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Differential Equations I - » Department of Mathematics

Differential Equations IMATB44H3 FVersion September 15, 2011-1949iiContents1 Preliminaries.. Sample Application of Differential Equations ..22 First Order Ordinary Differential Separable Equations .. Exact Differential Equations .. Integrating Factors.. Linear First Order Equations .. Substitutions.. Bernoulli Equation.. Homogeneous Equations .. Substitution to Reduce Second Order Equations to FirstOrder..203 Applications and Examples of First Orderode Orthogonal Trajectories.. Exponential Growth and Decay.. Population Growth.. Predator-Prey Models.. Newton s Law of Cooling.. Water Tanks.. Motion of Objects Falling Under Gravity with Air Resistance.. Escape Velocity.. Planetary Motion.. Particle Moving on a Curve..39iiiivCONTENTS4 Linear Differential Homogeneous Linear Equations .. Linear Differential Equations with Constant Nonhomogeneous Linear Equations ..545 Second Order Linear Reduction of Order.

1.2. SAMPLE APPLICATION OF DIFFERENTIAL EQUATIONS 3 Sometimes in attempting to solve a de, we might perform an irreversible step. This might introduce extra solutions.

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Transcription of Differential Equations I - » Department of Mathematics

1 Differential Equations IMATB44H3 FVersion September 15, 2011-1949iiContents1 Preliminaries.. Sample Application of Differential Equations ..22 First Order Ordinary Differential Separable Equations .. Exact Differential Equations .. Integrating Factors.. Linear First Order Equations .. Substitutions.. Bernoulli Equation.. Homogeneous Equations .. Substitution to Reduce Second Order Equations to FirstOrder..203 Applications and Examples of First Orderode Orthogonal Trajectories.. Exponential Growth and Decay.. Population Growth.. Predator-Prey Models.. Newton s Law of Cooling.. Water Tanks.. Motion of Objects Falling Under Gravity with Air Resistance.. Escape Velocity.. Planetary Motion.. Particle Moving on a Curve..39iiiivCONTENTS4 Linear Differential Homogeneous Linear Equations .. Linear Differential Equations with Constant Nonhomogeneous Linear Equations ..545 Second Order Linear Reduction of Order.

2 Undetermined Coefficients.. Shortcuts for Undetermined Coefficients.. Variation of Parameters..666 Applications of Second Order Differential Motion of Object Hanging from a Spring.. Electrical Circuits..757 Higher Order Linear Differential Undetermined Coefficients.. Variation of Parameters.. Substitutions: Euler s Equation..828 Power Series Solutions to Linear Differential Introduction.. Background Knowledge Concerning Power Series.. Analytic Equations .. Power Series Solutions: Levels of Success.. Level 1: Finding a finite number of coefficients.. Level 2: Finding the recursion relation.. Solutions Near a Singular Point.. Functions Defined via Differential Equations .. Chebyshev Equation.. Legendre Equation.. Airy Equation.. Laguerre s Equation.. Bessel Equation..1169 Linear Preliminaries.. ComputingeT.. The 2 2 Case in Detail.. The Non-Homogeneous Case.. Phase Portraits.

3 Real Distinct Eigenvalues.. Complex Eigenvalues.. Repeated Real Roots..14110 Existence and Uniqueness Picard s Method.. Existence and Uniqueness Theorem for First Order ODE s.. Existence and Uniqueness Theorem for Linear First Order ODE Existence and Uniqueness Theorem for Linear Systems..15611 Numerical Euler s Method.. Error Bounds.. Improved Euler s Method.. Runge-Kutta Methods..167viCONTENTSC hapter PreliminariesDefinition ( Differential equation)Adifferential equation(de) is an equation involving a function and its Equations are calledpartial Differential Equations (pde) oror-dinary Differential Equations (ode) according to whether or not they containpartial derivatives. Theorderof a Differential equation is the highest orderderivative occurring. Asolution(orparticular solution) of a Differential equa-tion of ordernconsists of a function defined andntimes differentiable on adomainDhaving the property that the functional equation obtained by substi-tuting the function and itsnderivatives into the Differential equation holds forevery point example of a Differential equation of order 4, 2, and 1 isgiven respectively by(dydx)3+d4ydx4+y= 2 sin(x) + cos3(x), 2z x2+ 2z y2= 0,yy = 1.

4 12 CHAPTER 1. INTRODUCTIONE xample functiony= sin(x) is a solution of(dydx)3+d4ydx4+y= 2 sin(x) + cos3(x)on domainR; the functionz=excos(y) is a solution of 2z x2+ 2z y2= 0on domainR2; the functiony= 2 xis a solution ofyy = 2on domain (0, ). Although it is possible for adeto have a unique solution, ,y= 0 is thesolution to (y )2+y2= 0, or no solution at all, , (y )2+y2= 1 has nosolution, mostde s have infinitely many functiony= 4x+Con domain ( C/4, ) is a solutionofyy = 2 for any constantC. Note that different solutions can have different domains. The set of allsolutions to adeis call itsgeneral Sample Application of Differential EquationsA typical application of Differential Equations proceeds along these lines:Real World Situation Mathematical Model Solution of Mathematical Model Interpretation of SAMPLE APPLICATION OF Differential EQUATIONS3 Sometimes in attempting to solve ade, we might perform an irreversiblestep.

5 This might introduce extra solutions. If we can get a short list whichcontains all solutions, we can then test out each one and throw out the invalidones. The ultimate test is this:does it satisfy the equation?Here is a sample application of Differential half-life of radium is 1600 years, , it takes 1600 years forhalf of any quantity to decay. If a sample initially contains 50 g, how long willit be until it contains 45 g? (t) be the amount of radium present at timetin years. The rateat which the sample decays is proportional to the size of the sample at thattime. Therefore we know thatdx/dt=kx. This Differential equation is ourmathematical model. Using techniques we will study in this course (see ,Chapter3), we will discover that the general solution of this equation is givenby the equationx=Aekt, for some constantA. We are told thatx= 50 whent= 0 and so substituting givesA= 50. Thusx= 50ekt. Solving fortgivest= ln(x/50)/k.

6 Withx(1600) = 25, we have 25 = 50e1600k. Therefore,1600k= ln(12)= ln(2),giving usk= ln(2)/1600. Whenx= 45, we havet=ln(x/50)k=ln(45/50) ln(2)/1600= 1600 ln(8/10)ln(2)= 1600 ln(10/8)ln(2) 1600 1600 , it will be approximately years until the sample contains 45 gof radium. Additional conditions required of the solution (x(0) = 50 in the above ex-ample) are calledboundary conditionsand a Differential equation together withboundary conditions is called aboundary-value problem(BVP). Boundary con-ditions come in many forms. For example,y(6) =y(22);y (7) = 3y(0);y(9) = 5are all examples of boundary conditions. Boundary-value problems, like the onein the example, where the boundary condition consists of specifying the valueof the solution at some point are also calledinitial-value problems(IVP).Example analogy from algebra is the equationy= y+ 2.( )4 CHAPTER 1. INTRODUCTIONTo solve fory, we proceed asy 2 = y,(y 2)2=y,(irreversible step)y2 4y+ 4 =y,y2 5y+ 4 = 0,(y 1) (y 4) = , the sety {1,4}contains all the solutions.

7 We quickly see thaty= 4satisfies Equation ( ) because4 = 4 + 2 = 4 = 2 + 2 = 4 = 4,whiley= 1 does not because1 = 1 + 2 = 1 = we accepty= 4 and rejecty= 1. Chapter 2 First Order OrdinaryDifferential EquationsThe complexity of solvingde s increases with the order. We begin with firstorderde Separable EquationsA first orderodehas the formF(x, y, y ) = 0. In theory, at least, the methodsof algebra can be used to write it in the form y =G(x, y). IfG(x, y) canbe factored to giveG(x, y) =M(x)N(y),then the equation is solve the separable equationy =M(x)N(y), we rewrite it in the formf(y)y =g(x). Integrating both sides gives f(y)y dx= g(x)dx, f(y)dy= f(y) 2xy+ 6x+(x2 4)y = 0. We use the notationdy/dx=G(x, y) anddy=G(x, y) 2. FIRST ORDER ORDINARY Differential , we have(x2 4)y = 2xy 6x,= 2xy 6x,y y+ 3= 2xx2 4, x6= 2ln(|y+ 3|) = ln( x2 4 )+C,ln(|y+ 3|) + ln( x2 4 )=C,whereCis an arbitrary constant. Then (y+ 3)(x2 4) =A,(y+ 3)(x2 4)=A,y+ 3 =Ax2 4,whereAis a constant (equal to eC) andx6= 2.

8 Alsoy= 3 is a solution(corresponding toA= 0) and the domain for that solution isR. Example theivpsin(x)dx+y dy= 0, wherey(0) = 1. : sin(x)dx+y dy= 0 is an alternate notation meaning the sameas sin(x) +y dy/dx= havey dy= sin(x)dx, y dy= sin(x)dx,y22= cos(x) +C1,y= 2 cos(x) +C2,whereC1is an arbitrary constant andC2= 2C1. Consideringy(0) = 1, we have1 = 2 +C2= 1 = 2 +C2= C2= ,y= 2 cos(x) 1 on the domain ( /3, /3), since we need cos(x) 1/2 and cos( /3) = 1 EXACT Differential EQUATIONS7An alternate method to solving the problem isy dy= sin(x)dx, y1y dy= x0 sin(x)dx,y22 122= cos(x) cos(0),y22 12= cos(x) 1,y22= cos(x) 12,y= 2 cos(x) 1,giving us the same result as with the first method. Example +y +x2+ 1 = 0. have(y4+ 1)y = x2 1,y55+y= x33 x+C,whereCis an arbitrary constant. This is an implicit solution which we cannoteasily solve explicitly foryin terms ofx. Exact Differential EquationsUsing algebra, any first order equation can be written in the formF(x, y)dx+G(x, y)dy= 0 for some functionsF(x, y),G(x, y).

9 DefinitionAn expression of the formF(x, y)dx+G(x, y)dyis called a(first-order) differ-ential form. A differentical formF(x, y)dx+G(x, y)dyis calledexactif thereexists a functiong(x, y) such thatdg=F dx+G =F dx+G dyis an exact Differential form, then = 0 is called anexactdifferential equation. Its solution isg=C, where = the following useful theorem fromMATB42:8 CHAPTER 2. FIRST ORDER ORDINARY Differential EQUATIONST heorem functions that are continuously differentiable throughout asimply connected region, thenF dx+G dyis exact if and only if G/ x= F/ is given inMATB42. Example (3x2y2+x2)dx+(2x3y+y2)dy= 0. Let =(3x2y2+x2) Fdx+(2x3y+y2) GdyThen note that G x= 6x2y= F , =dgfor someg. To findg, we know that g x= 3x2y2+x2,( ) g y= 2x3y+y2.( )Integrating Equation ( ) with respect toxgives usg=x3y2+x33+h(y).( )So differentiating that with respect toygives usEq. ( ) g y= 2x3y+dhdy,2x3y+y2= 2x3y+dhdy,dhdy=y2,h(y) =y33+ EXACT Differential EQUATIONS9for some arbitrary constantC.

10 Therefore, Equation ( ) becomesg=x3y2+x33+y33+ that according to our Differential equation, we haved(x3y2+x33+y33+C)= 0 which impliesx3y2+x33+y33+C=C for some arbitrary constantC . LettingD=C C, which is still an arbitraryconstant, the solution isx3y2+x33+y33=D. Example (3x2+ 2xy2)dx+(2x2y)dy= 0, wherey(2) = 3. have (3x2+ 2xy2)dx=x3+x2y2+Cfor some arbitrary constantC. SinceCis arbitrary, we equivalently havex3+x2y2=C. With the initial condition in mind, we have8 + 4 9 =C= C= ,x3+x2y2= 44 and it follows thaty= 44 with the restriction thaty(2) = 3, the only solution isy= 44 x3x2on the domain( 3 44,3 44)\{0}. Let =F dx+G dy. Lety=s(x) be the solution of thede = 0, ,F+Gs (x) = 0. Lety0=s(x0) and let be the piece of the graph ofy=s(x)from (x0, y0) to (x, y). this idea. Sincey=s(x) is a solutionto = 0, we must have = 0 along . Therefore, = 0. This can be seen10 CHAPTER 2. FIRST ORDER ORDINARY Differential Equations (x0,y) 1(x0,y0)(x,y) 2 y =s(x)yxFigure : The graph ofy=s(x) with connecting (x0, y0) to (x, y).


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