An Alternative Method to Solve Exact Differential Equations

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1 International Mathematical Forum, 5, 00, no. 54, An Alternative Method to Solve Eact Differential Equations Oswaldo González-Gaiola and S. Hernández Linares Departamento de Matemáticas Aplicadas y Sistemas Universidad Autónoma Metropolitana-Cuajimalpa Artificios 40, Col. Miguel Hidalgo, Delegación Álvaro Obregón Méico, D. F., C.P. 00. Méico Abstract. In this article form will be given an alternative form, more easy and direct, to solve an eact differential equation by obtaining information of two indefinite integrals. This will be done by taking special attention on a process of solution that is not normally eplored and this could be rich to acquire of concepts some time problematic to the students of regular courses of ordinary differential equations of engineering and science. Mathematics Subject Classification: Primary 97I50; Secondary 97I60, 97A0 Keywords: Differential equations, Eact equation, Line integral, Simple connected sets. Introduction In a normal curse of ordinary differential equations, it is common to find difficulties for the students to assimilate formal concepts of mathematics, in particular to solve an eact differential equation, in which they use the traditional method where they have to calculate at least two integrals that are a line integral in the real sense. This procedure is constructive but the eercises that the teacher usually proposed to the students are for memorize the algorithm and not for manipulate with a mathematical basis. As a consequence of this some students obtain correct solutions but they do not obtain maturity in the arguments behind the method as a students of engineering or science. In this article we will be seen an easy method which require to calculate two indefinite integrals but the justification require the elemental concept of line integral studied in the first course of integral calculus. ogonzalez@correo.cua.uam.m

2 690 O. González-Gaiola and S. Hernández Linares. Traditional Method In the first regular course of ordinary differential equation is is defined an eact differential equation as an equation of the form: ( M(, yd + N(, ydy =0, where the coefficients M,N :Ω R are of class C (Ω and Ω R is a simple connected domain which satisfies M = N. By this condition, the left side of equation (. correspond to the total differential of a function f :Ω R, f C (Ω, i.e. ( df = f f d + dy =0. By comparation of the equation (. and (. we obtain the equalities: (3 f = M(, y f = N(, y. The traditional method use the equation (.3 and in the first equality it is integrated with respect to, obtaining: (4 f(, y = M(, yd + g(y; where g is a function of the variable y to be determinate. Now, from the equation (.4, taking the derivative in bow side with respect to y and using the second equality of (.3 it is obtained: (5 (6 f = g(y = M(, yd + g(y =N(, y, ( N(, y M(, yd dy. Now, substituting g in (.4 and using that df = 0, the solution of the eact differential equation (., it is given implicitly by the family level curves f(, y =c, c R, i.e.: (7 M(, yd + ( N(, y M(, yd dy = c, c R.

3 An alternative method to solve eact differential equations 69 Let we saw some eamples with this method, and from these eamples we will conjecture an alternative method, more easy to solve an eact differential equation. 3. Eamples Eample 3.. Solve the differential equation d + ydy =0. Solution:: Clearly this is an eact differential equation, where M (, y = and N (, y =y. Then M (, y d =, N (, y dy = y, and [ ] M (, y d dy =0. From this, and using the equation (7, it is obtained the family of solutions + y = c, c R, which are in the implicit form. Eample 3.. Solve the net differential equation ( + y +d + ( y +3 dy =0. Solution:: Clearly this is an eact differential equation, where M (, y = + y + and N (, y = y +3. Then M (, y d = y + +, N (, y dy = y y3 3 +3y, and [ ] M (, y d dy = y. From this, using the equation (7, it is obtained the family of solutions (y (y y3 3 +3y y = c, c R, which ar given in implicit form; i.e., ( y ( y3 3 +3y = c, c R is the implicit family of solution of the differential equation.

4 69 O. González-Gaiola and S. Hernández Linares Eample 3.3. Solve the differential equation ( + y + d + ( y + y dy =0. Solution:: Clearly this is an eact differential equation, where M (, y = + and N (, y = y. Then +y +y M (, y d = ln ( + y + 3 3, and [ N (, y dy = ln ( + y, ] M (, y d dy = ln ( + y. From this, and using the equation (7 it is obtained the implicit family of solutions: ln ( ( + y 3 + = c, c R. 3 Eample 3.4. Solve the differential equation ( ( y + y + y + d + dy =0. + y Solution:: Clearly this is an eact differential equation, where M (, y = y + y, N (, y = + y + y. Then M (, y d = ln ( ( + y arctan, y N (, y dy = ln ( + y ( y + arctan, [ ] M (, y d dy = + y + y dy = ln ( + y ( y + arctan. From this, and using the equation (7, it is obtained the implicit family of solutions ln ( ( + y arctan = c, c R. y If we rewrite the integrals M (, y d and N (, y dy as (8 M (, y d = S (, y+g (+h (y and N (, y dy = P (, y+g (+h(y,

5 An alternative method to solve eact differential equations 693 where S (, y is the part which contain eclusively the mied variables and y in the integral M (, y d and, in the same way, P (, y is the part which contain eclusively the mied variables and y in the integral N (, y dy and with this notation it is obtained the net table: E. S (, y P (, y g ( h (y 0 0 y y + - y3 3 ln ( + y ln ( + y 3 ( ln ( + y arctan ln y ( + y + arctan ( y 0 0 E. Solución + y y + + y3 3 3 ln ( + y = c, c R. +3y = c, c R., c R. 3 (, c R. ln ( + y arctan y y 3 +3y From these tables we can observe that S = P and the solution of the differential equation is implicitly given by (9 S (, y+g (+h(y =c, c R; ecept, maybe, for the last eample, where, apparently, S is different to P, however S and P differed by a constant (the partial derivative of S (, y P (, y is zero. So, the equations (8 and (9 permit us to conjecture an alternative form of haw obtain the solutions of an eact differential equation. We can see that this conjecture depend strongly from the concept of the part which contain eclusively the mied variables and y of a given function R (, y. In the net section we will give a formalization of this concept and then we will establish the precise alternative algorithm (see the Theorem 4.5 for solve an eact differential equation; an finally we will give a proof of this result. 4. The Alternative Method In this section we will give the concept of essential part of a function of class C, which is basically the formalization of the concept of part of a function f(, y which contain eclusively the mied variables and y. Definition 4.. Let Ω be a domain of R and R :Ω R is a function of class C (Ω. We said that the function ( R S (, y := (, y d dy is the essential part of R.

6 694 O. González-Gaiola and S. Hernández Linares Note 4.. The integrals in the previous definition are indefinite integral, where we do not use integral constants. Lemma 4.3. Let Ω R be a simple connected domain and let M,N : Ω R be functions of class C (Ω such that M (0 = N en Ω, and moreover, S and P are the essential part of the indefinite integrals R (, y : = M (, y d, R (, y : = N (, y dy, respectively; then S and P differed by a constant. Proof. The proof follow from the definition 4. and the equation (0. It is known the following result for eact differential equation Proposition 4.4. If Ω R is a simple connected domain and M,N : Ω R are functions of class C (Ω such that ( M (u, v = N (u, v, (u, v Ω, then, for each ( 0,y 0 Ω the eact differential equation ( M (, y d + N (, y dy =0 has a unique maimal solution which pass throughout the point ( 0,y 0. Moreover, there eist a function f :Ω R of class C such that the solutions of ( are given implicitly by the equation f (, y =c, c R; i.e., f satisfies (3 f = M en Ω, f (4 = N en Ω. We will see a way to obtain this function f in terms of the following integrals: M (, y d = S (, y+g (+h (y, N (, y dy = P (, y+g (+h (y; where S and P are the essential part (see the definition 4. of the indefinite integrals M (, y d and N (, y dy respectively.

7 An alternative method to solve eact differential equations 695 In the section 3 we had conjectured that S and P are equals ecept for a constant; and this is a consequence of the lemma 4.3; and from this we obtain (5 M (, y d = S (, y+g (+h (y, (6 N (, y dy = S (, y+g (+h (y. Also we had conjectured that the solution of the eact differential equation (4 are given implicitly by the equation S (, y+g (+h (y =c,, c R. This is precisely true an this is the principal result of this work. The precise statement is the following: Theorem 4.5. Under the same hypothesis of the previous theorem it is obtained f (, y =S (, y+g (+h (y. For their proof we will need the lemma Proof of the Main Theorem Observe that from the lemma (4.3 we obtain the equations (5 y (6. We need the following lemma to proof the main theorem. Lemma 5.. Under the same hypothesis of the proposition 4.4 and if the points ( 0,y 0, (, y Ω could join: (a by a segment of horizontal line followed by a vertical laying in the domain Ω, then: f (, y f ( 0,y 0 =R (, y 0 R ( 0,y 0 +R (, y R (, y 0, or, (b by a segment of vertical line followed by a horizontal laying entirely in the domain Ω, then: f (, y f ( 0,y 0 =R (, y R ( 0,y+R ( 0,y R ( 0,y 0 ; where R and R are like in the previous lemma. Proof. By the Theorem 4.4 there eist f :Ω R of class C that satisfies the equations (3 and (4. Let C(t =(C (t,c (t, t [0, ] be a parametrization of the horizontal an vertical segment joining the point ( 0,y 0 to the point (, y. Then, using the fundamental theorem of calculus, the chain rule and

8 696 O. González-Gaiola and S. Hernández Linares the equations (3 and (4, we obtain: d f (, y f ( 0,y 0 = (f (C (t dt dt = 0 where we obtain the proof of (a. In the same way is obtained (b. 0 (M (C (t,n(c (t (C (t,c (t dt = R (, y 0 R ( 0,y 0 +R (, y R (, y 0 ; Corollary 5.. Under the same hypothesis of the previous lemma we have f (, y f ( 0,y 0 =R (, y R ( 0,y 0 ; where R (, y =S (, y +g ( +h (y; and g and h are given by the equations (5 and (6. Proof. In any case (a or (b is obtained f (, y f ( 0,y 0 = R (, y R ( 0,y 0. The Main Theorem: Using the fact that in any simple connected domain Ω we can join any two point by a polygonal formed by horizontal and vertical segments laying enterally in Ω and using the previous lemma, we can join the points ( 0,y 0, (,y,...,( m,y m, ( m,y m =(, y, by such polygonal laying in Ω (see [3] and, by previous corollary we obtain f (, y f ( 0,y 0 = = m k=0 (f ( k+,y k+ f ( k,y k, m (R ( k+,y k+ R ( k,y k, k=0 = R (, y R ( 0,y Summary From the eperiences with groups of students of engineering and science they find that the method given by the main theorem is more direct and help them to understand that integrating the equalities in (3 conduce at the same equivalent epressions. We recommend to the teacher to obtain the solution of any eact differential equation by the two procedure sowed in this work. ACKNOWLEDGEMENTS. We thank to the departments of mathematics of the Universidad Autónoma Metropolitana-Cuajimalpa for their hospitality to write this paper.

9 An alternative method to solve eact differential equations 697 References [] D. G. Zill, Differential Equations Fifth Edition, India, (009. [] J. E. Marsden, M. J. Hoffman, Vector Calculus Fifth Edition, New York, (003. [3] J. E. Marsden, M. J. Hoffman, Basic Comple Analysis 3rd Edition, New York, (999. [4] R. Remmert, Theory of Comple Functions, Springer-Verlag, New York, (989. Received: April, 00

Exact Equations. M(x,y) + N(x,y) y = 0, M(x,y) dx + N(x,y) dy = 0. M(x,y) + N(x,y) y = 0

Exact Equations. M(x,y) + N(x,y) y = 0, M(x,y) dx + N(x,y) dy = 0. M(x,y) + N(x,y) y = 0 Eact Equations An eact equation is a first order differential equation that can be written in the form M(, + N(,, provided that there eists a function ψ(, such that = M (, and N(, = Note : Often the equation