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Solution of Exact Equations Contents Solution of Exact Equations Contents • First order ordinary differential equation • Differential of a function of two variables • Short Notes on Partial Derivatives • Exact Equations • Criterion for Exactness • Examples • Method of Solution • Worked Example • Practice Problems • Solutions to practice problems First Order Ordinary differential equations • A differential equation having a first derivative as the highest derivative is a first order differential equation. • If the derivative is a simple derivative, as opposed to a partial derivative, then the equation is referred to as ordinary. Differential of a function of two variables If given a function 푧 = 푓(푥, 푦) Then its differential is defined as the following: 휕푓 휕푓 푑푧 = 푑푥 + 푑푦 휕푥 휕푦 The symbol 휕 represents the partial derivative of the function. This tells us that if we know the differential of a function, we can get back the original function under certain conditions. In a special case when 푓 푥, 푦 = 푐 then we have: 휕푓 휕푓 푑푥 + 푑푦 = 0 휕푥 휕푦 Short note on Partial derivatives For a function of two variables, a partial derivative with respect to a particular variable means differentiating the function with that variable while assuming the other variable to be fixed. Ex: if 푓 = 푥 + 푦 sin 푥 + 푦3푙푛푥 휕푓 1 then = 1 + 푦 cos 푥 + 푦3 휕푥 푥 휕푓 and = sin 푥 + 3푦2푙푛푥 휕푦 Exact Equation If given a differential equation of the form 푀 푥, 푦 푑푥 + 푁 푥, 푦 푑푦 = 0 Where M(x,y) and N(x,y) are functions of x and y, it is possible to solve the equation by separation of variables. However, another method can be used is by examining exactness. The whole idea is that if we know M and N are differentials of f, then it is possible to reconstruct the original function f. Criterion for Exactness This method can only be used if the differential given is exact. Specifically, 휕푀 휕푁 = 휕푦 휕푥 If the above condition is satisfied, then we can use the method. Remember, the function M(x,y) is the one that multiplies dx and the function N(x,y) is the one that multiplies dy. Look at some of the examples on the next slide Examples 2푥푦푑푥 + 푥2 − 1 푑푦 = 0 Here, 2 푀 푥, 푦 = 2푥푦 and 푁 푥, 푦 = 푥 − 1 휕푀 휕푁 Then, we have = 2푥 and = 2푥 휕푦 휕푥 Therefore, this particular equation is exact. Another Example 푒2푦 − 푦 cos 푥푦)푑푥 + (2푥 푒2푦 − 푥 cos 푥푦 + 2푦 푑푦 = 0 Here, 푀 푥, 푦 = 푒2푦 − 푦 cos 푥푦 ; 푁 푥, 푦 = 2푥푒2푦 − 푥 cos 푥푦 + 2푦 휕푀 휕푁 Therefore, = 2푒2푦 − cos 푥푦 + 푥푦 sin 푥푦 = 휕푦 휕푥 As we see, this equation is also exact. Method of Solution The basic idea behind the obtaining the solution is very simple really. It’s also logical. • If given a differential, we first test for exactness. If it is exact, 휕푓 we know that since = 푀 푥, 푦 then 푓 = 푀 푥, 푦 푑푥 + 휕푥 ℎ(푦) • Since integration yields a constant of integration, the most general constant would be a function of y, since on differentiating it with x, we get 0. Contd. • Since we know that the reconstructed function must satisfy the 휕푓 휕 condition = 푁(푥, 푦) then 푁 푥, 푦 = 푀 푥, 푦 푑푥 + ℎ 푦 휕푦 휕푦 • From this, we will be able to solve the above expression for h’(y). • Integrating that, we get h(y) and our solution is complete. • The complete and most general solution is then 푀 푥, 푦 푑푥 + ℎ 푦 = 푐 Worked Example 푑푦 푥푦2−sin 푥 cos 푥 Given = 푑푥 푦(1−푥2) As we can see, this would be extremely difficult to solve using separation of variables. However, let’s try and use exactness to solve the equation. Rearranging it, we get 푥푦2 − sin 푥 cos 푥 푑푥 − 푦 1 − 푥2 푑푦 = 0 Testing for exactness, 휕푀 휕푁 = 2푥푦 = 휕푦 휕푥 Thus, we know the equation is exact, so we proceed with our method. Contd. Now, we know that integrating M(x,y) with respect to x will give us the original function back. 2 2 2 푥 푦 cos 2푥 Thus, (푥푦 − sin 푥 cos 푥)푑푥 = + + ℎ(푦) 2 4 Furthermore, 휕 푥2푦2 cos 2푥 + + ℎ(푦) = 푥2푦 + ℎ′(푦) 휕푦 2 4 And this should be equal to N(x,y). Contd. Then we have, 푥2푦 + ℎ′ 푦 = −푦(1 − 푥2) Simplifying this, we get ℎ′ 푦 = −푦 Notice that if the equation was not exact to begin with, then h’(y) will not be a function of y alone in the step above. Solving for h(y), we have 푦2 ℎ 푦 = − 2 Hence, the complete solution is 푥2푦2 cos 2푥 푦2 + − =c 2 4 2 Where ‘c’ is an arbitrary constant. Practice Problems Determine Whether these equations are exact or not. If they are, solve them. 푑푦 푥 2 1. 푥 = 2푥푒 − 푦 + 6푥 푑푥 2. (푥2−푦2)푑푥 + (푥2 − 2푥푦)푑푦 = 0 3. 5푦 − 2푥 푦′ − 2푦 = 0 Answers to Practice Problems 1. 푥푦 − 2푥푒푥 + 2푒푥 − 2푥3 = 푐 휕푀 휕푁 2. Not Exact. ( = −2푦 ≠ 2푥 − 2푦 = ) 휕푦 휕푥 휕푀 휕푁 3. Not Exact ( = 5 ≠ 0 = ) 휕푦 휕푥 References • A first Course in Differential Equations 9th Ed., Dennis Zill. • Fundamentals of Differential Equations 3rd Ed, Nagle & Saff • Differential and Integral Calculus Vol 2, N. Piskunov Good Luck! .
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