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- 1. EXACT & NON EXACT DIFFERENTIAL EQUATION 8/2/2015 Differential Equation 1
- 3. EXACT DIFFERENTIAL EQUATION A differential equation of the form M(x, y)dx + N(x, y)dy = 0 is called an exact differential equation if and only if 8/2/2015 Differential Equation 3
- 4. SOLUTION OF EXACT D.E. • The solution is given by : ∫ 𝑦=𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑀𝑑𝑥 +∫ 𝑡𝑒𝑟𝑚𝑠𝑜𝑓𝑁𝑛𝑜𝑡𝑐𝑜𝑛𝑡𝑎𝑖𝑛𝑖𝑛𝑔𝑥𝑑𝑦 = 𝑐 8/2/2015 Differential Equation 4
- 5. Example : 1 Find the solution of differential equation Solution: 𝒚𝒆𝒙𝒅𝒙 + 𝟐𝒚 + 𝒆𝒙 𝒅𝒚 = 𝟎 . Let M(x, y)= 𝑦𝑒𝑥 and N(x, y)= 2𝑦 + 𝑒𝑥 Now, 𝜕 𝑦 𝜕𝑀 = 𝑒𝑥 𝜕 𝑥 , 𝜕𝑁 = 𝑒𝑥 ∴ 𝝏𝑴 = 𝝏𝑵 𝝏𝒚 𝝏𝒙 8/2/2015 Differential Equation 5
- 6. Example : 1 (cont.) The given differential equation is exact , ∫ 𝑦=𝑐𝑜𝑛𝑠𝑡𝑎𝑛 𝑡 𝑀𝑑𝑥 +∫ 𝑡𝑒𝑟𝑚𝑠𝑜𝑓 𝑁𝑛𝑜𝑡𝑐𝑜𝑛𝑡𝑎𝑖𝑛𝑖𝑛𝑔𝑥 𝑑𝑦 = 𝑐 𝑦=𝑐𝑜𝑛𝑠𝑡𝑎𝑛 𝑡 ∫ 𝑦𝑒𝑥𝑑𝑥 + ∫ 2𝑦 𝑑𝑦 = 0 8/2/2015 Differential Equation 6 ⇒ ⇒ ⇒ 𝒚𝒆𝒙+ 𝒚𝟐= c
- 8. NON EXACT DIFFERENTIAL EQUATION • For the differential equation 𝑀 𝑥, 𝑦 𝑑𝑥 + 𝑁 𝑥, 𝑦 𝑑𝑦 = 0 IF 𝝏𝑴 ≠ 𝝏𝑵 𝝏𝒚 𝝏𝒙 then, 𝑫𝒊𝒇𝒇𝒆𝒓𝒆𝒏𝒕𝒊𝒂𝒍 𝑬𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝒊𝒔 𝒔𝒂𝒊𝒅 𝒕𝒐 𝒃𝒆 𝑵𝑶𝑵𝑬𝑿𝑨𝑪𝑻 • If the given differential equation is not exact then make that equation exact by finding INTEGRATING FACTOR. 8/2/2015 Differential Equation 8
- 9. INTEGRATING FACTOR • In general, for differential equation M(x, y)dx + N(x, y)dy = 0 is not exact. In such situation, we find a function 𝜆 such that by multiplying𝜆 to the equation, it becomes an exact equation. So, 𝝀M(x, y)dx +𝝀N(x, y)dy = 0 becomes exact equation Here the function 𝜆 = 𝜆(𝑥, 𝑦) is then called an Integrating Factor 8/2/2015 Differential Equation 9
- 10. Methods to find an INTEGRATING FACTOR (I.F.) for given non exact equation: M(x, y)dx + N(x, y)dy = 0 CASES: CASE I CASE II CASE III CASE IV
- 11. CASE I : If 1 (𝜕𝑀 + 𝜕𝑁 )f 𝑥 𝑁 𝜕𝑦 𝜕𝑥 {i.e.functionofxonly} Then I.F. = 𝒆∫ 𝒇 𝒙 .𝒅𝒙 8/2/2015 Differential Equation 11
- 12. Example : 2 Solve : (𝒙𝟐+𝒚𝟐 + 𝟑)𝒅𝒙 − 𝟐𝒙𝒚. 𝒅𝒚 = 𝟎 Solution: Let M(x, y)= 𝑥2+𝑦2 + 3 and N(x, y)= 2𝑥𝑦. 𝑑𝑦 𝜕 𝑦 Now, 𝜕𝑀 = 2𝑦 , 𝜕 𝑥 𝜕𝑁 =-2𝑦 ∴ 𝝏𝑴 ≠ 𝝏𝑵 𝝏𝒚 𝝏𝒙 8/2/2015 Differential Equation 12
- 13. 𝑁 𝜕 𝑦 1 (𝜕𝑀 + 𝜕𝑁 ) = 1 −2𝑥 𝑦 2𝑦 + 2𝑦 𝜕𝑥 f 𝑥 = −2 𝑥 Now, I.F. = 𝑒∫ 𝑓 𝑥 .𝑑𝑥 = 𝑒∫ 𝑥 −2 𝑑𝑥 = 𝑒−2𝑙𝑜𝑔𝑥 = 𝑒𝑙𝑜𝑔𝑥−2 =𝑥−2 ∴ 𝐼. 𝐹. = 𝑓 𝑥, 𝑦 = 1 𝑥2 Example : 2 (cont.) 8/2/2015 Differential Equation 13
- 14. 𝒙𝟐 Multiply both side by I.F. (i.e. 1 ), we get 𝑥2 1 (𝑥2 + 𝑦2 + 3)𝑑𝑥 − 2𝑥𝑦. 𝑑𝑦 = 0 𝑦2 3 [( 1 + + )𝑑𝑥 − 2𝑥𝑦. 𝑑𝑦] = 0 𝑥2 𝑥2 Example : 2 (cont.) 8/2/2015 Differential Equation 14
- 15. Example : 2 (cont.) 2 Let M(x, y)=1 + 𝑦 + 3 𝑥2 𝑥2 and N(x, y)= 2𝑥𝑦 Now, 𝜕𝑦 𝜕 𝑥 𝜕𝑀 = 2𝑦 , 𝜕𝑁 = 2𝑦 𝑥2 𝑥2 ∴ 𝝏𝑴 = 𝝏𝑵 𝝏𝒚 𝝏𝒙 8/2/2015 Differential Equation 15
- 16. Example : 2 (cont.) 2 [( 1 + 𝑦 + 3 )𝑑𝑥 − 2𝑥𝑦. 𝑑𝑦] = 0, 𝑥2 𝑥2 which is exact differential equation. It’s solution is : ∫ 𝑀𝑑𝑥+∫ 𝑡𝑒𝑟𝑚𝑠𝑜𝑓𝑁𝑛𝑜𝑡𝑐𝑜𝑛𝑡𝑎𝑖𝑛𝑖𝑛𝑔𝑥𝑑𝑦 = 𝑐 𝑦=𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 ∫ 𝑦=𝑐𝑜𝑛𝑠𝑡𝑎𝑛 𝑡 𝑦2 3 1 + 𝑥2 + 𝑥2)𝑑𝑥 +∫ (0)𝑑𝑦= 𝑐 x − 3 8/2/2015 Differential Equation 16 𝑦2 𝑥 𝑥 − = 𝑐
- 17. CASE II : 𝑀 𝜕𝑥 𝜕 𝑦 If 1 (𝜕𝑁 − 𝜕𝑀 ) is a function of y only , say g(y), then 𝑒∫ 𝑔 𝑦 𝑑𝑦 is an I.F.(Integrating Factor). 8/2/2015 Differential Equation 17
- 18. Example : 3 Solve 𝑦4 + 2y dx + 𝑥𝑦3 + 2𝑦4 − 4𝑥 𝑑𝑦=0 Solution: 𝜕 𝑦 Here M=𝑦4 + 2𝑦 and so 𝜕𝑀 = 4𝑦3+2 𝜕 𝑥 N=𝑥𝑦3 + 2𝑦4 − 4𝑥 and so 𝜕𝑁 = 𝑦3 − 4 𝜕𝑌 𝜕 𝑥 8/2/2015 Differential Equation 18 Thus, 𝜕𝑀 ≠ 𝜕𝑁 and so the given differential equation is non exact.
- 19. Example : 3 (cont.) Now, 1 𝜕𝑁 − 𝜕𝑀 𝑀 𝜕𝑥 𝜕𝑦 = 1 𝑦4+2𝑦 𝑦3 − 4 − 4𝑦3 − 2 −(3𝑦3+6) = 𝑦(𝑦3+2) 3 = - , 𝑦 which is a function of y only . Therefore I.F.=𝑒∫−3/𝑦=𝑒−3𝑙𝑛𝑦 = 1 8/2/2015 Differential Equation 19 𝑦3
- 20. Example : 3 (cont.) Multiplying the given differential equation by 1 𝑦3 ,we have 1 𝑦4 + 2𝑦 𝑑𝑥 + 𝑥𝑦3 + 2𝑦4 − 4𝑥 𝑑𝑦 ⇒ 𝑦 + 𝑦3 2 𝑦2 𝑑𝑥 + 𝑥 + 2𝑦 − 4 𝑥 𝑦3 𝑑𝑦 = 0 ----------------(i) Now here, M=𝑦 + 2 𝑦2 𝜕 𝑥 and so 𝜕𝑀 = 1 − 4 𝑦3 N=𝑥 + 2𝑦 − 4 𝑥 𝑦3 𝜕 𝑥 and so 𝜕𝑁 = 1 − 4 𝑦3 𝜕𝑦 𝜕 𝑥 8/2/2015 Differential Equation 20 Thus, 𝜕𝑀 = 𝜕𝑁 and hence 𝑒𝑞𝑛(i) is an exact differential equatio
- 21. Example : 3 (cont.) Therefore , General Solution is ∫ 𝑀𝑑𝑥 + ∫ 𝑦𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑖𝑛𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡 𝑜𝑓 𝑥) (𝑇𝑒𝑟𝑚𝑠 𝑖𝑛 𝑁 𝑤ℎ𝑖𝑐ℎ 𝑎𝑟𝑒 𝑑𝑦 = 𝑐 𝑦 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 ⇒ ∫ 𝑦 + 2 𝑦2 𝑑𝑥 + ∫ 2𝑦𝑑𝑦 = 𝑐 ⇒𝒙𝒚 + 𝟐𝒙 + 𝒚𝟐=c 8/2/2015 Differential Equation 21 𝒚𝟐 where c is an arbitrary constant.
- 22. CASE III : If the given differential equation is homogeneous with 𝑀𝑥 + 𝑁𝑦 ≠ 0 then 1 8/2/2015 Differential Equation 22 𝑀𝑥+𝑁 𝑦 is an I.F.
- 23. Example : 4 Solve 𝑥2𝑦𝑑𝑥 − 𝑥3 + 𝑥𝑦3 𝑑𝑦 = 0 Solution: 𝜕 𝑦 Here M=𝑥2𝑦 and so 𝜕𝑀 = 𝑥2 N=−𝑥3 − 𝑥𝑦2 and so 𝜕𝑁 = −3𝑥2 − 𝑦2 ∴ 𝜕𝑀 ≠ 𝜕𝑥 𝜕𝑁 8/2/2015 Differential Equation 23 𝜕𝑦 𝜕𝑥 ∴ The given differential equation is non exact.
- 24. Example : 4 (cont.) The given differential equation is homogeneous function of same degree=3. [ 𝑀 𝑡𝑥, 𝑡𝑦 = (𝑡𝑥)2(𝑡𝑦) =𝑡3 𝑥2𝑦 =𝑡3 𝑀 𝑥, 𝑦 𝑁 𝑥, 𝑦 = −3 𝑡𝑥 3 − (𝑡𝑥) (𝑡𝑦)2 =𝑡3 −3𝑥2 − 𝑥𝑦2 =𝑡3 𝑁 𝑥, 𝑦 8/2/2015 Differential Equation 24
- 25. Example : 4 (cont.) Now, 𝑀𝑥 + 𝑁𝑦 = 𝑥2𝑦 𝑥 − 𝑥3 + 𝑥𝑦2 𝑦 =𝑥3𝑦 − 𝑥3𝑦 − 𝑥𝑦3 =−𝑥𝑦3 ≠ 0 1 1 Thus, I.F.= = 𝑀𝑥+𝑁𝑦 𝑥𝑦3 Now, multiplying given differential equation by 1 𝑥𝑦3 we have − 1 𝑥𝑦3 𝑥2𝑦𝑑𝑥 − 𝑥3 + 𝑥𝑦2 𝑑𝑦 = 0 8/2/2015 Differential Equation 25
- 26. Example : 4 (cont.) 𝑥 ⇒ − 𝑑𝑥 + 𝑦2 𝑥3 1 𝑦3 + 𝑦 𝑑𝑦 = 0 −−−−−− −(i) −𝑥 𝜕𝑀 2𝑥 Here, M= and so = 𝑦2 𝜕𝑦 𝑦3 𝑥2 1 𝜕𝑁 2𝑥 N= + and so = 𝑦3 𝑦 𝜕𝑥 𝑦3 𝜕𝑦 𝜕 𝑥 8/2/2015 Differential Equation 26 Thus, 𝜕𝑀 = 𝜕𝑁 and hence 𝑒𝑞𝑛(i) is an exact differential equation.
- 27. ∫ 𝑀𝑑𝑥 + ∫ 𝑦𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑖𝑛𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡 𝑜𝑓 𝑥) Example : 4 (cont.) Therefore , General Solution is (𝑇𝑒𝑟𝑚𝑠 𝑖𝑛 𝑁 𝑤ℎ𝑖𝑐ℎ 𝑎𝑟𝑒 𝑑𝑦 = 𝑐 ⇒ ∫ 𝑦 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 −𝑥 𝑦2 𝑑𝑥 + ∫ 𝑦 1 𝑑𝑦 = 𝑐 𝟏 𝒙𝟐 ⇒ − + 𝒍𝒏𝒚 = 𝒄 𝟐 𝒚𝟐 where c is an arbitrary constant. 8/2/2015 Differential Equation 27
- 28. CASE IV : If the given differential equation is of the form 𝑓1 𝑥𝑦 𝑦𝑑𝑥 + 𝑓2 𝑥𝑦 𝑥𝑑𝑦 = 1 8/2/2015 Differential Equation 28 𝑀𝑥−𝑁𝑦 is an 0 𝑤𝑖𝑡ℎ 𝑀𝑥 − 𝑁𝑦 ≠ 0, 𝑡ℎ𝑒𝑛 I.F.
- 29. Example : 4 (cont.) Solve (𝑥2𝑦 + 2)𝑦𝑑𝑥 + 2 − 𝑥2𝑦2 𝑥𝑑𝑦 = 0 Solution: 𝜕 𝑦 Here, M=(𝑥2𝑦3 + 2𝑦) and so 𝜕𝑀 = 3𝑥2𝑦2 + 2 𝜕 𝑥 N=(𝑥 − 𝑥3𝑦2) and so 𝜕𝑁 = 2 − 3𝑥𝑦2 ∴ ≠ 𝜕𝑀 𝜕𝑁 8/2/2015 Differential Equation 29 𝜕𝑦 𝜕𝑥 ∴ The given differential equation is non exact.
- 30. Example : 4 (cont.) Now, 𝑀𝑥 − 𝑁𝑦 = 𝑥2𝑦2 + 2 𝑦𝑥 + 2 − 𝑥2𝑦2 𝑥𝑦 =𝑥3𝑦3 + 2𝑥𝑦 − 2𝑥𝑦 + 𝑥3𝑦3 =2𝑥3𝑦3 So, I.F.= = 1 1 𝑀𝑥−𝑁𝑦 2𝑥3𝑦3 Multiplying the given equation by 1 2𝑥3𝑦3 , we have 1 2𝑥3𝑦3 [ 𝑥2 + 2 𝑦𝑑𝑥 + 2 − 𝑥2𝑦2 𝑥𝑑𝑦] 8/2/2015 Differential Equation 30
- 31. Example : 4 (cont.) ⇒ 1 2𝑥 + 1 1 𝑥3𝑦3 𝑥2𝑦3 𝑑𝑥 + − 1 2 𝑦 𝑑𝑦 = 0----------(i) 2𝑥 1 Here, M= + 1 𝑥3𝑦3 𝜕 𝑦 and so 𝜕𝑀 = − 2 𝑥2𝑦3 1 N= − 1 2𝑦 𝜕 𝑥 and so 𝜕𝑁 = − 𝑥3𝑦3 2 𝑥3𝑦3 𝜕𝑦 𝜕 𝑥 8/2/2015 Differential Equation 31 Thus, 𝜕𝑀 = 𝜕𝑁 and hence 𝑒𝑞𝑛(i) is an exact differential equation.
- 32. Example : 4 (cont.) ∫ 𝑀𝑑𝑥 + ∫ 𝑦𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑖𝑛𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡 𝑜𝑓 𝑥) Therefore , General Solution is (𝑇𝑒𝑟𝑚𝑠 𝑖𝑛 𝑁 𝑤ℎ𝑖𝑐ℎ 𝑎𝑟𝑒 𝑑𝑦 = 𝑐 ⇒ ∫ 𝑦 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 1 1 + 2𝑥 𝑥3𝑦3 1 𝑑𝑥 + ∫ − 𝑑𝑦 = 𝑐 2𝑦 𝟏 𝟏 𝟏 ⇒ 𝒍𝒐𝒈𝒙 − − 𝒍𝒐𝒈𝒚 = 𝒄 𝟐 𝟐𝒙𝟐𝒚𝟐 𝟐 where c is an arbitrary constant. 8/2/2015 Differential Equation 32