Lecture notes in Ordinary Differential Equations

lecture notes on ordinary differential equations pdf and lecture notes in applied differential equations of mathematical physics, lecture notes on introduction to partial differential equations pdf free download
Prof.WilliamsHibbs Profile Pic
Prof.WilliamsHibbs,United States,Teacher
Published Date:28-07-2017
Your Website URL(Optional)
Comment
Chapter 1 Basic Concepts and Classifying Differential Equations In This Chapter: ✔ Differential Equations ✔ Notation ✔ Solutions ✔ Initial-Value and Boundary-Value Problems ✔ Standard and Differential Forms ✔ Linear Equations ✔ Bernoulli Equations ✔ Homogeneous Equations ✔ Separable Equations ✔ Exact Equations 1 Copyright 2003 by The McGraw-Hill Companies, Inc. Click Here for Terms of Use.2 DIFFERENTIALEQUATIONS Differential Equations A differential equation is an equation involving an unknown function and its derivatives. Example 1.1: The following are differential equations involving the un- known function y. dy =+ 53 x (1.1) dx 2 2 dy dy   y (1.2) e +21 = 2   dx dx 3 2 dy dy (1.3) 45 ++ (sin x) xy=0 3 2 dx dx 3 2 7 2   dy dy dy     3 +35 y + y = x (1.4)   2     dx dx dx   2 2 ∂ y ∂ y −40 = (1.5) 2 2 ∂t ∂x Adifferential equation is an ordinary differential equation if the unknown function depends on only one independent variable. If the unknown func- tion depends on two or more independent variables, the differential equa- tion is a partial differential equation. In this book we will be concerned solely with ordinary differential equations. Example 1.2: Equations 1.1 through 1.4 are examples of ordinary differ- ential equations, since the unknown function y depends solely on the vari- able x. Equation 1.5 is a partial differential equation, since y depends on both the independent variables t and x.CHAPTER 1: Basic Concepts and Classification 3 Note The order of a differential equation is the order of the highest derivative appearing in the equation. Example 1.3: Equation 1.1 is a first-order differential equation; 1.2, 1.4, and 1.5 are second-order differential equations. (Note in 1.4 that the or- der of the highest derivative appearing in the equation is two.) Equation 1.3 is a third-order differential equation. Notation () 4 (n) The expressions yy′′ , ′′ ,y′′,y ,...,y are often used to represent, re- spectively, the first, second, third, fourth, . . ., nth derivatives of y with re- spect to the independent variable under consideration. Thus, repre- y′′ 22 22 sents dy /dx if the independent variable is x, but represents dy /dp (n) if the independent variable is p. Observe that parenthesis are used in y n to distinguish it from the nth power, y . If the independent variable is time, usually denoted by t, primes are often replaced by dots. Thus, 22 3 3 ˙ ˙˙ ˙˙˙ yy , , andy represent, dy/, dt d y/ dt, and d y/ dt, respectively. Solutions A solution of a differential equation in the unknown function y and the independent variable x on the interval  is a function y(x) that satisfies the differential equation identically for all x in . Example 1.4: Is yx ( )=+ c sin22 x c cos x, where c and c are arbi- 12 1 2 trary constants, a solution ofyy ′′+= 40? Differentiating y, we find y = 2c cos2x − 2c sin2x and y = 1 2 −− 42 cx sin 4c cos2x . Hence, 124 DIFFERENTIALEQUATIONS y′′+=44 y (−cx sin2−4c cos2x)+4(cx sin2+c cos2x) 12 1 2 =− (44 cc + )sin2x+ (−4c+4c )cos2x 11 2 2 = 0 Thus, satisfies the differential equation for all yc=+ sin22 x c cos x 12 values of x and is a solution on the interval (, −∞ ∞). 2 42 Example 1.5: Determine whether yx=−1 is a solution of ()yy ′ + =−1. Note that the left side of the differential equation must be nonnegative for every real function y(x) and any x, since it is the sum of terms raised to the second and fourth powers, while the right side of the equation is neg- ative. Since no function y(x) will satisfy this equation, the given differ- ential equation has no solutions. We see that some differential equations have infinitely many solu- tions (Example 1.4), whereas other differential equations have no solu- tions (Example 1.5). It is also possible that a differential equation has ex- 42 actly one solution. Consider ()yy ′ += 0, which for reasons identical to those given in Example 1.5 has only one solution y ≡ 0. You Need to Know A particular solution of a differential equation is any one solution. The general solution of a differential equation is the set of all solutions. Example 1.6: The general solution to the differential equation in Ex- ample 1.4 can be shown to be (see Chapters Four and Five) y = cx sin22 +c cos x. That is, every particular solution of the differential 12 equation has this general form. A few particular solutions are: (a) y = 52 sinxx −3cos2 (choose c = 5 and c =−3), (b) yx = sin 2 (choose 2 1 c = 1 and c = 0), and (c) y ≡ 0 (choose cc== 0). 1 2 12CHAPTER 1: Basic Concepts and Classification 5 The general solution of a differential equation cannot always be ex- pressed by a single formula. As an example consider the differential equa- 2 tion yy ′+= 0 , which has two particular solutions yx =1/ and y ≡ 0. Initial-Value and Boundary-Value Problems A differential equation along with subsidiary conditions on the unknown function and its de- rivatives, all given at the same value of the in- dependent variable, constitutes an initial-value problem. The subsidiary conditions are initial conditions. If the subsidiary conditions are giv- en at more than one value of the independent variable, the problem is a boundary-value prob- lem and the conditions are boundary conditions. x Example 1.7: The problem yy ′′ +21 ′== e ;(ypp ) ,y′( ) =2 is an initial value problem, because the two subsidiary conditions are both given at x x = p . The problem yy ′′ +20 ′== e ;(y )1,y(1)=1 is a boundary-value problem, because the two subsidiary conditions are given at x = 0 and x = 1. A solution to an initial-value or boundary-value problem is a func- tion y(x) that both solves the differential equation and satisfies all given subsidiary conditions. Standard and Differential Forms Standard form for a first-order differential equation in the unknown func- tion y(x) is yf ′ = (,xy) (1.6) where the derivative y′ appears only on the left side of 1.6. Many, but not all, first-order differential equations can be written in standard form by algebraically solving for y′ and then setting f(x,y) equal to the right side of the resulting equation.6 DIFFERENTIALEQUATIONS The right side of 1.6 can always be written as a quotient of two oth- er functions M(x,y) and −N(x,y). Then 1.6 becomes dy/( dx = M x, y) / −Nx (,y), which is equivalent to the differential form Mx (,y)dx+= N(, x y)dy 0 (1.7) Linear Equations Consider a differential equation in standard form 1.6. If f(x,y) can be writ- ten as fx (,y)=−p(x)y+q(x) (that is, as a function of x times y, plus an- other function of x), the differential equation is linear. First-order linear differential equations can always be expressed as yp ′+= ()xy q()x (1.8) Linear equations are solved in Chapter Two. Bernoulli Equations A Bernoulli differential equation is an equation of the form n yp ′+= ()xy q()xy (1.9) where n denotes a real number. When n = 1 or n = 0, a Bernoulli equation reduces to a linear equation. Bernoulli equations are solved in Chapter Two. Homogeneous Equations A differential equation in standard form (1.6) is homogeneous if ft(,xty) = f(x,y) (1.10) for every real number t. Homogeneous equations are solved in Chapter Two.CHAPTER 1: Basic Concepts and Classification 7 Note In the general framework of differential equations, the word “homogeneous” has an entirely different meaning (see Chapter Four). Only in the context of first-order differential equations does “homoge- neous” have the meaning defined above. Separable Equations Consider a differential equation in differential form (1.7). If M(x,y) = A(x) (a function only of x) and N(x,y) = B(y) (a function only of y), the differ- ential equation is separable, or has its variables separated. Separable equations are solved in Chapter Two. Exact Equations A differential equation in differential form (1.7) is exact if ∂Mx (,y) ∂Nx (,y) = (1.11) ∂y ∂x Exact equations are solved in Chapter Two (where a more precise defini- tion of exactness is given).Chapter 2 Solutions of First-Order Differential Equations In This Chapter: ✔ Separable Equations ✔ Homogeneous Equations ✔ Exact Equations ✔ Linear Equations ✔ Bernoulli Equations ✔ Solved Problems Separable Equations General Solution The solution to the first-order separable differential equation (see Chap- ter One). A() x dx+= B(y)dy 0 (2.1) 8 Copyright 2003 by The McGraw-Hill Companies, Inc. Click Here for Terms of Use.CHAPTER 2: Solutions of First-Order Differential Equations 9 is A() x dx+= B(y)dy c (2.2) ∫ ∫ where c represents an arbitrary constant. (See Problem 2.1) The integrals obtained in Equation 2.2 may be, for all practical pur- poses, impossible to evaluate. In such case, numerical techniques (see Chapter 14) are used to obtain an approximate solution. Even if the indi- cated integrations in 2.2 can be performed, it may not be algebraically possible to solve for y explicitly in terms of x. In that case, the solution is left in implicit form. Solutions to the Initial-Value Problem The solution to the initial-value problem (2.3) A() x dx+= B(y)dy 0; y(x )= y 00 can be obtained, as usual, by first using Equation 2.2 to solve the differ- ential equation and then applying the initial condition directly to evalu- ate c. Alternatively, the solution to Equation 2.3 can be obtained from x y A() s ds+= B(t)dt 0 (2.4) ∫∫ x y 00 where s and t are variables of integration. Homogeneous Equations The homogeneous differential equation dy =fx (,y) (2.5) dx having the property f(tx, ty) = f(x, y) (see Chapter One) can be trans- formed into a separable equation by making the substitution10 DIFFERENTIALEQUATIONS y = xv (2.6) along with its corresponding derivative dy dv =+vx (2.7) dx dx The resulting equation in the variables v and x is solved as a separable differential equation; the required solution to Equation 2.5 is obtained by back substitution. Alternatively, the solution to 2.5 can be obtained by rewriting the dif- ferential equation as dx 1 = (2.8) dy f(, x y) and then substituting x = yu (2.9) and the corresponding derivative dx du =+ uy (2.10) dy dy into Equation 2.8. After simplifying, the resulting differential equation will be one with variables (this time, u and y) separable. Ordinarily, it is immaterial which method of solution is used. Occa- sionally, however, one of the substitutions 2.6 or 2.9 is definitely superi- or to the other one. In such cases, the better substitution is usually appar- ent from the form of the differential equation itself. (See Problem 2.2) Exact Equations Defining Properties A differential equation M(x, y)dx + N(x, y)dy = 0 (2.11)CHAPTER 2: Solutions of First-Order Differential Equations 11 is exact if there exists a function g(x, y) such that dg(x, y) = M(x, y)dx + N(x, y)dy (2.12) Note Test for exactness: If M(x,y) and N(x,y) are con- tinuous functions and have continuous first partial derivatives on some rectangle of the xy-plane, then Equation 2.11 is exact if and only if ∂Mxy (, ) ∂N(x,y) = (2.13) ∂y ∂x Method of Solution To solve Equation 2.11, assuming that it is exact, first solve the equations ∂gxy (, ) =Mx (,y) (2.14) ∂x ∂gxy (, ) =Nx (,y) (2.15) ∂y for g(x, y). The solution to 2.11 is then given implicitly by g(x, y) = c (2.16) where c represents an arbitrary constant. Equation 2.16 is immediate from Equations 2.11 and 2.12. If 2.12 is substituted into 2.11, we obtain dg(x, y(x)) = 0. Integrating this equation (note that we can write 0 as 0dx), we have dgx(,y x( )) = 0dx , which, ∫ ∫ in turn, implies 2.16. Integrating Factors In general, Equation 2.11 is not exact. Occasionally, it is possible to trans- form 2.11 into an exact differential equation by a judicious multiplica- tion. A function I(x, y) is an integrating factor for 2.11 if the equation12 DIFFERENTIALEQUATIONS I(x, y)M(x, y)dx + N(x, y)dy = 0 (2.17) is exact. A solution to 2.11 is obtained by solving the exact differential equation defined by 2.17. Some of the more common integrating factors are displayed in Table 2.1 and the conditions that follow: 1  ∂M ∂N  If − ≡gx ( ), a function of x alone, then   N  ∂y ∂x  gxd () x ∫ Ix (,y) =e (2.18) 1  ∂M ∂N  If − ≡hy ( ), a function of y alone, then   M  ∂y ∂x  − hy ()dy ∫ Ix (,y) =e (2.19) If M = yf(xy) and N = xg(xy), then 1 Ix (,y) = (2.20) xM − yN In general, integrating factors are difficult to uncover. If a differential equation does not have one of the forms given above, then a search for an integrating factor likely will not be successful, and other methods of solution are recommended. (See Problems 2.3–2.6) Linear Equations Method of Solution A first-order linear differential equation has the form (see Chapter One) yp ′+= ()xy q()x (2.21) An integrating factor for Equation 2.21 is px()dx ∫ Ix () =e (2.22)CHAPTER 2: Solutions of First-Order Differential Equations 13 Table 2.1 which depends only on x and is independent of y. When both sides of 2.21 are multiplied by I(x), the resulting equation I() xy′+= p() xI() xy I() x q() x (2.23) is exact. This equation can be solved by the method described previous- ly. A simpler procedure is to rewrite 2.23 as dy() I = Iq() x dx14 DIFFERENTIALEQUATIONS integrate both sides of this last equation with respect to x, and then solve the resulting equation for y. The general solution for Equation 2.21 is I() x q() x dx + c ∫ y = Ix () where c is the constant of integration. (See Problem 2.7) Bernoulli Equations A Bernoulli differential equation has the form n yp ′+= ()xy q()xy (2.24) where n is a real number. The substitution 1−n zy = (2.25) transforms 2.24 into a linear differential equation in the unknown func- tion z(x). (See Problem 2.8) Solved Problems 2 dy x + 2 Solved Problem 2.1 Solve = . dx y This equation may be rewritten in the differential form 2 () x+−20 dx ydy= 2 which is separable with A(x) = x + 2 and B(y) =−y. Its solution is 2 () x+− 2 dx ydy= c ∫∫ or 1 1 32 xx +− 2 y=c 3 2CHAPTER 2: Solutions of First-Order Differential Equations 15 Solving for y, we obtain the solution in implicit form as 2 23 yx=+ 4x+k 3 with k = −2c. Solving for y explicitly, we obtain the two solutions 2 2 33 y=+ x 4xk+ and y=− x+ 4xk+ 3 3 yx + Solved Problem 2.2 Solve y′ = . x This differential equation is not separable. Instead it has the form yf = (,xy), ′ with yx + fx (,y) = x where ty + tx ty() +x yx + ft(,xty) = = = =fx (,y) tx tx x so it is homogeneous. Substituting Equations 2.6 and 2.7 into the equa- tion, we obtain dv xv + x vx+= dx x which can be algebraically simplified to dv 1 x=− 1 or dx dv= 0 dx x This last equation is separable; its solution is 1 dx−= dv c ∫∫ x vx=− ln c, which, when evaluated, yields or vk = ln x (2.26) where we have set ck =− ln and have noted that ln xk += ln ln xk . Finally, substituting v = y/x back into 2.26, we obtain the solution to the given differential equation as yx = ln kx .16 DIFFERENTIALEQUATIONS 2 Solved Problem 2.3 Solve 21 xydx++() x dy=0. This equation has the form of Equation 2.11 with M(x, y) = 2xy and 2 N(x, y) = 1 + x . Since ∂∂ My// =∂N∂x= 2x, the differential equation is exact. Because this equation is exact, we now determine a function g(x, y) that satisfies Equations 2.14 and 2.15. Substituting M(x, y) = 2xy ∂∂ gxx /. = 2 y into 2.14, we obtain Integrating both sides of this equation with respect to x, we find ∂g dx = 2xydx ∫∫ ∂x or 2 gx (,y)=+ x y hy () (2.27) Note that when integrating with respect to x, the constant (with respect to x) of integration can depend on y. We now determine h(y). Differentiating 2.27 with respect to y, we 2 obtain ∂∂ gyx / = +h ′(y) Substituting this equation along with N(x, y) = 2 1 + x into 2.15, we have 22 xh + ′()y =+11 x orh′()y = Integrating this last equation with respect to y, we obtain h(y) = y + c 1 (c = constant). Substituting this expression into 2.27 yields 1 2 g(x, y) = x y + y + c 1 The solution to the differential equation, which is given implicitly by 2.16 as g(x, y) = c, is 2 x y + y = c (c = c − c ) 2 2 1 2 Solving for y explicitly, we obtain the solution as y = c /(x + 1). 2 Solved Problem 2.4 Determine whether the differential equation ydx − xdy = 0 is exact.CHAPTER 2: Solutions of First-Order Differential Equations 17 This equation has the form of Equation 2.11 with M (x, y) = y and N (x, y) =−x. Here ∂M ∂N =11 and =− ∂y ∂x which are not equal, so the differential equation is not exact. 2 Solved Problem 2.5 Determine whether −1/x is an integrating factor for the differential equation ydx − xdy = 0. It was shown in Problem 2.4 that the differential equation is not exact. 2 Multiplying it by −1/x , we obtain −1 −y 1 () ydx−= xdy 0or dx+= dy 0 (2.28) 22 x x x 2 Equation 2.28 has the form of Equation 2.11 with M (x, y) =−y/x and N (x, y) = 1/x.Now ∂M ∂ −y −11 ∂ ∂N     = = = = 22     ∂yy ∂ ∂xx ∂x xx 2 so 2.28 is exact, which implies that −1/x is an integrating factor for the original differential equation. Solved Problem 2.6 Solve ydx − xdy = 0. Using the results of Problem 2.5, we can rewrite the given differential equation as xdy − ydx = 0 2 x which is exact. Equation 2.28 can be solved using the steps described in Equations 2.14 through 2.16. Alternatively, we note from Table 2.1 that 2.28 can be rewritten as d (y/x) = 0. Hence, by direct integration, we have y / x = c, or y = xc,as the solution. 4 Solved Problem 2.7 Solve yx ′+= (/ 4 )yx .18 DIFFERENTIALEQUATIONS The differential equation has the form of Equation 2.21, with p(x) = 4/ x 4 and q(x) = x , and is linear. Here 4 4 pxd ( ) x== dx 4 ln x = ln x ∫∫ x so 2.22 becomes 4 px()dx ln x 4 ∫ Ix () =e == e x (2.29) Multiplying the differential equation by the integrating factor de- fined by 2.29, we obtain d 43 8 4 8 xy′+= 4x y x or (yx )= x dx Integrating both sides of this last equation with respect to x, we obtain 1 c 1 49 5 yx=+ x c or y= + x 4 9 x 9 2 Solved Problem 2.8 Solve yx ′+=y xy . This equation is not linear. It is, however, a Bernoulli differential equa- tion having the form of Equation 2.24 with p(x) = q(x) = x, and n = 2. We 1−2 −1 make the substitution suggested by 2.25, namely z = y = y , from which follow 1 z′ y = and y′=− 2 z z Substituting these equations into the differential equation, we obtain z′ x x − += orzx ′−=z −x 22 z z z This last equation is linear for the unknown function z(x). It has the form of Equation 2.21 with y replaced by z and p(x) = q(x) =−x. The integrat- ing factor is 2 () −xdx −x /2 ∫ Ix () =e =e Multiplying the differential equation by I(x), we obtainCHAPTER 2: Solutions of First-Order Differential Equations 19 22 2 dz −− xx// 22 −x/2 e −= xe z−xe dx or 22 d −− xx// 22 ze =−xe () dx Upon integrating both sides of this last equation, we have 22 −− xx// 22 ze =+ e c whereupon 2 x /2 zx ()=+ ce 1 The solution of the original differential equation is then 11 y== 2 x /2 z ce +1Chapter 3 Applications of First-Order Differential Equations In This Chapter: ✔ Growth and Decay Problems ✔ Temperature Problems ✔ Falling Body Problems ✔ Dilution Problems ✔ Electrical Circuits ✔ Orthogonal Trajectories ✔ Solved Problems Growth and Decay Problems Let N(t) denote the amount of substance (or popula- tion) that is either growing or decaying. If we assume that dN/dt, the time rate of change of this amount of substance, is proportional to the amount of substance present, then dN/dt = kN, or 20 Copyright 2003 by The McGraw-Hill Companies, Inc. Click Here for Terms of Use.

Advise: Why You Wasting Money in Costly SEO Tools, Use World's Best Free SEO Tool Ubersuggest.