1 Di Erential Equations - University Of Texas At Dallas
1Differential EquationsBy: Patrick BourqueDesigned for students of MATH 2420 at The University of Texas at Dallas.
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Contents1 First Order Equations.51.1Separable Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .51.2First Order Linear Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .141.3Bernoulli Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .231.4Homogenous Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .271.5Shift to Homogenous . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .31α1.6The z Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .341.7Equations of the form: y’ G(ax by c) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .361.8Exact Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .401.9Integrating Factors for non-exact Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .431.10 Orthogonal Trajectories . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .482 Second Order Equations.34552.1Wronskian, Fundamental Sets and Able’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .552.2Reduction of Order . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .622.3Equations of the form y” f(x,y’) and y” f(y,y’) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .652.4Homogenous Linear Equations with Constant Coefficients. . . . . . . . . . . . . . . . . . . . . . . . . . . .702.5The Method of Undetermined Coefficients. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .742.6Variation of Parameters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .792.7Cauchy Euler Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .852.8Everyone Loves a Slinky: Springs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .912.9Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .99Series Solution1013.1Series Solutions Around Ordinary Points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1013.2Method of Frobenius: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1063.3The Gamma Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1133.4Bessel’s Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115Laplace Transform1194.1Calculating Laplace and Inverse Laplace Transforms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1194.2Solving Initial Value Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1254.3Unit Step Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1314.4Convolution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1343
4CONTENTS4.5Delta Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1405 First Order Systems of Differential Equations1435.1Linear Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1435.2Locally Linear Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1495.3Linear Systems and the Laplace Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153
Chapter 1First Order Equations.1.1Separable EquationsA Differential Equation is Separable if it can be written as:f (x)dx g(y)dyThe Solution is found by integrating both sides.An Example: Solve:ex ydx (e2x 1)dyy(0) 1Solution:exdx 2xe 1ZZdyyUsing the substitution u ex , du ex dx on the integral on the leftZdu 2u 1Zdyyarctan(ex ) C ln y arctan(u) C ln y Applying our initial conditionsπ C 04arctan(1) C ln(1)arctan(ex ) π ln y 4Solving for y gives the solution to the differential equation:5C π4
6CHAPTER 1. FIRST ORDER EQUATIONS.xy earctan(e) π4 An Application:A parachutist falling toward Earth is subject to two forces: the parachutist weight (w 32m) and the drag of theparachute. The drag of the parachute the drag is proportional to the velocity of the parachute and in this case is equalto 8 v . The parachutist weight is 128lb and initial velocity is zero. Find formulas for the parachutists velocity v(t) anddistance x(t).Since x(t) increases as the parachutist falls, the downward direction is the positive direction. The force from theparachutist’s weight acts in the positive direction while the drag from the parachute acts in the negative direction. Sincethe parachutist falls down (the positive direction) velocity is always positive so v v. The resultant force will be theforce of the weight of the parachutist minus the force of the drag of the parachute:F 128 8vAnd the mass of the parachutist is:128 32mm 4By Newton’s second law:F ma 4dvdtsincea dvdtThis gives a differential equation:4dv 128 8vdtThis equation is Separable:dv dt32 2vIntegrating 1ln(32 2v) t C2orln(32 2v) 2t CApplying the initial condition v(0) 0C ln(32)Solving for v(t) gives:v(t) 16 16e 2tThe parachutist terminal velocity is given by:
1.1. SEPARABLE EQUATIONS7lim v(t) lim 16 16e 2t 16ft/sect t We can now find an equation: x(t) for how far the parachutist has fallen:Zx(t) Zv(t)dt (16 16e 2t )dt 16t 8e 2t KSince x(0) 0 we see that K 8. Thus,x(t) 16t 8e 2t 81.Showy cosh(x) ex e x2is a solution to1y2K where K is the curvature y 00 K 1 (y 0 )2 232.Find the values of r so that y erx is a solution toy 000 6y 00 11y 0 6y 03.Find the values of k so that y sin(kx) is a solution toy 00 100y 04.Find the values of n so that y xn is a solution tox2 y 00 7xy 0 8y 05.The Clairaut Equation isy xy 0 f (y 0 )Show y kx f (k) is a solution for some constant k.Use the above result to find a solution to
8CHAPTER 1. FIRST ORDER EQUATIONS.y xy 0 (y 0 )36.Show(y ex 11 ex 0 xx 0is a solution toy 0 y 1Remember, you must use the definition of the derivative to calculate y 0 (0).7.Solve the differential equation(x2 1)dy (4x xy 2 )dxy(0) 28.Solve the differential equation2(2xy 2x)dx e x dypy( ln(5)) 09.Solve the differential equationpdy 16x2 y 4x2 y 2dxy(2) 110.Solve the differential equation ypdy2 ln x x 1dxy(1) e11.Solve the differential equationxdy (y 2 4y 5)px3 1dxy(0) 212.Solve the differential equation(x3 x2 x 1)dy (3x2 y 2xy y)dxy(0) e13.Solve the differential equation y xydy arcsin(x)dxy(0) 1
1.1. SEPARABLE EQUATIONS14.9ASolve the differential equationcos(2x)dy (1 sin(2x))(cos(x) sin(x))(1 y 2 )dxy15.Solve the initial value problempdy1 x2 2xydxy(1) 216.Solve the initial value problem4x ln(x) 4xy ln(x) ydydxy(e) 017.Solve the initial value problem15e y sin3 (x) cos6 (x)dydxy(0) ln(2)18.Solve the initial value problem1 y (1 sin(x))dydxy(0) 019.Solve(ex e x )(y 2 1) (2yex 2ye x )20.Solve35(y 2 1)dx (x 4 x 4 )dy21.Solve:dydx sin(x)cos2 (x) cos2 (x) y22.Solve:x2 (y 23.Solve: y)dx (x4 2x2 1)dydydx π 04
10CHAPTER 1. FIRST ORDER EQUATIONS.dyxex (y 2 4y 5) dxx2 2x 1y(0) 124.Solvexe3xdy 2(3x 1)2ydx25.Solve the differential equation y 1x 2dxdy y 1x 226.Solve the differential equation(xy 2x y 2)dx (x2 y 2 2xy 2 )dy27.Solve the differential equation4dy 1xy xy 2 x2 y x2 y 2 dx28.Use the Second Fundamental Theorem of Calculus to verifyy Ce Rxag(u)duis a solution toy 0 g(x)y 029.Use the substitution u yex to transform the equation into a separable equation and then solve itydx (1 y 2 e2x )dy 030.Use the substitution y zex to transform the equation into a separable equation and then solve itpdy y e2x y 2dx31.Use the substitution z y 2 x 1 to transform the equation into a separable equation and then solve it2ydy y2 x 1dx
1.1. SEPARABLE EQUATIONS11Some times it is useful to convert a differential equation to polar coordinates before solving it. The conversions topolar coordinates is:x r cos θy r sin θCalculating the total differential of both x and y we get:dx cos θdr r sin θdθdy sin θdr r cos θdθMakingdysin θdr r cos θdθ dxcos θdr r sin θdθUse this conversion to polar coordinates to solve the next two problems:32.Solve by converting to polar coordinatesx y xdydx33.Solve by converting to polar coordinates(2xy 3y 2 )dx (2xy x2 )dy34.Solve by converting to polar coordinatesdyy 3 x2 y x y 3dxx xy 2 x y35.Salt water containing .25 pounds of salt per gallon is being pumped into a tank initially containing 100 gallons ofwater and 10 pounds of salt at a rate of 4 gallons per minute. The mixture in the tank is kept well stirred and fluid flowsout of the tank at a rate of 4 gallons per minute. Find a formula that represents the amount of salt in the tank at anytime.36.The logistic differential equation that models the size of a population of species in an environment of fixed size is givenby the following differential equation:dP kP (M P )dtwhere M is the carrying capacity of the environment: that is, M is the maximum population of the species that can fitin the environment and k 0 is a constant depending on the reproduction rate of the species. For example if P representthe population of a bacteria in a petri dish then M would be the maximum population of the bacteria in the dish. Wealso see from the differential equation if a population P is less than M thenapproach the carrying capacity and if P is greater than M thenthe carrying capacity.dPdtdPdt 0 and the population will increase and 0 and the population will decrease and approach
12CHAPTER 1. FIRST ORDER EQUATIONS.Show the population is increasing fastest when the population is half the carrying capacity then solve this differentialequation for the population P (t) as a function of time and show:lim P (t) Mt 37.It has been calculated that the world population cannot exceed 20 billion people. In 1970 the population was 3.7 billionand in 2014 the population grew to 6.8 billion. Write a logistic differential equation representing the world populationand solve it for the world population as a function of time. When will the population exceed 10 billion?38.Another type of population model is the Gopertz growth model. It is similar to the logistic equation in that the modelassumes the population will increase at a rate proportional to the size of the population.That means the population willincrease a a rate of kP (t). The like the logistic model the Gopertz growth model also takes into account the maximumpopulation a species can have in an environment of fixed size and resources. Instead of using (M P (t)) as a factor likethe logistic model does the Gopertz growth model uses ln PM(t) as a factor, with M being the maximum population(carrying capacity). The Gopertz growth model isdP kP lndt MP k 0We also see from the differential equation if a population P is less than M thenincrease and approach the carrying capacity and if P is greater than M thenand approach the carrying capacity.Show the population is increasing fastest when the population isMedPdtdPdt 0 and the population will 0 and the population will decreaseand then solve this differential equation for thepopulation P (t) as a function of time and show:lim P (t) Mt 39.200 fish of a particular species are introduced to a lake which can sustain no more than 3000 fish. After 2 years the fishpopulation had increased to 800 fish. If the population follows the Gopertz growth model how long after the introductionof the fish to the lake will the population reach 2000 fish. Repeat the calculation using the logistic model and comparethe results.40.Differential equations can also be used to model the genetic change or evolution of a species. A commonly used hybridselection model isdy ky(1 y)(a by)dtWith y represents the portion of a population that has a certain characteristic and a, b, k constants and t is timemeasured in generations.At the beginning of a study of a population of a particular species it is found that half population had the advantageoustrait T and three generations later 60 percent of the population had trait T. Use the hybrid selection model with a 2and b 1 to determine the number of generations it will take until more than 80 percent of the population has trait T.
1.1. SEPARABLE EQUATIONS1341.Newton’s law of cooling states that an object with temperature T in a medium of constant temperature M willexperience a change in temperature proportional to the difference in the temperature of the object and the medium(M T ). This gives the differential equation:dT k(M T )dtA cup of 170 coffee is place in a 75 room. After 10 minutes the coffee is measured to have a temperature of 150 .How long will it take for the coffee to cool to 120 ?42.In the study of learning it has been shown that a person’s ability to learn a task is governed by the differential equation33dy2py 2 (1 y) 2 dtnWhere y represents the level that a student has learned a skill as a function of time and n and p are constantsdepending on the person learning the skill and the diffuculty of learning the skill. Solve this differential equation withthe initial condition y(0) 0 and p 1, n 4.43.Torricelli’s Law states that water draining from a tank of volume V (t) through a hole of area a in the bottom willhave an exiting velocity ofv(t) p2gy(t)where y(t) measures the height of the water level above the hole in the tank. The change in volume in the tank isgiven bypdV av(t) a 2gy(t)dtIf A(y) denotes the area of the cross section of the tank at height y then for any slice of water at a height of y andthickness dy will have volumeZV (y) yA(y)dy0Using the second fundamental theorem of calculus to differentiate this integral givesdydV A(y)dtdtequating this result to the previous formula fordVdtA(y)gives Torricelli’s Law:pdy a 2gy(t)dtUse the above results to find how long it takes a spherical tank with radius of 60 inches to be drained through a 1inch hole in the bottom.44.Show that if y1 and y2 are solutions to
14CHAPTER 1. FIRST ORDER EQUATIONS.y 0 P (x)y Q1 (x)andy 0 P (x)y Q2 (x)respectively then y y1 y2 is a solution toy 0 P (x)y Q1 (x) Q2 (x)45.Show that if y1 and y2 are solutions toy 0 P (x)y 2 Q1 (x)andy 0 P (x)y 2 Q2 (x)respectively then y y1 y2 is not a solution toy 0 P (x)y 2 Q1 (x) Q2 (x)46.There are about 3300 families of human languages spoken in the world. Assuming that all languages have evolvedfrom a single language and that one family of languages evolves into 1.5 families of language every 6000 years how longago was the original language spoken?47.For every point P(x, y) on a curve in the first quadrant, the rectangle containing the points O(0, 0) and P(x, y) asvertices is divided by the curve into two regions: upper region A and lower region B. If the curve contains the point Q(1,3), and region A always has twice the area of region B, find the equation of the curve.48.Find a function f (x) with the following properties: The average value of f on [1, x] is equal to twice the functionsvalue at x and f (2) 1.1.2First Order Linear EquationsA Differential Equation is First Order Linear if it has the form:dy P (x)y Q(x)dxTo solve this equation we recognize the left hand side:dydx P (x)y looks close to the derivative of the product ofsome function times y. Idea: multiply both sides of the equation by some function I(x) to make the left hand side thederivative of the product of I(x) times y. Multiplying both sides by I(x) gives:I(x)dy I(x)P (x)y I(x)Q(x)dx
1.2. FIRST ORDER LINEAR EQUATIONS15If the left hand side is the derivative of the product I(x) · y: ddyI(x) · y I(x) · I 0 (x) · ydxdxThen:I(x)dydy I(x)P (x) · y I(x) · I 0 (x) · ydxdxSoI(x)P (x) · y I 0 (x) · yI 0 (x) P (x)I(x)Integrating gives:Zln I(x) P (x)dxSolving for the Integrating Factor I(x) gives:RI(x) eP (x)dxAfter multiplying both sides of the original differential equation by I(x) the left hand side is the derivative of theproduct I(x) · y so the equation:I(x)dy I(x)P (x)y I(x)Q(x)dxBecomes: dI(x) · y I(x)Q(x)dxIntegrating gives: I(x) · yZ I(x)Q(x)dxAnd the solution is given by:y 1I(x) Z I(x)Q(x)dx CAn Example: Solve:dy sin(x)y sec2 (x)dxWriting the differential equation in standard form:cos(x)dy tan(x)y sec3 (x)dxCreating the integrating factor
16CHAPTER 1. FIRST ORDER EQUATIONS.I eRtan(x)dx eln(sec(x)) sec(x)Our Solution is:y 1sec(x) Zsec(x) sec3 (x)dx C Z Zsec2 (x)(1 tan2 (x))dx Csec4 (x)dx C cos(x)y cos(x) Z Z222y cos(x)sec (x)dx tan (x) sec (x)dx CUsing the substitution u tan(x), du sec2 (x)dx on the second integral ZZsec2 (x)dx u2 du Cy cos(x)u3y cos(x) tan(x) C3 13y cos(x) tan(x) tan (x) C3 ExampleIn this next example we will transform a nonlinear differential equation into a linear equation by converting it to polarcoordinates.(x2 y 2 x)dy ydxLet x r cos θ, y r sin θTherefore dx cos θdr r sin θdθ and dy sin θdr r cos θdθUnder this substitution our differential equation becomes:(r2 r cos θ)(sin θdr r cos θdθ) r sin θ(cos θdr r sin θdθ)Multiplying things out givesr2 sin θdr r3 cos θdθ r cos θ sin θdr r2 cos2 θdθ r sin θ cos θdr r2 sin2 θdθWhich reduces tor cos θdθ sin θdr dθ 0
1.2. FIRST ORDER LINEAR EQUATIONS17dr cot θr csc θdθThis is first order linear with integrating factorRI e cot θdθ sin θAnd the solution is.r 1sin θ Z sin θ csc θdθ Cr C θsin θwhich reduces toOrr sin θ C θConverting back to rectangular coordinate system gives the solution:y C arctan yx49.Solve:xy 0 3y x4y(1) 1y 0 ex y exy(0) 2e50.Solve:51.Solve:0y tan(x)y tan(x) πy 1452.Solve:p1 x2dy y 1dxy(0) 453.Solve:3dy6x2 4x 8ex 12x 3y dx x x2 4x 4(x 1)2 (x2 4)
18CHAPTER 1. FIRST ORDER EQUATIONS.54.Solve:cos(x) sin(x)dy y sec3 (x)dx cos(x) sin(x)y(0) 455.Solve:(1 x2 )dy xy 1dxy 31 2256.Solve:dy4x 1 y exdxxy(1) 057.Solve:(1 x2 )dy (4x2 4x 2)y 9 ln(x)dxy(1) 058.Solve:dy sin(x)y sin(2x)dx59.Solve:dyy1 2xdx 1 e xe 2xex x2y(0) 060.Solve:dy 2xy (2 x 2 )dxy(1) 061.Solve:(1 x2 )dy 2xy (1)dxy(0) 162.Solve:cos2 (x)y 0 y 163.Solve:y(0) 3
1.2. FIRST ORDER LINEAR EQUATIONS19sin(x)dy cos(x)y ln(x)dx64.Solve:(ex e x )dy dx (ex e x )2y 1(ex e x )65.Solve:(1 x4 )dy 4x3 y (x5 x) arctan(x2 )dxy(1) π66.Solve:(1 x2 )5dy xy (x2 1) 2dxy(0) 167.Solve:(1 x2 )dy 4xy x2dx68.Solve:6xdy y xdx x4 5x2 469.Solve:y xdydy y 2 eydxdx70.Solve:dy2 tan2 (x) y cos(x)dxx tan(x)71.Solve:dy (3x2 2x)y 3x5 5x4 2x3dx72.dy1 cos3 (x) y cos2 (x)dx sin(x) cos(x)(1 cos(x))73.Find all values of k so that the solution y approaches 0 as x approaches
20CHAPTER 1. FIRST ORDER EQUATIONS.y0 ky x2x74.Express the solution tody 1 2xydxin terms of the error function:2erf(x) πZx2e t dt075.The solution to the differential equationdy P (x)y Q(x)dxisy(x) Ce 2x t 1Find functions P (x) and Q(x).76.Salt water containing .25 pounds of salt per gallon is being pumped into a t
8 CHAPTER 1. FIRST ORDER EQUATIONS. y xy0 (y0)3 6. Show y (ex 1 x 0 1 xe x 0 is a solution to y0 jyj 1 Remember, you must use the de nition of the derivative to calculate y0(0). 7.
1.2. Rewrite the linear 2 2 system of di erential equations (x0 y y0 3x y 4et as a linear second-order di erential equation. 1.3. Using the change of variables x u 2v; y 3u 4v show that the linear 2 2 system of di erential equations 8 : du dt 5u 8v dv dt u 2v can be rewritten as a linear second-order di erential equation. 5
69 of this semi-mechanistic approach, which we denote as Universal Di erential 70 Equations (UDEs) for universal approximators in di erential equations, are dif- 71 ferential equation models where part of the di erential equation contains an 72 embedded UA, such as a neural network, Chebyshev expansion, or a random 73 forest. 74 As a motivating example, the universal ordinary di erential .
General theory of di erential equations of rst order 45 4.1. Slope elds (or direction elds) 45 4.1.1. Autonomous rst order di erential equations. 49 4.2. Existence and uniqueness of solutions for initial value problems 53 4.3. The method of successive approximations 59 4.4. Numerical methods for Di erential equations 62
Unit #15 - Di erential Equations Some problems and solutions selected or adapted from Hughes-Hallett Calculus. Basic Di erential Equations 1.Show that y x sin(x) ˇsatis es the initial value problem dy dx 1 cosx To verify anything is a solution to an equation, we sub it in and verify that the left and right hand sides are equal after
Ordinary Di erential Equations by Morris Tenenbaum and Harry Pollard. Elementary Di erential Equations and Boundary Value Problems by William E. Boyce and Richard C. DiPrima. Introduction to Ordinary Di erential Equations by Shepley L. Ross. Di
First Order Linear Di erential Equations A First Order Linear Di erential Equation is a rst order di erential equation which can be put in the form dy dx P(x)y Q(x) where P(x);Q(x) are continuous functions of x on a given interval. The above form of the equation is called the Standard Form of the equation.
1 Introduction XPP (XPPAUT is another name; I will use the two interchangeably) is a tool for solving di erential equations, di erence equations, delay equations, functional equations, boundary value problems, and stochastic equations. It evolved from a chapter written by
1 Introduction XPP(XPPAUT is anothername; I will use the twointerchangeably)is a tool for solving di erential equations, di erence equations, delay equations, functional equations, boundary value problems, and stochastic equations. It evolved from a chapter written by
