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# Exact & non differential equation

exact & non exact differential equations, integrating factor

exact & non exact differential equations, integrating factor

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## Weitere Verwandte Inhalte

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### Exact & non differential equation

1. 1. EXACT & NON EXACT DIFFERENTIAL EQUATION 8/2/2015 Differential Equation 1
2. 2.  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
3. 3.  SOLUTION OF EXACT D.E. • The solution is given by : 𝑦=𝑐𝑜𝑛𝑠𝑡 𝑎 𝑛𝑡 𝑀𝑑𝑥 + 𝑡𝑒𝑟𝑚𝑠𝑜𝑓𝑁𝑛𝑜𝑡𝑐𝑜𝑛𝑡 𝑎𝑖 𝑛𝑖 𝑛 𝑔𝑥 𝑑𝑦 = 𝑐 8/2/2015 Differential Equation 4
4. 4.  Example : 1 Find the solution of differential equation 𝒚𝒆 𝒙 𝒅𝒙 + 𝟐𝒚 + 𝒆 𝒙 𝒅𝒚 = 𝟎 . Solution: Let M(x, y)= 𝑦𝑒 𝑥 and N(x, y)= 2𝑦 + 𝑒 𝑥 Now, 𝜕𝑀 𝜕𝑦 = 𝑒 𝑥 , 𝜕𝑁 𝜕𝑥 = 𝑒 𝑥 ∴ 𝝏𝑴 𝝏𝒚 = 𝝏𝑵 𝝏𝒙 8/2/2015 Differential Equation 5
5. 5.  Example : 1 (cont.) The given differential equation is exact , 𝑦=𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑀𝑑𝑥 + 𝑡𝑒𝑟𝑚𝑠 𝑜𝑓 𝑁 𝑛𝑜𝑡 𝑐𝑜𝑛𝑡𝑎𝑖 𝑛𝑖 𝑛 𝑔 𝑥 𝑑𝑦 = 𝑐 𝑦=𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑦𝑒 𝑥 𝑑𝑥 + 2𝑦 𝑑𝑦 = 0 8/2/2015 Differential Equation 6 ⇒ ⇒ ⇒ 𝒚𝒆 𝒙+ 𝒚 𝟐= c
6. 6.  NON EXACT DIFFERENTIAL EQUATION • For the differential equation 𝑀 𝑥, 𝑦 𝑑𝑥 + 𝑁 𝑥, 𝑦 𝑑𝑦 = 0 IF 𝝏𝑴 𝝏𝒚 ≠ 𝝏𝑵 𝝏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
7. 7.  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
8. 8.  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
9. 9.  CASE I : If 1 𝑁 ( 𝜕𝑀 𝜕𝑦 + 𝜕𝑁 𝜕𝑥 )f 𝑥 {i.e. function of x only} Then I.F. = 𝒆 𝒇 𝒙 .𝒅𝒙 8/2/2015 Differential Equation 11
10. 10.  Example : 2 Solve : (𝒙 𝟐 +𝒚 𝟐 + 𝟑)𝒅𝒙 − 𝟐𝒙𝒚. 𝒅𝒚 = 𝟎 Solution: Let M(x, y)= 𝑥2 +𝑦2 + 3 and N(x, y)= 2𝑥𝑦. 𝑑𝑦 Now, 𝜕𝑀 𝜕𝑦 = 2𝑦 , 𝜕𝑁 𝜕𝑥 =-2𝑦 ∴ 𝝏𝑴 𝝏𝒚 ≠ 𝝏𝑵 𝝏𝒙 8/2/2015 Differential Equation 12
11. 11. 1 𝑁 ( 𝜕𝑀 𝜕𝑦 + 𝜕𝑁 𝜕𝑥 ) = 1 −2𝑥𝑦 2𝑦 + 2𝑦 f 𝑥 = −2 𝑥 Now, I.F. = 𝑒 𝑓 𝑥 .𝑑𝑥 = 𝑒 −2 𝑥 𝑑𝑥 = 𝑒−2𝑙𝑜𝑔𝑥 = 𝑒 𝑙𝑜𝑔𝑥−2 =𝑥−2 ∴ 𝐼. 𝐹. = 𝑓 𝑥, 𝑦 = 1 𝑥2 8/2/2015 Differential Equation 13  Example : 2 (cont.)
12. 12. Multiply both side by I.F. (i.e. 1 𝒙 𝟐), we get 1 𝑥2 (𝑥2 + 𝑦2 + 3)𝑑𝑥 − 2𝑥𝑦. 𝑑𝑦 = 0 [( 1 + 𝑦2 𝑥2 + 3 𝑥2 )𝑑𝑥 − 2𝑥𝑦. 𝑑𝑦] = 0 8/2/2015 Differential Equation 14  Example : 2 (cont.)
13. 13.  Example : 2 (cont.) Let M(x, y)=1 + 𝑦2 𝑥2 + 3 𝑥2 and N(x, y)= 2𝑥𝑦 Now, 𝜕𝑀 𝜕𝑦 = 2𝑦 𝑥2, 𝜕𝑁 𝜕𝑥 = 2𝑦 𝑥2 ∴ 𝝏𝑴 𝝏𝒚 = 𝝏𝑵 𝝏𝒙 8/2/2015 Differential Equation 15
14. 14.  Example : 2 (cont.) [( 1 + 𝑦2 𝑥2 + 3 𝑥2)𝑑𝑥 − 2𝑥𝑦. 𝑑𝑦] = 0, which is exact differential equation. It’s solution is : 𝑦=𝑐𝑜𝑛𝑠𝑡 𝑎 𝑛𝑡 𝑀𝑑𝑥 + 𝑡𝑒𝑟𝑚𝑠𝑜𝑓𝑁𝑛𝑜𝑡𝑐𝑜𝑛𝑡𝑎𝑖 𝑛𝑖 𝑛 𝑔𝑥 𝑑𝑦 = 𝑐 𝑦=𝑐𝑜𝑛𝑠𝑡 𝑎 𝑛𝑡 1 + 𝑦2 𝑥2 + 3 𝑥2 )𝑑𝑥+ (0)𝑑𝑦 = 𝑐 8/2/2015 Differential Equation 16 x − 𝑦2 𝑥 − 3 𝑥 = 𝑐
15. 15.  CASE II : If 1 𝑀 ( 𝜕𝑁 𝜕𝑥 − 𝜕𝑀 𝜕𝑦 ) is a function of y only , say g(y), then 𝑒 𝑔 𝑦 𝑑𝑦 is an I.F.(Integrating Factor). 178/2/2015 Differential Equation
16. 16.  Example : 3 Solve 𝑦4 + 2y dx + 𝑥𝑦3 + 2𝑦4 − 4𝑥 𝑑𝑦=0 Solution: Here M=𝑦4 + 2𝑦 and so 𝜕𝑀 𝜕𝑦 = 4𝑦3+2 N=𝑥𝑦3 + 2𝑦4 − 4𝑥 and so 𝜕𝑁 𝜕𝑥 = 𝑦3 − 4 Thus, 𝜕𝑀 𝜕𝑌 ≠ 𝜕𝑁 𝜕𝑥 and so the given differential equation is non exact. 188/2/2015 Differential Equation
17. 17.  Example : 3 (cont.) Now, 1 𝑀 𝜕𝑁 𝜕𝑥 − 𝜕𝑀 𝜕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 𝑦3 198/2/2015 Differential Equation
18. 18.  Example : 3 (cont.) Multiplying the given differential equation by 1 𝑦3 ,we have 1 𝑦3 𝑦4 + 2𝑦 𝑑𝑥 + 𝑥𝑦3 + 2𝑦4 − 4𝑥 𝑑𝑦 ⇒ 𝑦 + 2 𝑦2 𝑑𝑥 + 𝑥 + 2𝑦 − 4 𝑥 𝑦3 𝑑𝑦 = 0 ----------------(i) Now here, M=𝑦 + 2 𝑦2 and so 𝜕𝑀 𝜕𝑥 = 1 − 4 𝑦3 N=𝑥 + 2𝑦 − 4 𝑥 𝑦3 and so 𝜕𝑁 𝜕𝑥 = 1 − 4 𝑦3 Thus, 𝜕𝑀 𝜕𝑦 = 𝜕𝑁 𝜕𝑥 and hence 𝑒𝑞 𝑛 (i) is an exact differential equatio 208/2/2015 Differential Equation
19. 19.  Example : 3 (cont.) Therefore , General Solution is 𝑦𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑀𝑑𝑥 + (𝑇𝑒𝑟𝑚𝑠 𝑖𝑛 𝑁 𝑤ℎ𝑖𝑐ℎ 𝑎𝑟𝑒 𝑖𝑛𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡 𝑜𝑓 𝑥) 𝑑𝑦 = 𝑐  ⇒ 𝑦 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑦 + 2 𝑦2 𝑑𝑥 + 2𝑦𝑑𝑦 = 𝑐 ⇒𝒙𝒚 + 𝟐𝒙 𝒚 𝟐 + 𝒚 𝟐 =c where c is an arbitrary constant. 218/2/2015 Differential Equation
20. 20.  CASE III : If the given differential equation is homogeneous with 𝑀𝑥 + 𝑁𝑦 ≠ 0 then 1 𝑀𝑥+𝑁𝑦 is an I.F. 228/2/2015 Differential Equation
21. 21.  Example : 4  Solve 𝑥2 𝑦𝑑𝑥 − 𝑥3 + 𝑥𝑦3 𝑑𝑦 = 0 Solution: Here M=𝑥2 𝑦 and so 𝜕𝑀 𝜕𝑦 = 𝑥2 N=−𝑥3 − 𝑥𝑦2 and so 𝜕𝑁 𝜕𝑥 = −3𝑥2 − 𝑦2 ∴ 𝜕𝑀 𝜕𝑦 ≠ 𝜕𝑁 𝜕𝑥 ∴ The given differential equation is non exact. 238/2/2015 Differential Equation
22. 22.  Example : 4 (cont.) The given differential equation is homogeneous function of same degree=3. [ 𝑀 𝑡𝑥, 𝑡𝑦 = (𝑡𝑥)2(𝑡𝑦) =𝑡3 𝑥2 𝑦 =𝑡3 𝑀 𝑥, 𝑦 𝑁 𝑥, 𝑦 = −3 𝑡𝑥 3 − (𝑡𝑥) (𝑡𝑦)2 =𝑡3 −3𝑥2 − 𝑥𝑦2 =𝑡3 𝑁 𝑥, 𝑦 248/2/2015 Differential Equation
23. 23.  Example : 4 (cont.) Now, 𝑀𝑥 + 𝑁𝑦 = 𝑥2 𝑦 𝑥 − 𝑥3 + 𝑥𝑦2 𝑦 =𝑥3 𝑦 − 𝑥3 𝑦 − 𝑥𝑦3 =−𝑥𝑦3 ≠ 0 Thus, I.F.= 1 𝑀𝑥+𝑁𝑦 = 1 𝑥𝑦3 Now, multiplying given differential equation by 1 𝑥𝑦3 we have − 1 𝑥𝑦3 𝑥2 𝑦𝑑𝑥 − 𝑥3 + 𝑥𝑦2 𝑑𝑦 = 0 258/2/2015 Differential Equation
24. 24.  Example : 4 (cont.) ⇒ − 𝑥 𝑦2 𝑑𝑥 + 𝑥3 𝑦3 + 1 𝑦 𝑑𝑦 = 0 −−−−−− −(i) Here, M= −𝑥 𝑦2 and so 𝜕𝑀 𝜕𝑦 = 2𝑥 𝑦3 N= 𝑥2 𝑦3 + 1 𝑦 and so 𝜕𝑁 𝜕𝑥 = 2𝑥 𝑦3 Thus, 𝜕𝑀 𝜕𝑦 = 𝜕𝑁 𝜕𝑥 and hence 𝑒𝑞 𝑛 (i) is an exact differential equation. 268/2/2015 Differential Equation
25. 25.  Example : 4 (cont.) Therefore , General Solution is 𝑦𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑀𝑑𝑥 + (𝑇𝑒𝑟𝑚𝑠 𝑖𝑛 𝑁 𝑤ℎ𝑖𝑐ℎ 𝑎𝑟𝑒 𝑖𝑛𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡 𝑜𝑓 𝑥) 𝑑𝑦 = 𝑐 ⇒ 𝑦 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 −𝑥 𝑦2 𝑑𝑥 + 1 𝑦 𝑑𝑦 = 𝑐 ⇒ − 𝟏 𝟐 𝒙 𝟐 𝒚 𝟐 + 𝒍𝒏𝒚 = 𝒄 where c is an arbitrary constant. 278/2/2015 Differential Equation
26. 26.  CASE IV : If the given differential equation is of the form 𝑓1 𝑥𝑦 𝑦𝑑𝑥 + 𝑓2 𝑥𝑦 𝑥𝑑𝑦 = 0 𝑤𝑖𝑡ℎ 𝑀𝑥 − 𝑁𝑦 ≠ 0, 𝑡ℎ𝑒𝑛 1 𝑀𝑥−𝑁𝑦 is an I.F. 288/2/2015 Differential Equation
27. 27.  Example : 4 (cont.)  Solve (𝑥2 𝑦 + 2)𝑦𝑑𝑥 + 2 − 𝑥2 𝑦2 𝑥𝑑𝑦 = 0 Solution: Here, M=(𝑥2 𝑦3 + 2𝑦) and so 𝜕𝑀 𝜕𝑦 = 3𝑥2 𝑦2 + 2 N=(𝑥 − 𝑥3 𝑦2) and so 𝜕𝑁 𝜕𝑥 = 2 − 3𝑥𝑦2 ∴ 𝜕𝑀 𝜕𝑦 ≠ 𝜕𝑁 𝜕𝑥 ∴ The given differential equation is non exact. 298/2/2015 Differential Equation
28. 28.  Example : 4 (cont.) Now, 𝑀𝑥 − 𝑁𝑦 = 𝑥2 𝑦2 + 2 𝑦𝑥 + 2 − 𝑥2 𝑦2 𝑥𝑦 =𝑥3 𝑦3 + 2𝑥𝑦 − 2𝑥𝑦 + 𝑥3 𝑦3 =2𝑥3 𝑦3 So, I.F.= 1 𝑀𝑥−𝑁𝑦 = 1 2𝑥3 𝑦3 Multiplying the given equation by 1 2𝑥3 𝑦3 , we have 1 2𝑥3 𝑦3 [ 𝑥2 + 2 𝑦𝑑𝑥 + 2 − 𝑥2 𝑦2 𝑥𝑑𝑦] 308/2/2015 Differential Equation
29. 29.  Example : 4 (cont.) ⇒ 1 2𝑥 + 1 𝑥3 𝑦3 𝑑𝑥 + 1 𝑥2 𝑦3 − 1 2𝑦 𝑑𝑦 = 0----------(i) Here, M= 1 2𝑥 + 1 𝑥3 𝑦3 and so 𝜕𝑀 𝜕𝑦 = − 2 𝑥3 𝑦3 N= 1 𝑥2 𝑦3 − 1 2𝑦 and so 𝜕𝑁 𝜕𝑥 = − 2 𝑥3 𝑦3 Thus, 𝜕𝑀 𝜕𝑦 = 𝜕𝑁 𝜕𝑥 and hence 𝑒𝑞 𝑛 (i) is an exact differential equation. 318/2/2015 Differential Equation
30. 30.  Example : 4 (cont.) Therefore , General Solution is 𝑦𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑀𝑑𝑥 + (𝑇𝑒𝑟𝑚𝑠 𝑖𝑛 𝑁 𝑤ℎ𝑖𝑐ℎ 𝑎𝑟𝑒 𝑖𝑛𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡 𝑜𝑓 𝑥) 𝑑𝑦 = 𝑐 ⇒ 𝑦 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 1 2𝑥 + 1 𝑥3 𝑦3 𝑑𝑥 + − 1 2𝑦 𝑑𝑦 = 𝑐 ⇒ 𝟏 𝟐 𝒍𝒐𝒈𝒙 − 𝟏 𝟐𝒙 𝟐 𝒚 𝟐 − 𝟏 𝟐 𝒍𝒐𝒈𝒚 = 𝒄 where c is an arbitrary constant. 328/2/2015 Differential Equation
31. 31. 8/2/2015 Differential Equation 33