Lecture notes Ordinary Differential equations

lecture notes on numerical solution of ordinary differential equations and nonlinear ordinary differential equations lecture notes, ordinary differential equations with application
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ORDINARYDIFFERENTIALEQUATIONS FOR ENGINEERS THE LECTURE NOTES FOR MATH- 263 (2011)Chapter 1 INTRODUCTION 1. Definitions and Basic Concepts 1.1 Ordinary Differential Equation (ODE) An equation involving the derivatives of an unknown function y of a single variable x over an interval x ∈ (I). More clearly and precisely speaking, a well defined ODE must the following features: It can be written in the form: ′ ′′ n Fx,y,y,y ,···,y =0; (1.1) where the mathematical expression on the right hand side contains (1). variable x, (2). function y of x, and (3). some derivatives of y with respect to x; The values of variables x, y must be specified in a certain number field, such asN,R, orC; The variation region of variable x of Eq. must be specified, such as x∈(I)=(a,b). 1.2 Solution Any function y = f(x) which satisfies this equation over the interval (I) is called a solution of the ODE. More clearly speaking, functionϕ(x) is called a solution of the give EQ. (1.1), if the following requirements are satisfies: The function ϕ(x) is defined in the region x∈(I); ′ (n) The function ϕ(x) is differentiable, hence, ϕ(x),···,ϕ (x) all exit, in the region x∈(I); 12 ORDINARY DIFFERENTIAL EQUATIONS FOR ENGINEERS ′ (n) With the replacements of the variables y,y,···,y in 1.1 by the ′ (n) functions ϕ(x),ϕ(x),···,ϕ (x), the EQ. (1.1) becomes an identity overx∈(I). Inotherwords,therighthandsideofEq. (1.1)becomes to zero for all x∈(I). 2x For example, one can verify that y =e is a solution of the ODE ′ y =2y, x∈(−∞,∞), 2 and y =sin(x ) is a solution of the ODE ′′ ′ 3 xy −y +4x y =0, x∈(−∞,∞). 1.3 Order n of the DE (n) An ODE is said to be order n, if y is the highest order derivative ′ occurring in the equation. The simplest first order ODE is y =g(x). Note that the expression F on the right hand side of an n-th order ′ (n) ODE: Fx,y,y,...,y = 0 can be considered as a function of n+2 variables (x,u ,u ,...,u ). Namely, one may write 0 1 n F(x,u ,u ,···,u )=0. 0 1 n Thus, the equations ′′ 3 ′ 2 ′′′ ′ xy +y =x , y +y =0, y +2y +y =0 whichareexamplesofODE’sofsecondorder, firstorderandthirdorder respectively, can be in the forms: 3 F(x,u ,u ,u )=xu +u −x , 0 1 2 2 0 2 F(x,u ,u )=u +u , 0 1 1 0 F(x,u ,u ,u ,u )=u +2u +u . 0 1 2 3 3 1 0 respectively. 1.4 Linear Equation: If the functionF is linear in the variablesu ,u ,...,u , which means 0 1 n every term in F is proportional to u ,u ,...,u , the ODE is said to be 0 1 n linear. If, in addition, F is homogeneous then the ODE is said to be homogeneous. The first of the above examples above is linear are linear, the second is non-linear and the third is linear and homogeneous. The general n-th order linear ODE can be written n n1 d y d y dy a (x) +a (x) +···+a (x) n n n−1 n1 1 dx dx dx +a (x)y =b(x). 0INTRODUCTION 3 1.5 Homogeneous Linear Equation: The linear DE is homogeneous, if and only if b(x)≡0. Linear homo- geneousequationshavetheimportantpropertythatlinearcombinations of solutions are also solutions. In other words, if y ,y ,...,y are solu- 1 2 m tions and c ,c ,...,c are constants then 1 2 m c y +c y +···+c y 1 1 2 2 m m is also a solution. 1.6 Partial Differential Equation (PDE) An equation involving the partial derivatives of a function of more than one variable is called PED. The concepts of linearity and homo- geneitycanbeextendedtoPDE’s. ThegeneralsecondorderlinearPDE in two variables x,y is 2 2 2 ∂ u ∂ u ∂ u ∂u a(x,y) +b(x,y) +c(x,y) +d(x,y) 2 2 ∂x ∂x∂y ∂y ∂x ∂u +e(x,y) +f(x,y)u=g(x,y). ∂y Laplace’s equation 2 2 ∂ u ∂ u + =0 2 2 ∂x ∂y 2 2 isalinear,homogeneousPDEoforder2. Thefunctionsu=log(x +y ), 2 2 u=xy, u=x −y are examples of solutions of Laplace’s equation. We will not study PDE’s systematically in this course. 1.7 General Solution of a Linear Differential Equation It represents the set of all solutions, i.e., the set of all functions which satisfy the equation in the interval (I). For example, given the differential equation ′ 2 y =3x . Its general solution is 3 y =x +C where C is an arbitrary constant. To select a specific solution, one needs to determine the constant C with some additional conditions. For instance, the constant C can be determined by the value of y at x=0. This condition is called theinitial condition, which completely4 ORDINARY DIFFERENTIAL EQUATIONS FOR ENGINEERS determines the solution. More generally, it will be shown in the future that given a,b there is a unique solution y of the differential equation with the initial condition y(a) = b. Geometrically, this means that the 2 one-parameter family of curves y =x +C do not intersect one another 2 and they fill up the planeR . 1.8 A System of ODE’s The normal form of system of ODE’s is ′ y =G (x,y ,y ,...,y ) 1 1 2 n 1 ′ y =G (x,y ,y ,...,y ) 2 1 2 n 2 . . . ′ y =G (x,y ,y ,...,y ). n 1 2 n n The number of the equations is called the order of such system. Ann-th (n) ′ n−1 order ODE of the form y = G(x,y,y,...,y ) can be transformed in the form of then-th order system of DE’s. If we introduce dependant variables ′ n−1 y =y,y =y,...,y =y 1 2 n we obtain the equivalent system of first order equations ′ y =y , 2 1 ′ y =y , 3 2 . . . ′ y =G(x,y ,y ,...,y ). 1 2 n n ′′ For example, the ODE y =y is equivalent to the system ′ y =y , 2 1 ′ y =y . 1 2 In this way the study of n-th order equations can be reduced to the studyofsystemsofnfirstorderequations, orsay,n-thorderofsystemof ODE’s. Systemsofequationsariseinthestudyofthemotionofparticles. For example, if P(x,y) is the position of a particle of mass m at time t, moving in a plane under the action of the force field (f(x,y),g(x,y), we have 2 d x m =f(x,y), 2 dt 2 d y m =g(x,y). 2 dt This is a system of two second order Eq’s, it can be easily transformed into a normal form of 4-th order system of ODE’s, by introducing the ′ ′ new unknown functions: x =x,x =x and y =y,y =y . 1 2 1 2INTRODUCTION 5 The general first order ODE in normal form is ′ y =F(x,y). ∂F If F and are continuous one can show that, given a,b, there is a ∂y unique solution with y(a) = b. Describing this solution is not an easy task and there are a variety of ways to do this. The dependence of the solution on initial conditions is also an important question as the initial values may be only known approximately. ′ The non-linear ODE yy = 4x is not in normal form but can be brought to normal form 4x ′ y = . y by dividing both sides by y. 2. The Approaches of Finding Solutions of ODE 2.1 Analytical Approaches Analytical solution methods: finding the exact form of solutions; Geometrical methods: finding the qualitative behavior of solutions; Asymptotic methods: finding the asymptotic form of the solution, which gives good approximation of the exact solution. 2.2 Numerical Approaches Numerical algorithms — numerical methods; Symbolic manipulators — Maple, MATHEMATICA, MacSyma. This course mainly discuss the analytical approaches and mainly on analytical solution methods.Chapter 2 FIRSTORDERDIFFERENTIALEQUATIONS In this chapter we are going to treat linear and separable first order ODE’s. 1. Linear Equation ′ The general first order ODE has the form F(x,y,y ) = 0 where y = y(x). If it is linear it can be written in the form ′ a (x)y +a (x)y =b(x) 0 1 where a (x), a x), b(x) are continuous functions of x on some interval 0 ( (I). ′ Tobringittonormalformy =f(x,y)wehavetodividebothsidesof theequationbya (x). Thisispossibleonlyforthosexwherea (x)= ̸ 0. 0 0 After possibly shrinking (I) we assume that a (x) ̸= 0 on (I). So our 0 equation has the form (standard form) ′ y +p(x)y =q(x) with p(x)=a (x)/a (x), q(x)=b(x)/a (x), 1 0 0 ′ both continuous on (I). Solving for y we get the normal form for a linear first order ODE, namely ′ y =q(x)−p(x)y. 78 ORDINARY DIFFERENTIAL EQUATIONS FOR ENGINEERS 1.1 Linear homogeneous equation Let us first consider the simple case: q(x)=0, namely, dy +p(x)y =0. (2.1) dx To find the solutions, we proceed in the following three steps: Assume that the solution exists and in the form y =y(x); Find the necessary form of the function y(x). In doing so, by the definition of the solution, one substitute the functiony(x) in the Eq., and try to transform the EQ. in such a way that its LHS of EQ. is a d complete differentiation ....., while its RHS is a known function. dx Verify y(x) is indeed a solution. In the step 2, with the chain law of derivative, one may transform Eq. (2.1) into the following form: ′ y (x) d = lny(x)=−p(x). y dx Its LHS is now changed a complete differentiation with respect to x, while its RHS is a known function. By integrating both sides, we derive ∫ lny(x)=− p(x)dx+C, or ∫ − p(x)dx y =±C e , 1 C whereC,aswellasC =e 0,isarbitraryconstant. AsC 0,y(x)=0 1 = is a trivial solution, we derive the necessary form of solution: ∫ − p(x)dx y =Ae . (2.2) Asthefinalstepofderivation, onecanverifythefunction(2.2)isindeed asolutionforanyvalueofconstantA. Insummary,wededucethat(2.2) is the form of solution, that contains all possible solution. Hence, one can call (2.2) as the general solution of Eq. (2.1). 1.2 Linear inhomogeneous equation We now consider the general case: dy +p(x)y =q(x). dxFIRST ORDER DIFFERENTIAL EQUATIONS 9 We still proceed in the three steps as we did in the previous subsection. However, now in the step 2 one cannot directly transform the LHS of Eq. in the complete differential form as we did for the case of homoge- neousEq. Forthispurpose,wemultiplythebothsidesofourdifferential equation with a factor µ(x)̸= 0. Then our equation is equivalent (has the same solutions) to the equation ′ µ(x)y (x)+µ(x)p(x)y(x)=µ(x)q(x). We wish that with a properly chosen function µ(x), d ′ µ(x)y (x)+µ(x)p(x)y(x)= µ(x)y(x). dx For this purpose, the function µ(x) must has the property ′ µ(x)=p(x)µ(x), (2.3) and µ(x) ̸= 0 for all x. By solving the linear homogeneous equation (2.3), one obtain ∫ p(x)dx µ(x)=e . (2.4) Withthisfunction, whichiscalledanintegratingfactor, ourequation is now transformed into the form that we wanted: d µ(x)y(x)=µ(x)q(x), (2.5) dx Integrating both sides, we get ∫ µ(x)y = µ(x)q(x)dx+C with C an arbitrary constant. Solving for y, we get ∫ 1 C y = µ(x)q(x)dx+ =y (x)+y (x) (2.6) P H µ(x) µ(x) as the general solution for the general linear first order ODE ′ y +p(x)y =q(x). In solution (2.6): the first part, y (x): a particular solution of the inhomogeneous P equation,10 ORDINARY DIFFERENTIAL EQUATIONS FOR ENGINEERS the second part, y (x): the general solution of the associate H homogeneous equation. Note that for any pair of scalars a,b with a in (I), there is a unique scalarC such thaty(a)=b. Geometrically, this means that the solution curves y = ϕ(x) are a family of non-intersecting curves which fill the region I×R. ′ Example 1: y +xy = x. This is a linear first order ODE in standard form with p(x)=q(x)=x. The integrating factor is ∫ 2 xdx x /2 µ(x)=e =e . Hence, after multiplying both sides of our differential equation, we get d 2 2 x /2 x /2 (e y)=xe dx which, after integrating both sides, yields ∫ 2 2 2 x /2 x /2 x /2 e y = xe dx+C =e +C. 2 −x /2 Hence the general solution is y =1+Ce . The solution satisfying the initial condition: y(0)=1: y =1, (C =0); and the solution satisfying I.C., y(0)=a: 2 −x /2 y =1+(a−1)e , (C =a−1). ′ 3 Example 2: xy −2y =x sinx, (x 0). We bring this linear first order equation to standard form by dividing by x. We get 2 ′ 2 y − y =x sinx. x The integrating factor is ∫ −2dx/x −2lnx 2 µ(x)=e =e =1/x . 2 After multiplying our DE in standard form by 1/x and simplifying, we get d 2 (y/x )=sinx dxFIRST ORDER DIFFERENTIAL EQUATIONS 11 2 from which y/x =−cosx+C, and 2 2 y =−x cosx+Cx . (2.7) Note that (2.7) is a family of solutions to the DE ′ 3 xy −2y =x sinx. To determine a special solution, one needs to impose an IC. For this problem, let us exam the following IC’s: 1 For given IC:y(0)=0, from the general solution (2.7) we derive that 0=y(0)=0+c∗0=0. So that, the IC is satisfied for any constant C. The problem has infinitely many solutions. 2 For given IC: y(0)=b= ̸ 0, from the general solution (2.7) we derive that b=y(0)=0+c∗0=0. Sothat,theICcannotbesatisfiedwithanyconstantC. Theproblem has no solution. 3 For given IC: y(a) = b where a ̸= 0 and b is any number, from the general solution (2.7) we derive that 2 2 2 b=y(0)=−a cosa+c∗a =a (C−cosa), 2 so that, C =cosa+b/a is uniquely determined. The problem has a unique solution. This example displays a complicated situation, when the IC is imposed at x = 0. Why does this happen? The simple explanation to such abnormalityisthatbecausex=0isasingularpoint,wherep(0)=∞. Suchanabnormalsituationwillbediscussedmoredeeplyinthefuture. 2. Nonlinear Equations (I) 2.1 Separable Equations. ′ The first order ODE y =f(x,y) is said to be separable if f(x,y) can be expressed as a product of a function of x times a function of y. The DE then has the form: ′ y =g(x)h(y), x∈(I).12 ORDINARY DIFFERENTIAL EQUATIONS FOR ENGINEERS Weapplytheprocedureofthreestepsforthesolutionsasbefore. Assume that the solution y = y(x) exists, and h(y) ̸= 0 as x ∈ (I). Then by dividing both sides by hy(x), it becomes ′ y (x) =g(x). (2.8) hy(x) Of course this is not valid for those solutions y = y(x) at the points wherehy(x)=0. Furthermore,weassumethattheLHScanbewritten d in the form of complete differentiation: Hy(x), where Hy(x) is a dx composite function of x to be determined. Once we find the function H(y), we may write ∫ ∫ ′ y (x) Hy(x)= dx= g(x)dx+C. hy(x) However, by chain rule we have d ′ ′ Hy(x)=H (y)y (x). dx By comparing the above with the LHS of (3.33), it follows that 1 ′ H (y)= . h(y) Thus, we derive that ∫ ∫ dy H(y)= = g(x)dx+C, (2.9) h(y) This gives the implicit form of the solution. It determines the value of y implicitlyintermsofx. Thefunctiongivenin(2.9)canbeeasilyverified as indeed a solution. Note that with the assumption h(y) = ̸ 0 at the beginning of the derivation, some solution may be excluded in (2.9). As a matter of fact, one can verify that the Eq. may allow the constant solutions, y =y , (2.10) ∗ as h(y )=0. ∗ ′ x−5 Example 1: y = . 2 y To solve it using the above method we multiply both sides of the 2 equation by y to get 2 ′ y y =(x−5).FIRST ORDER DIFFERENTIAL EQUATIONS 13 3 2 Integrating both sides we get y /3=x /2−5x+C. Hence, 1/3 2 y = 3x /2−15x+C . 1 y−1 ′ Example 2: y = (x −3). By inspection, y = 1 is a solution. x+3 Dividing both sides of the given DE by y−1 we get ′ y 1 = . y−1 x+3 This will be possible for those x where y(x)̸=1. Integrating both sides we get ∫ ∫ ′ y dx dx= +C , 1 y−1 x+3 C 1 from which we get lny−1=ln(x+3)+C . Thusy−1=e (x+3) 1 C C 1 1 from which y−1=±e (x+3). If we let C =±e , we get y =1+C(x+3) . Since y = 1 was found to be a solution by inspection the general solution is y =1+C(x+3), whereC can be any scalar. For any (a,b) witha̸=−3, there is only one member of this family which passes through (a,b). However, it is seen that there is a family of lines passing through (−3,1), whilenosolutionlinepassingthrough(−3,b)withb̸=1). Here, x=−3 is a singular point. ′ ycosx Example 3: y = . Transforming in the standard form then 2 1+2y integrating both sides we get ∫ ∫ 2 (1+2y ) dy = cosxdx+C, y from which we get a family of the solutions: 2 lny+y =sinx+C, where C is an arbitrary constant. However, this is not the general solu- tion of the equation, as it does not contains, for instance, the solution: y =0. With I.C.: y(0)=1, we get C =1, hence, the solution: 2 lny+y =sinx+1.14 ORDINARY DIFFERENTIAL EQUATIONS FOR ENGINEERS 2.2 Logistic Equation ′ y =ay(b−y), where a,b 0 are fixed constants. This equation arises in the study of the growth of certain populations. Since the right-hand side of the equation is zero for y = 0 and y =b, the given DE has y = 0 and y =b ′ as solutions. More generally, if y = f(t,y) and f(t,c) = 0 for all t in some interval (I), the constant function y = c on (I) is a solution of ′ ′ y =f(t,y) since y =0 for a constant function y. To solve the logistic equation, we write it in the form ′ y =a. y(b−y) Integrating both sides with respect to t we get ∫ ′ y dt =at+C y(b−y) ′ which can, since y dt=dy, be written as ∫ dy =at+C. y(b−y) Since, by partial fractions, 1 1 1 1 = ( + ) y(b−y) b y b−y we obtain 1 (lny−lnb−y)=at+C. b Multiplying both sides by b and exponentiating both sides to the base e, we get y bC abt abt =e e =C e , 1 b−y bC where the arbitrary constant C = e 0 can be determined by the 1 initial condition (IC): y(0)=y as 0 y 0 C = . 1 b−y 0 Two cases need to be discussed separately.FIRST ORDER DIFFERENTIAL EQUATIONS 15 y y 0 0 Case (I), y b: one has C = = 0. So that, 0 1 b−y b−y 0 0 ( ) y y 0 abt = e 0, (t∈(I)). b−y b−y 0 From the above we derive abt y/(b−y)=C e , 1 abt y =(b−y)C e . 1 This gives ( ) y 0 abt abt b e bC e b−y 1 0 ( ) y = = . abt y 0 abt 1+C e 1 1+ e b−y 0 It shows that if y = 0, one has the solution y(t) = 0. However, if 0 0y b, one has the solution 0y(t)b, and as t→∞, y(t)→b. 0 y y 0 0 Case (II), y b: one has C = =− 0. So that, 0 1 b−y b−y 0 0 ( ) y y 0 abt = e 0, (t∈(I)). b−y y −b 0 From the above we derive ( ) y 0 abt y/(y−b)= e , y −b 0 ( ) y 0 abt y =(y−b) e . y −b 0 This gives ( ) y 0 abt b e y −b 0 ( ) y = . y 0 abt e −1 y −b 0 It shows that if y b, one has the solution y(t) b, and as t→∞, 0 y(t)→b. It is derived that y(t)=0 is an unstable equilibrium state of the system; y(t)=b is a stable equilibrium state of the system.16 ORDINARY DIFFERENTIAL EQUATIONS FOR ENGINEERS 2.3 Fundamental Existence and Uniqueness Theorem Ifthefunctionf(x,y)togetherwithitspartialderivativewithrespect to y are continuous on the rectangle (R):x−x ≤a, y−y ≤b 0 0 there is a unique solution to the initial value problem ′ y =f(x,y), y(x )=y 0 0 defined on the intervalx−x h where 0 h=min(a,b/M), M =maxf(x,y), (x,y)∈(R). Note that this theorem indicates that a solution may not be defined for all x in the intervalx−x ≤a. For example, the function 0 abx bCe y = abx 1+Ce ′ abx is solution to y = ay(b−y) but not defined when 1+Ce = 0 even though f(x,y)=ay(b−y) satisfies the conditions of the theorem for all x,y. The next example show why the condition on the partial derivative in the above theorem is important sufficient condition. ′ 1/3 Considerthedifferentialequationy =y . Againy =0isasolution. Separating variables and integrating, we get ∫ dy =x+C 1 1/3 y which yields 2/3 y =2x/3+C and hence 3/2 y =±(2x/3+C) . Taking C =0, we get the solution 3/2 y =±(2x/3) ,(x≥0) which along with the solution y = 0 satisfies y(0) = 0. Even more, 3/2 Taking C =−(2x /3) , we get the solution: 0FIRST ORDER DIFFERENTIAL EQUATIONS 17 0 (0≤x≤x ) 0 y = 3/2 ±2(x−x )/3 , (x≥x ) 0 0 which also satisfies y(0)=0. So the initial value problem ′ 1/3 y =y , y(0)=0 does not have a unique solution. The reason this is so is due to the fact that ∂f 1 (x,y)= 2/3 ∂y 3y is not continuous when y =0. Many differential equations become linear or separable after a change of variable. We now give two examples of this. 2.4 Bernoulli Equation: ′ n y =p(x)y+q(x)y (n̸=1). Note that y =0 is a solution. To solve this equation, we set α u=y , where α is to be determined. Then, we have ′ α−1 ′ u =αy y, hence, our differential equation becomes ′ α+n−1 u/α=p(x)u+q(x)y . (2.11) Now set α =1−n, Thus, (2.11) is reduced to ′ u/α=p(x)u+q(x), (2.12) whichislinear. Weknowhowtosolvethisforufromwhichwegetsolve 1−n u=y to get 1/(1−n) y =u . (2.13)18 ORDINARY DIFFERENTIAL EQUATIONS FOR ENGINEERS 2.5 Homogeneous Equation: ′ y =F(y/x). To solve this we let u=y/x, so that ′ ′ y =xu, and y =u+xu. ′ Substituting for y,y in our DE gives ′ u+xu =F(u) which is a separable equation. Solving this for u gives y via y =xu. Note that u≡a is a solution of ′ xu =F(u)−u whenever F(a)=a and that this gives y =ax as a solution of ′ y =f(y/x). ′ Example. y =(x−y)/x+y. This is a homogeneous equation since x−y 1−y/x = . x+y 1+y/x Setting u=y/x, our DE becomes 1−u ′ xu +u= 1+u so that 2 1−u 1−2u−u ′ xu = −u= . 1+u 1+u √ Notethattheright-handsideiszeroifu=−1± 2. Separatingvariables and integrating with respect to x, we get ∫ (1+u)du =lnx+C 1 2 1−2u−uFIRST ORDER DIFFERENTIAL EQUATIONS 19 which in turn gives 2 (−1/2)ln1−2u−u =lnx+C . 1 Exponentiating, we get 1 C 1 √ =e x. 2 1−2u−u Squaring both sides and taking reciprocals, we get 2 2 u +2u−1=C/x 2C 1 withC =±1/e . Thisequationcanbesolvedforuusingthequadratic formula. If x ,y are given with 0 0 x ̸=0, and u =y /x = ̸ −1 0 0 0 0 thereis,bythefundamental,existenceanduniquenesstheorem,aunique solution with I.C. y(x )=y . 0 0 For example, if x = 1,y = 2, we have, u(x ) = 2, so, C = 7 and 0 0 0 hence 2 2 u +2u−1=7/x Solving for u, we get √ 2 u=−1+ 2+7/x where the positive sign in the quadratic formula was chosen to make u=2,x=1 a solution. Hence √ √ 2 2 y =−x+x 2+7/x =−x+ 2x +7 is the solution to the initial value problem x−y ′ y = , y(1)=2 x+y forx0andonecaneasilycheckthatitisasolutionforallx. Moreover, using the fundamental uniqueness theorem, it can be shown that it is the only solution defined for all x.20 ORDINARY DIFFERENTIAL EQUATIONS FOR ENGINEERS 3. Nonlinear Equations (II)— Exact Equation and Integrating Factor 3.1 Exact Equations. By a region of the (x,y)-plane we mean a connected open subset of the plane. The differential equation dy M(x,y)+N(x,y) =0 dx is said to be exact on a region (R) if there is a function F(x,y) defined on (R) such that ∂F ∂F =M(x,y); =N(x,y) ∂x ∂y In this case, if M,N are continuously differentiable on (R) we have ∂M ∂N = . (2.14) ∂y ∂x Conversely,itcanbeshownthatcondition(2.14)isalsosufficientforthe exactnessofthegivenDEon(R)providingthat(R)issimplyconnected, .i.e., has no “holes”. The exact equations are solvable. In fact, supposey(x) is its solution. Then one can write: dy ∂F ∂F dy Mx,y(x)+Nx,y(x) = + dx ∂x ∂y dx d = F x,y(x)=0. dx It follows that F x,y(x)=C, whereC isanarbitraryconstant. Thisisanimplicitformofthesolution y(x). Hence, the function F(x,y), if it is found, will give a family of the solutions of the given DE. The curves F(x,y)=C are calledintegral curves of the given DE. dy 2 2 Example 1. 2x y +2xy +1=0. Here dx 2 2 M =2xy +1, N =2x y 2 2 and R =R , the whole (x,y)-plane. The equation is exact onR since 2 R is simply connected and ∂M ∂N =4xy = . ∂y ∂x

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