider the differential equation y 41y 1 y I2 C2 1 is a oneparameter family of solutions verify this In Section 212 you will learn how to solve this equation Also it is easy to see that the constant function y E 1 is a solution of the equation y E 1 implies y E 0i and 0 411 7 1 2 0 the equation is satis ed This solution is not included in the general solution because there is no value that you can assign to C that will produce the solution y E 1 y E 1 is a singular solutioni Additional examples of differential equations having singular solutions are given in the Exercises 13 PARTICULAR SOLUTION 1f speci c values are assigned to the arbitrary constants in the general solution of a differential equation then the resulting solution is called a particular solution of the equation Example 6 a y C65 is the general solution of the rst order differential equation y 5y see Example 3 y 20065 is a particular solution of the equation b y C112 C21 213 is the general solution of the second order differential equation 12y 7 21y 2y 413i Setting C1 2 and C2 3 we get the particular solution y 212 31 213 I The Differential Equation of an nParameter Family If we are given an nparameter family of curves then we can regard the family as the general solution of an nthorder differential equation and attempt to nd the equation The equation that we search for called the di erential equation of the family should be free of the parameters arbitrary constants and its order should equal the number of parameters The strategy for nding the differential equation of a given nparameter family is to differentiate the equation n times This produces a system of n 1 equations which can be used to eliminate the parameters Example 7 Given the oneparameter family y CI2 3 Find the differential equation of the family SOLUTION Since we have a oneparameter family we are looking for a rst order equationi Dif ferentiating the given equation we obtain y 2C1 We can solve this equation for C to get C Substituting this into the given equation gives 1 y 12 3 which simpli es to my 7 2y 6 0 z This is the differential equation of the given oneparameter family I Example 8 Find the differential equation of the twoparameter family C yi1C2i z SOL UTION We are looking for a second order differential equation Differentiating twice we obtain the equations C1 2C1 if and H 7 y 12 y 13 Solving each of these equations for C1 we get C1 712M and C1 zgy Therefore zgy 712M which simpli es to my Zy 0 This is the differential equation of the given familyi Example 9 Find the differential equation of the twoparameter family y C1 cos 21 C2 sin 21 SOLUTION We differentiate twice y C1 cos 21 C2 sin 21 y 72C1 sin 21 2C2 cos 21 y 74C1 cos 21 7 4C2 sin 21 Multiplying the rst equation by 4 and adding it to the third equation7 we get y 4y 0 This is the differential equation of the given family I Remark These examples illustrate that there is no 77general method77 for nding the differential equation of a given nparameter family of functions You can only follow the strategy and try to nd some way to eliminate the parameters from the system of equations I 13 Initial Conditions InitialValue Problems As we noted in the preceding section we can obtain a particular solution of an nth order differential equation simply by assigning speci c values to the n constants in the general solution However in typical applications of differential equations you will be asked to nd a solution of a given equation that satis es certain preassigned conditions Example 1 Find a solution of 2 y 31 7 21 that passes through the point l 3 SOLUTION In this case we can nd the general solution by integrating y31272z dzzg712Ci The general solution is y 13 7 12 C To nd a solution that passes through the point 2 6 we set I 2 and y 6 in the general solution and solve for C 623722C874C whichimplies 02 3 Thus y z 7 12 2 is a solution of the differential equation that satis es the given condition In fact it is the only solution that satis es the condition since the general solution represented all solutions of the equation and the constant C was uniquely determined I Example 2 Find a solution of 12y 7 21y 2y 413 which passes through the point 1 4 with slope 2 SOLUTION As shown in Example 4 in the preceding section the general solution of the differential equation is y C112 C21 213 Setting 1 l and y 4 in the general solution yields the equation C1 C2 2 4 which implies C1 C2 2 The second condition slope 2 at z l is a condition on y We want y l 2 We calculate 1 y 2C11 C2 612 and then set I l and y 2 This yields the equation 201 C2 6 2 which implies 2C1 02 4 Now we solve the two equations simultaneously C1 C2 2 2C1 C2 74 We get C1 76 C2 8 A solution of the differential equation satisfying the two conditions is 7612 81 213 It will follow from our work in Chapter 3 that this is the only solution of the differential equation that satis es the given conditions INITIAL CONDITIONS Conditions such as those imposed on the solutions in Examples 1 and 2 are called initial conditions This term originated with applications where processes are usually observed over time starting with some initial state at time t 0 Example 3 The position yt of a weight suspended on a spring and oscillating up and down is governed by the differential equation y 9y 0 a Show that the general solution of the differential equation is yt C1 sin 3t C2 cos 3t b Find a solution that satis es the initial conditions y0 l y 0 72 SOL UTION a y C1 sin 3t C2 cos 3t y 3C1 cos 3t 7 SC sin 3t y 79C1 sin 3t 7 9C2 cos 3t Substituting into the differential equation we get y 9y 79C1 sin 3t 7 9C2 cos 3t 9 C1 sin 3t C2 cos 3t 0 Thus yt C1 sin 3t C2 cos St is the general solution b Applying the initial conditions we obtain the pair of equations y0 l C1 sin 0 C2 cos 0 C2 which implies C2 l yO 72 3C1 cos 0 7 SC sin 0 which implies C1 7 A solution which satis es the initial conditions is yt 7 sin 3t cos St I Any nth order differential equation with independent variable I and unknown function y can be written in the form F y39vy wym lhymd 7 0 1 12 by moving all the nonzero terms to the lefthand side Since we are talking about an nth order equation gm must appear explicitly in the expression F Each of the other arguments may or may not appear explicitly For example the thirdorder differential equation IQyH 7 gxyn yQExy written in the form of equation 1 is IQyH 7 gxyn 7 yQExy 0 and Fzyy y ym 12y 7 21y 7 erTy Note that y does not appear explicitly in the equation However it is there implicitly For example y y y y nth ORDER INITIALVALUE PROBLEM An nth order initialvalue problem consists of an n th order differential equation FIyy7y 7y WW 0 together with n initial conditions of the form MC 1607 40 k1 y 6 k2 WAVE 167171 where c and kg 161 16771 are given numbers It is important to understand that to be an nth order initialvalue problem there must be n conditions same n of exactly the form indicated in the de nition For example the problem 0 Find a solution of the differential equation y 9y 0 satisfying the conditions y0 0 y7r 0 is not an initialvalueproblem the two conditions are not of the form in the de nition Similarly the problem 0 Find a solution of the differential equation y 7 3y 7 y 0 satisfying the conditions y0 l y 0 2 is not an initialvalue problem a third order equation requires three conditions yc k0 y c k1 y c k2 EXISTENCE AND UNIQUENESS The fundamental questions in any course on differential equations are 1 Does a given initialvalue problem have a solution That is do solutions to the problem exist 2 If a solution does exist is it unique That is is there exactly one solution to the problem or is there more than one solution The initialvalue problems in Examples 17 2 and 3 each had a unique solution values for the arbitrary constants in the general solution were uniquely determined Example 4 The function y 12 is a solution of the differential equation y Z and y0 0 Thus the initialvalue problem y 2W y0 0 has a solution However7 y E 0 also satis es the differential equation and y0 0i Thus7 the initialvalue problem does not have a unique solution In fact7 for any positive number a the yaz 07 13a 17a7 zgta function is a solution of the initialvalue problemi I Y Example 5 The oneparameter family of functions y CI is the general solution of There is no solution that satis es y0 l the initialvalue problem 9 y g7 90 1 does not have a solution I The questions of existence and uniqueness of solutions Will be addressed in the speci c cases of interest to us A general treatment of existence and uniqueness of solutions of initialvalue problems is beyond the scope of this course Exercises 13 1 a Show that each member of the oneparameter family of functions y 065x is a solution of the differential equation y 7 5y 0 Find a solution of the initial value problem y 7 5y 07 y0 2 Show that each member of the twoparameter family of functions y 01621 02671 is a solution of the differential equation y 7 y 7 2y 0 Find a solution of the initial value problem y 7 y 7 2y 0 y0 2 y 0 1 Show that each member of the oneparameter family of functions 7 l iCeerl y is a solution of the differential equation y y y2l Find a solution of the initial value problem y y y2 yl 71 Show that each member of the threeparameter family of functions y 0212 C11 C0 is a solution of the differential equation y 0 Find a solution of the initial value problem y 0 yl l7 y l 4 y l 2 Find a solution of the initial value problem y 0 y2 y 2 y 2 0 Show that each member of the twoparameter family of functions y C1 sin 31 C2 cos 31 is a solution of the differential equation y 9y 0 Find a solution of the initial value problem y 9y 0 y7r2 y 7r2 1 Show that each member of the twoparameter family of functions y C112 C212 ln 1 is a solution of the differential equation 12y 7 3Iy 4y 0 Find a solution of the initial value problem 12y 7 3Iy 4y 0 yl 07 y l 1 Is there a member of the twoparameter family Which satis es the initial condition y0 y 0 0

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