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MTH102 EXACT DIFFERENTIAL EQUATION SUBMITTED TO : MR.GURPREET SIR SUBMITTED BY : AVINASH RAJPUT ACKNOWLEDGMENT I would express my gratitude to all those who gave me the possibility to complete this term work. I want to thank the department of MTH102 for giving me permission to commence this work in the first instance and to use the research data. I am deeply indebted to my teacher gurpreet sir whose suggestions and encouragement helped me in all time of research for writing this term work. CONTANT: EXACT DIFFERENTIAL EQUATIONS EQUATION REDUCIBLE TO EXACT EQUATION: I.F FOUND BY INSPECTION INTEGRATING FACTORS SEPARABLE EQUATIONS(I.F OF HOMOGENEOUS) PROBLEM BILIOGRPHY: EXACT DIFFERENTIAL EQUATIONS A differential equation obtained from its primitive directly by differentiation without any operation of multiplication, elimination or reduction is said to be an exact differential equation. Thus a differential equation of the form M(X,Y)dx+N(X,Y)dy=0is an exact differential equation if it can be obtained directly by differentiating the equation u(x,y)=c which is primitive. Du=Mdx+Ndy Mdx+Ndy=0is integ.Mdx+integ(terms of N not containing X) Theorem: Solutions to Exact Differential Equations Let M, N, My, and Nx be continuous with My = Nx Then there is a function f with fx = M and fy = N such that f(x,y) = C is a solution to the differential equation M(x,y) + N(x,y)y' = 0 A first-order differential equation is one containing a first—but no higher—derivative of the unknown function. For virtually every such equation encountered in practice, the general solution will contain one arbitrary constant, that is, one parameter, so a first-order IVP will contain one initial condition. There is no general method that solves every first-order equation, but there are methods to solve particular typesGiven a function f( x, y) of two variables, its total differential df is defined by the equation The DE's that come up in Calculus are Separable. As we just saw this means they can be and . This means that so that . Such a du is called an "Exact", "Perfect" or "Total" differential. As we will see in Orthogonal Trajectories (1.8), the expression represents a one-parameter family of curves in the plane. For example, is a family of circles of radius parabolas. Let us find the differential du for . Calculate (solution) and ended with and is a family of 1st order D. E. of the form This D. E. is called exact if there is some function u(x, y) so that and, of course, . since . Since and , and we make the mild assumption that a freebie that , If are continuous then we get as Now we can carry out two "partial" integrations: so so Notice that the integration so-called constants each depend on one of the variables. Now we do some "criss-crossing" to get our solution First, get by solving for dk/dy in and then carry out a ("partial') integration: . . EQUATION REDUCIBLE TO EXACT EQUATION: Differential equation which are not exact can be made exact after multiplying by a suitable factor called the integrating factor. Ydx-xdy=0 ,M=y ,N=-x the equation is not exact. Multiplying the equation 1/y2 it ydx-xdy/y2 =0 Multiplying the equation 1/x2 it d(y/x)=0 Multiplying the equation 1/xy it dx/x-dy/y=0 Which is exact -1/y2,1/x2,1/xy are integrating factors. I.F found by inspection :in a number of problem a little analyisis helps to find the ingrting factor . M ( x , y)dx y 1. Ydx+xdy=d(xy) Set F ( x, y ) = F( x, y) M ( x , y) . x N ( x , y)dy x Integrate with respect to x to get F ( x, y ) M ( x , y)dx + ( y) Differentiate with respect to y to get ( y) ]= N ( x , y) ( y) N ( x , y) F ( x, y ) = g(x,y) + Integrating Factors What if = M( x, y) N( x, y) y x ( y) , y [ M ( x , y)dx + if no boundary value is given. Definition If M(x, y)dx N(x, y)dy 0 is not exact, but (x, y)M(x, y)dx (x, y) N(x, y) 0 is exact, then ( x , y) is called an integrating factor. Example : Show that such that yn i 2 y 2xy 2 y 2x y y 2xy dx x 2 dy 0 is not exact, then find n is an integrating factor. 2 ( x 2 ) 2 x therefore x the DE is not exact. ii. Multiply the DE by yn, then solve. y n ( y 2 2xy )dx y n x 2 dy 0 equal y n x 2 2 y n x x 2n 1xy n must equal n2 y 2xy n 1 n 2y n 1 2n 1xy n y which means 2xy n must n 2y n 1 must equal 0 and for this to be so n must equal -2. Separable Equations(I.F OF HOMOGENEOUS) F( x )G ( y)dx f ( x )g( x )dy 0 This type of DE is called separable because it can be written in the form (variables can be separable) M(x )dx N( y)dy 0 The first equation is usually not exact but multiplying it by the appropriate integrating factor will make it exact, but use of an integrating factor may eliminate solutions or may lead to extraneous solutions.After multiplying by the integrating factor equation becomes: F( x ) g ( y) dx dy 0 where f (x) G ( y) M( x ) F( x ) f (x) 1 f ( x )G ( y ) and the N( x ) where g ( y) . G ( y) Solutions are of the form M(x)dx N( y)dy c f ( x ) 0 & G ( y) 0 . Problem 4xy 3 6 x sin x sin y dx cos x cos y 12 y 2 6x2 y 2 dy 0 , y 0 Step 0: Put the equation into standard form. 4xy 3 6 x sin x sin y cos x cos y 12 y 2 6x 2 y 2 y 0 Then M x, y 4 xy 3 6 x sin x sin y and N x, y cos x cos y 12 y 2 6 x 2 y 2 . Step 1: Compute M y x, y and N x x, y . M y x, y 12 xy 2 sin x cos y N x x, y sin x cos y 12 xy 2 Thus M y x, y N x x, y . Step 2: Make a choice between two methods of solving the equation. Choose one of the two methods. Below each method is presented, but you need only one of them. The purpose of presenting both methods is simply to demonstrate that either one will provide the answer. [First method: x, y M x, y dx ] Step 3: Perform the integration according to the method chosen in Step 2. Since the equation is exact, there is a function for which x x, y M x, y 4 xy 3 6 x sin x sin y and y x, y N x, y cos x cos y 12 y 2 6 x 2 y 2 . Then x, y M x, y dx 4xy 3 6x sin x sin y dx 2 x 2 y 3 3x 2 cos x sin y g y . Now it is given that N x, y cos x cos y 12 y 2 6 x 2 y 2 ; and from above y x, y 2 x 2 y 3 3x 2 cos x sin y g y 6 x 2 y 2 cos x cos y g y . y Since y x, y N x, y , 6 x 2 y 2 cos x cos y g y cos x cos y 12 y 2 6 x 2 y 2 g y 12 y 2 g y 4 y3 So the one parameter family of functions that define the solutions of the equations is 2 x2 y3 3x2 cos x sin y 4 y3 C . [Second method: x, y N x, y dy ] Step 3: Perform the integration according to the method chosen in Step 2. Since the equation is exact, there is a function for which x x, y M x, y 4 xy 3 6 x sin x sin y and y x, y N x, y cos x cos y 12 y 2 6 x 2 y 2 . Then x, y N x, y dy cos x cos y 12 y 2 6 x2 y 2 dy cos x sin y 4 y 3 2 x 2 y 3 f x . Now it is given that M x, y 4 xy 3 6 x sin x sin y ; and from above x x, y cos x sin y 4 y 3 2 x 2 y 3 f x sin x sin y 4 xy 3 f x x Since x x, y M x, y , sin x sin y 4 xy 3 f x 4 xy 3 6 x sin x sin y f x 6 x f x 3x 2 So the one parameter family of functions that define the solutions of the equations is cos x sin y 4 y3 2 x2 y3 3x2 C . Step 4: Evaluate C using the initial condition. Using the initial condition y 0 , C 2 2 0 3 2 cos sin0 4 0 3 2 . 3 3 So the solution of the is 2x2 y3 3x2 cos x sin y 4 y3 3 2 REFRANCE: www.wolframe.com www.math.fsu.edu/~fusaro/EngMath/Ch1/SEDE.htm www.ltcconline.net/greenl/courses/204/.../exactDiffEQs.html www.cliffsnotes.com/.../Exact-Equations.topicArticleId19736,articleId-19710.htm chapter form n.p.bali/b.s Grewal. Submitted for – www.mycollegebag.in