Lesson 29. Exact Differential Equations: Integrating Factors

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1 Module 3: Ordinar Differential Equations Lesson 9 Eact Differential Equations: Integrating Factors In general, equations of the tpe M,)d + N,)d = 0 are not eact. However, it is sometimes possible to transform the equation into an eact differential equation multipling it b a suitable function I,). That is, if I,) is an integrating factor then the differential equation I,)M,)d+I,)N,)d = 0 becomes eact. A solution to the above equation is obtained b solving the eact differential equation as in the previous lesson. Note that the given equation ma have several integrating factors. This is eactl the procedure we have used for solving linear differential equations in earlier lesson. Here we deal with more general differential equation. 9. Rule I: B Inspection There is not much theor behind finding integrating factor b inspection. This method works based on recognition of some standard eact differentials that occur frequentl in practice. The following list of eact differentials would be quite useful in solving eact differential equations: i) d) = d+d d d ii) d = ) iii) d ln ) = iv) d arctan ) = v) dln) = d+d ) or d = d d d d or d ln ) = d d d d + or d arctan ) = d d Eample Solve the differential equation +)d+ )d.

2 Solution: The given equation can be rewritten as d+d)+d d This is further rewritten as ) d d d+d)+ = 0 Using standard differential forms given above we get ) d)+d = 0 Integrating the above equation, the desired solution is given as Here c is an arbitrar constant. + = c 9. Rule II: Md+Nd = 0 is homogeneous and M+N 0 If the equationmd+nd = 0 is homogeneous andm+n 0, theni,) = is an integrating factor. In order to prove the result, we need to show that M+N) Md+Nd M+N = dsome function and ) Rewriting Md+Nd as Md+Nd = { d M+N) + d ) d +M N) d )} Multipling b proposed integrating factor we get Md+Nd M+N = { d + d ) + M N) d M+N) d )} 9.) Given thatm,) andn,) are homogeneous functions of some degreen, i.e.,mt,t) = t n M,) and N,) = t n N,). Then ) M, = M, ) = ) nm,) M,) = n M,

3 Similarl, we get N,) = n N Now consider ) M N) n M,) n N, M+N) = ) ) = n M, + n N, Going back to the Equation 9.), we have Md+Nd M+N = ), { dln))+f ) M,) N, ) ) = f M, +N, ) d ln )} ) Rewriting f /) = f epln/))) and defining g) := fep)), the above equation becomes Md+Nd M+N = { dln))+gln/))d ln )} Hence, we have shown that Md+Nd M+N = d [ ln)+ g ln ) d ln )] Thus M+N Nd = 0. is an integrating factor of the homogenous differential equation Md Eample Solve the differential equation )d 3 3 )d = 0 Solution: The given equation is a homogeneous differential equation. Comparing it with Md+Nd = 0, we have M = and N = 3 3 ). Since the integrating factor is M+N = ) 3 3 ) = 0, M+N) = Multipl b the integrating factor, the given differential equation becomes / /)d / 3/)d = 0 3

4 This is now eact and can be rewritten as d d d+ 3 d = 0 d ) d+ 3 d = 0 Integrating the above equation we obtain the desired solution as ln+3ln = c 9.3 Rule III: Md+Nd = 0 is of the form f )d+f )d = 0 If the equationmd+nd = 0 is of the form f )d+f )d = 0, then M N) is an integrating factor provided M N 0. Similar to rule II we now show that Md+Nd M N = dsome function and ) Again, rewriting Md + Nd as Md+Nd = { d M+N) + d ) d +M N) d )} Now dividing b M N we get { Md+Nd M+N) d M N = M N + d ) d + d )} Using M = f ) and N = f ) we obtain Md+Nd M N = { f )+f ) f ) f ) dln)+d ln )} Let f) := f )+f ) and g) := fep)), the above equation reduces to f ) f ) Md+Nd M N = { f)dln)+d ln )} = { gln)dln)+d ln )} This shows that [ Md+Nd M N = d gln)dln)+ ln )] 4

5 9.3. Eample Solve +)d+ )d = 0. Solution: Comparing withmd+nd = 0, we havem = +) andn = ). The given equation is of the form f )d+f )d = 0 and we have M N = +) ) = Therefore, multipling the equation b /3 3 3, we obtain /3+/3 3 ))d+/3 3 ) /3)d = 0 This is an eact differential equation which can be solved with the technique discussed in previous lesson. 9.4 Rule IV: Most general approach Now we discuss the most general approach of finding integrating function. The idea is to multipl the given differential equation M,)d+N,)d = 0 9.) b a function I,) and then tr to choose I,) so that the resulting equation I,)M,)d+I,)N,)d = 0 9.3) becomes eact. The above equation is eact if and onl if IM) = IN) 9.4) If a function I, ) satisfing the partial differential Equation 9.4) can be found, then 9.3) will be eact. Unfortunatel, solving Equation 9.4), is as difficult to solve as the original Equation 9.) b some other methods. Therefore, while in principle integrating factors are powerful tools for solving differential equations, in practice the can be found 5

6 onl in special cases. The cases we will consider are: i) an integrating factor I that is either as function of onl, or ii) a function of onl. Let us determine necessar conditions on M and N so that 9.) has an integrating factor I that depends on onl. Assuming that I is a function of onl, then Equation 9.4) reduces to IM = IN +N di d di d = IM IN N 9.5) If M N )/N is a function of onl, sa f), then there is an integrating factor I that also depends onl on which can be found b solving 9.5) as I) = e f)d. A similar procedure can be used to determine a condition under which Equation 9.) has an integrating factor depending onl on. To conclude, we have: If M N N ) is function ofalone saf), then I) = f)d e is an I.F. If N M M ) is function of alone saf), then I) = f)d e is an I.F. 9.5 Eample Problems 9.5. Problem Find an integrating factor of + + )d + d = 0 Solution: Comparing with Md+Nd = 0, we have Further, note that M = + +) and N = N M N ) = is a function of alone. Hence, the integrating factor of the given problem is e /d = Problem Find an integrating factor of 4 e + 3 +)d+ 4 e 3)d = 0 Solution: Compare with Md + Nd = 0, we get M = 4 e + 3 +) and N = 4 e 3) 6

7 Also, note that N M M ) = 4 is a function of alone. Hence the integrating factor of the given problem ise 4/d = / 4. Suggested Readings Boce, W.E. and DiPrima, R.C. 00). Elementar Differential Equations and Boundar Value Problems. Seventh Edition, John Wille & Sons, Inc., New York. Dube, R. 00). Mathematics for Engineers Volume II). Narosa Publishing House. New Delhi. McQuarrie, D.A. 009). Mathematical Methods for Scientist and Engineers. First Indian Edition. Viva Books Pvt. Ltd. New Delhi. Raisinghania, M.D. 005). Ordinar & Partial Differential Equation. Eighth Edition. S. Chand & Compan Ltd., New Delhi. Kreszig, E. 993). Advanced Engineering Mathematics. Seventh Edition, John Wille & Sons, Inc., New York. Arfken, G.B. 00). Mathematical Methods for Phsicists. Fifth Edition, Harcourt Academic Press, San Diego. Grewal, B.S. 007). Higher Engineering Mathematics. Fourteenth Edition. Khanna Publishilers, New Delhi. Piskunov, N. 996). Differential and Integral Calculus Volume - ). First Edition. CBS Publisher, Moscow. 7

Particular Solutions

Particular Solutions Our eamples so far in this section have involved some constant of integration, K. We now move on to see particular solutions, where we know some boundar conditions and we substitute