Week 4 [compatibility mode]

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Week 4 [compatibility mode]

  1. 1. KNF1023 Engineering Mathematics II First Order ODEs Prepared By Annie ak JosephPrepared ByAnnie ak Joseph Session 2007/2008
  2. 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. 3. Integrating Factor The idea of the method in this section is quite simple. If given an equation P(x, y)dx + Q(x, y)dy = 0 −−− (1) that is not exact, but if multiply it by a suitable function F ( x, y) , the new equation FPdx + FQdy = 0 − − − (2) is exact, so that it can be solved, the function F ( x, y ) is then called an integrating factor of equation (1).
  4. 4. How to Find Integrating Factors Equation (2) with M=FP, N=FQ is exact by the definition of an integrating factor. Hence ∂M ∂N ∂ ∂ = now is (FP) = (FQ) −−− (3) ∂y ∂x ∂y ∂x That is F y P + FPy = Fx Q + FQ x (subscripts denoting partial derivatives) which is complicated and useless. So we follow the Golden Rule: If you cannot solve your problem, try to solve a simpler one—the result may be useful (and may also help you later on). Hence we look for an integrating factor depending only on one variable;
  5. 5. How to Find Integrating Factors fortunately, in many practical cases, there are such factors, as we shall see. Thus, let F = F (x). Then F y = 0 and F x = F = dF so that (3) becomes dx FP = F Q + FQ y x Dividing by F and reshuffling terms, we have FPy = F Q + FQ x F Q FQ x Py = + F F
  6. 6. How to Find Integrating Factors 1 dF Py = Q + Qx F dx 1 dF Py − Q x = Q F dx 1 1 dF ( Py − Q x ) = Q F dx 1 dF 1  ∂P ∂Q  =  −   ∂y ∂x  F dx Q  
  7. 7. How to Find Integrating Factors Thus, we can write it as 1 dF 1  ∂P ∂Q  =R R =  −  −−− (4) F dx Q  ∂y ∂x  This prove the following theorem.
  8. 8. How to Find Integrating Factors dF* If we assume F = F ( y ). Then Fx = 0 and F y = F = so that (3) becomes dy F P + FPy = FQ xDividing by F and reshuffling terms, we have
  9. 9. How to Find Integrating Factors F P FPy Qx = + F F 1 dF Qx = P + Py F dy 1 dF Qx − Py = P − − − (4*) F dy 1 1 dF P ( Qx − Py ) = F dy 1 dF 1 = R, R = ( Qx − Py ) F dy P
  10. 10. Theorem 1 (Integrating factor F(x)) If (1) is such that the right side of (4), call it R depends only on x, then (1) has an integrating factor F=F(x), which is obtained by integrating (4) and taking exponents on both sides, F ( x ) = exp ∫ R ( x ) dx − − − ( 5 ) Similarly, if F=F(y), then instead of (4) we get 1 dF 1  ∂Q ∂P  =  −  − − − (6 ) F dy P  ∂x ∂y  Here 1  ∂Q ∂ P  And have the R =   ∂ x − ∂ y  companion  p  
  11. 11. THEOREM 2 [Integrating factor F(y)] If (1) is such that the right side R of (6) depends only on y, then (1) has an integrating factor F=F(y), which is obtained from (4*) in the form F ( y) = exp ∫ R( y)dy − − − ( 7 )
  12. 12. Example 1 (Integrating factor F(x)) Solve 2 sin( y 2 )dx + xy cos( y 2 )dy = 0 by Theorem 1. We have P = 2 sin( y 2 ), Q = xy cos( y 2 ) hence (4) on the right, 1 3 R= 2 xy cos( y ) [ 2 2 4 y cos( y ) − y cos( y ) = x ] And thus F ( x) = exp ∫ (3 / x )dx = x 3
  13. 13. Example 2 Solve the initial value problem 2 xydx + (4 y + 3 x 2 )dy = 0, y (0.2) = −1.5 2Here P = 2xy , Q = 4 y + 3 x , the equation is not exact, the right side of (4) depends on both x and y (verify!), but the right side of (6) is 1 2 R= (6 x − 2 x ) = 2 xy y ThusF ( y ) = y 2
  14. 14. Continue... Is an integrating factor by (7). Multiplication 2 by y gives the exact equation 3 3 2 2 2xy dx + (4 y + 3x y )dy = 0 Which we can write as 3 M = 2 xy 3 2 2 N = 4y + 3x y
  15. 15. Continue... As we know that u ( x, y ) = c So to get u (x, y ) = c , we use u = ∫ Mdx + k ( y ) u = ∫ 2 xy dx + k ( y ) 3 u = x y + k(y) 2 3
  16. 16. Continue... ( ) To get k y , we differential u with respect to y ,from there we get ∂u 2 2 dk 3 2 2 = 3x y + = N = 4 y + 3x y ∂y dy dk so = 4y3 k = y4 + c * dy 2 3 4 * u=x y + y +c =c 2 3 4 u=x y +y =c
  17. 17. Continue... x = 0.2, y = −1.5 4 2 3 y + x y = 4.9275
  18. 18. 1st Order ODEs In Linear Form A first order differential equation is linear if it has the form dy + f ( x) y = r ( x) dx If the right side r(x) is zero for all x in the interval in which we consider the equation (written r(x)≡0), the equation is said to be homogeneous other it is said to be nonhomogeous.
  19. 19. Homogeneous 1st Order ODEs In Linear Form Linear ODE is said to be homogeneous if the function r(x) is given by r(x)=0 for all x. That is, a homogeneous 1st order ODE is given by dy + f ( x) y = 0 dx dy = − f ( x) y dx 1 dy = − f ( x)dx y
  20. 20. Homogeneous 1st Order ODEs In Linear Form 1 ∫ y dy = −∫ f ( x)dxGives us the general solution of the homogeneous 1st orderODE above. ln( y ) = − ∫ f ( x)dx y = exp(− ∫ f ( x)dx)
  21. 21. Inhomogeneous 1st Order ODEs in LinearForm (Method 1:use of integrating factor) dy1. + f ( x ) y = r ( x ) is a general form of the dx linear DE.2. Here f and r are function of x or constants3. ∫ fdx e ∫ fdx4. Integrating factor =I.f=5. Solution is y (I . f ) = ∫ r (I . f )dx + C
  22. 22. Example dy 2x + 5y = e dx dy The above DE is of the form + f (x ) y = r (x ) dx 2x f = 5, r = e I. f = e ∫ fdx =e 5x y (I . f ) = ∫ r (I . f )dx + C
  23. 23. Continue... ( )= ∫ e ye 5x 2x 5x e dx + C ye 5x = ∫ e dx + C 7x 7x e 5x ye = +C 7 1 2x −5 x y = e + Ce 7
  24. 24. Method 2: Variation Parameter In the 1st step, we solve the corresponding homogeneous ODE, i.e dy + f ( x) y = 0 dx Let us say that we obtain y = y h (x) as particular solution for the above homogeneous ODE. We will use it in the 2nd step below to construct a general solution for the original inhomogeneous ODE.
  25. 25. Method 2: Variation Parameter In the 2nd step, for the general solution of the () inhomogeneous ODE, we let y ( x) = y h x . v(x) and substitute it into ODE to obtain a 1st order separable ODE in v(x).
  26. 26. Example dy 2x + 5y = e dx dy + 5y = 0 dx dy = −5 y dx dy ∫ y = −5∫ dx
  27. 27. Continue... ln y = −5 x −5 x y=e y=e −5 x .v ( x ) dy = e .v − 5e .v ( x ) −5 x −5 x dx
  28. 28. Continue… e −5 x .v − 5e −5 x .v ( x ) + 5.e −5 x .v = e 2x −5 x 2x e v =e dv −5 x 2x e =e dx 2x dv e = −5 x dx e
  29. 29. Continue… dv 7x =e dx 7x e v= +c 7 7x −5 x e y=e ( + c) 7 2x e −5 x y= + ce 7
  30. 30. Prepared By Annie ak JosephPrepared ByAnnie ak Joseph Session 2007/2008

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