Week 3 [compatibility mode]
- 1. KNF1023 Engineering Mathematics II First Order ODEs Prepared By Annie ak Joseph Prepared By Annie ak Joseph Session 2008/2009
- 2. Learning Objectives Demonstrate how to find integrating factor for non-exact differential equation Demonstrate the solution of Homogeneous 1st order ODE in linear form Demonstrate the solution of inhomogeneous 1st order ODE in linear form
- 3. Integrating Factor If a function u ( x, y ) has continuous partial derivatives, its total or exact differential is ∂u ∂u du = dx + dy ∂x ∂y From this it follows that if u ( x, y ) =c=const, then du = 0 A first-order differential equation of the form M ( x, y )dx + N ( x, y )dy = 0 − −− → (1) Is called exact if its left side is the total or exact differential
- 4. Exact Differential Equation ∂u ∂u du = dx + dy − −− → (2) ∂x ∂y of some function u ( x, y ) . Then the differential equation (1) can be written du = 0 By integration we immediately obtain the general solution of (1) in the form u ( x, y ) = C − −− → (3)
- 5. Exact Differential Equation Comparing (1) and (2), we see that (1) is exact if there is some function u ( x, y ) such that ∂u ∂u a) =M b) =N (4) ∂x ∂y ∂M ∂u 2 ∂N ∂ 2u = = ∂y ∂y∂x ∂x ∂x∂y By the assumption of continuity the two second derivatives are equal. Thus ∂M ∂N = − −− → (5) ∂y ∂x
- 6. Exact Differential Equation This condition is not only necessary but also sufficient for Mdx + Ndy to be an exact differential. If (1) is exact, the function u ( x, y ) can be found by guessing or in the following systematic way. From 4(a) we have by integration with respect to x u = ∫ Mdx + k ( y ) − −− → ( 6 )
- 7. Exact Differential Equation in this integration, y is to be regarded as a constant, and k(y) plays the role of a “constant” of integration. To determine k(y), we derive∂u / ∂y from (6), use (4b) to get ∂k / ∂y and integrate ∂k / ∂y to get k. Formula (6) was obtained from (4a). Instead of 4(a) we may equally well use (4b). Then instead of (6) we first have u = ∫ Ndy + l ( x) − −− → (*6 )
- 8. Exact Differential Equation To determine l(x) we derive ∂u / ∂x from (6*), use (4a) to get ∂l / ∂x and integrate it to get l .
- 9. Example 1: An exact equation 3 2 2 3 Solve ( x + 3 xy )dx + (3x y + y )dy = 0 (7) Solution: 1st step: Test for exactness. Our equation is of the form (1) with M = x 3 + 3 xy 2 N = 3x 2 y + y 3 ∂M = 6 xy ∂y ∂N = 6 xy ∂x exact solution
- 10. Continue… 2nd Step: Implicit solution. From (6) we obtain u = ∫ Mdx + k ( y ) = ∫ ( x 3 + 3 xy 2 )dx + k ( y ) 1 4 3 2 2 = x + x y + k ( y) (8) 4 2 To find k(y), we differentiate this formula with respect to y and use formula (4b), obtaining ∂u 2 dk = 3x y + = N = 3x 2 y + y3 ∂y dy
- 11. Continue… dk 3 y4 * Hence dy = y , so that k = 4 + c . Inserting this into (8) we get the answer 1 4 u ( x, y ) = ( x + 6 x 2 y 2 + y 4 ) = c (9) 4 3rd step: Checking. For checking, we can differentiate u ( x, y ) = c implicitly and see whether this leads to dy = − M dx N or Mdx+ Ndy= 0 , the given equation.
- 12. Continue… In the present case, differentiating (9) implicitly with respect to x, we obtain 1 (4 x3 + 12 xy 2 + 12 x 2 yy '+ 4 y 3 y ') = 0 4 Collecting terms, we see that this equals M + Ny’=0 with M and N as in (7); thus Mdx + Ndy = 0 .This completes the check.
- 13. Example 2 ( ) Solve cos( x + y)dx + 3y 2 + 2 y + cos(x + y) dy = 0 Solution: M = cos( x + y ) N = 3 y 2 + 2 y + cos( x + y ) Thus ∂M = − sin( x + y ) ∂y exact ∂N = − sin( x + y ) ∂x
- 14. Continue… Step 2: Implicit general solution. u = ∫ Mdx + k ( y ) = ∫ cos( x + y )dx + k ( y ) = sin( x + y ) + k ( y ) To find k(y), we differentiate this formula with respect to y and obtain ∂u dk = cos( x + y ) + = N = 3 y 2 + 2 y + cos( x + y ) ∂y dy 2 Hence dk / dy = 3 y + 2 y By integration, k = y 3 + y 2 + c * . Inserting this result u ( x, y ) = sin( x + y ) + y 3 + y 2 = c
- 15. Continue… Step 3: Checking an implicit solution. We can check by differentiating the implicit solution u(x,y) = c implicitly and see whether this leads to the given ODE: ∂u ∂u du = dx + dy = cos( x + y )dx + (cos( x + y ) + 3 y 2 + 2 y )dy = 0 ∂x ∂y This completes the check.
- 17. Identify whether the ODE is homogeneous or not??? dy Differential equation ( x , y ) is call = f dx Homogeneous equation if f ( λ x, λ y ) = f ( x, y ) for every real value of λ Example 3: Identify whether dy = y − x is dx y + x homogeneous or not.
- 18. Continue… y−x f ( x, y ) = y+x λ y − λx f ( λ x, λ y ) = λ y + λx y−x = y+x = f ( x, y ) This equation is homogeneous
- 19. Example 4 Check whether the equation given is homogeneous or not? dy = x− y dx f ( x, y ) = x − y f ( λ x, λ y ) = λ x − λ y = λ ( x − y) = λ f ( x, y ) This equation is not homogeneous.
- 20. Example 5 dy x + 3 y Solve = dx 2x Solution: The above equation is a homogeneous y = ux dy du =u+x dx dx du x + 3(ux) x(1 + 3u ) 1 + 3u u+x = = = dx 2x 2x 2
- 21. Continue… dv 1 + 3u u+x = dx 2 du 1 + 3u 1 + 3u − 2u 1 + u x = −u = = dx 2 2 2 du 1 + u 1 = = (1 + u )( ) dx 2x 2x du dx = u +1 2x du 1 dx ∫ u +1 = 2 ∫ x
- 22. Continue… 1 ln ( u + 1) = ln x + ln c 2 1 ln ( u + 1) = ln x + ln c 2 1 ln ( u + 1) = ln( x c) 2 1 u + 1 = x 2c
- 23. Continue… y 1 + 1 = x 2c x 3 y + x = x 2c
- 24. Example 6 dy xy Solve = 2 dx x + y 2 subject y(0)=2 Let y = xu , rewrite this equation become du x ( xu ) u+x = 2 dx x + ( xu )2 u = 1+ u2 du u x = 2 −u dx 1 + u u3 =− 1+ u2
- 25. Continue… 1+ u2 dx 3 du = − u x 1 1 dx 3 + du = − u u x 1 1 dx ∫ u 3 + u du = − ∫ x 1 − 2 + ln u = − ln x + c 2u
- 26. Continue… 1 ln u + ln x = c + 2 2u 1 ln xu = c + 2 2u y Substitute u= x x2 ln y = c + 2y2 x2 2 y2 y = Ae
- 27. Continue… y ( 0) = 2 0 2 = Ae x2 2 y2 y = 2e
- 28. Summary on solving Homogeneous equations 1. Identify whether the equation is homogeneous or not. dy du 2. Use substitution of y = xu and dx = u + x dx in the original equation. 3. Separate the variable x and u in 2. 4. Integral both sides with respect to the related variables and put only one constant, say A. y 5. Substitute back u = . x 6. If the equation has subject to any value, substitute it to get the constant value A.
- 29. Prepared By Annie ak Joseph Prepared By Annie ak Joseph Session 2008/2009