Variation of Parameters in Solving Linear Differential Equations

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Explore the method of variation of parameters to find particular solutions for linear differential equations, even with non-constant coefficients and general forms of the forcing function. Understand how to approach second-order linear DEs and associated homogeneous equations, with step-by-step processes outlined. Dive into the standard form, determinants, and solution assumptions to enhance your understanding of solving DEs.

  • Linear Equations
  • Differential Equations
  • Variation of Parameters
  • Second Order
  • Homogeneous Equations
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  1. 231 4-6 Variation of Parameters 4-6-1 The method can solve the particular solution for any linear DE (1) May not have constant coefficients (2) g(x) may not be of the special forms ( ) x y ( ) x ( ) x y ( ) x ( ) ( ) x y ( ) ( ) + + + + = ( ) n ( 1) n a a a x y x a g x 1 1 0 n n

  2. 232 4-6-2 Case of the 2ndorder linear DE ( ) x y x ( ) ( ) x y ( ) ( ) ( ) + + = a a x y x a g x 2 1 0 ( ) x y x ( ) ( ) x y ( ) ( ) + + = associated homogeneous equation: 0 n a a x y x a 1 0 Suppose that the solution of the associated homogeneous equation is + ( ) ( ) 1 1 c y x c y x 2 2 Then the particular solution is assumed as: = + ( ) ( ) u x y x ( ) ( ) p y u x y x 1 1 2 2 ( )

  3. 233 = + ( ) ( ) u x y x ( ) ( ) p y u x y x 1 1 2 2 2 = + + + p y 1 1 u y 1 1 u y 2 u y u y u y 2 2 2 1 1 2 2 2 = + + + + + 2 p y 1 1 u y 1 1 u y u y u y u y 2 2 2 ( ) (standard form) ( ) ( ) f x ( ) ( ) + + = y x P x y x Q x y ( ) ( ) ( ) ( ) ( ) ( ) a x a x a x a x g x a x ( ) ( ) ( ) f x = = = , , P x Q x 0 1 2 2 2 ( ) ( ) p p 2 2 + 2 + + = + + + + + 2 2 y P x y Q x y 1 1 u y P u y + 1 1 u y u y + 1 1 u y u y + u y + u y u y 2 2 2 p ( ) ( ) 1 1 2 2 + u y Q u y u y 1 1 2 2 1 1 2 2 zero zero P yu ( ) ( ) p p 1 1 2 2 + + = + + + + + + + y P x y Q x y u y Py Qy + u y Py Qy 1 1 yu 1 u y 1 2 2 p 2 + + + 2 2 2 2 y u u y y u 1 1 2 2 1 1 2

  4. 234 ( ) ( ) ( ), f x p p = + + + = p y 1 1 u y u y y P x y Q x y 2 2 p yu ( ) ( ) p p 1 1 u y 2 2 2 + + = + + + + + 2 2 y P x y Q x y y u u y P yu y u 1 1 2 2 1 1 2 p 1 1 u y 2 2 + + + 1 1 yu y u u y 2 2 d dx ( ) f x 2 2 1 1 yu 2 + + + + + = 1 1 yu y u P yu y u y u 2 1 1 2 2 + = 0 1 1 yu ( ) f x 2 2 y u 1 1 y u 2 + = y u 2 + + = = 0 1 1 1 yu yu 2 2 y u y u 0 y y y y u u = 1 2 ( ) f x ( ) f x 1 1 2 2 1 2 2

  5. 235 ( ) y f x W W W ( ) = 1 u u x dx 1 = = 2 u 1 + + = = 0 1 1 yu yu 2 2 y u y u 1 ( ) f x ( ) ( ) x dx = 2 y f x W W W 1 1 2 2 u u 2 = = 1 u 2 2 0 ( ) f x 0 ( ) f x y y y y y y y y 1 2 2 1 = = = where W W W 1 2 1 2 2 1 | |: determinant ( ) x ( ) ( ) ( ) x y ( ) x = + p y u x y x u 1 1 2 2 1storder case (page 63)

  6. 236 4-6-3 Process for the 2ndOrder Case Step 2-1 standard form y x ( ) ( ) ( ) f x ( ) ( ) + + = P x y x Q x y Step 2-2 0 ( ) f x 0 ( ) f x y y y y y y y y 1 2 2 1 = = = W W W 1 2 1 2 2 1 W W W W 1 = 2 = u u 1 2 Step 2-3 ( ) ( ) x dx = = Step 2-4 1 2 u u x dx u u 1 2 ( ) x ( ) ( ) ( ) x y ( ) x = + Step 2-5 p y u x y x u 1 1 2 2

  7. 237 4-6-4 Examples Example 1 (text page 162) + = + 2 x 4 4 ( 1) y y y x e + = 4 4 0: Step 1: solution of y y y = + 2 2 x x cy = ce c xe 1 2 + = = 2 2 x x 2, , y 1 1 u y u y y e y x e Step 2-2: 2 1 2 p 2 2 x x e xe 2 x 0 1) xe = = 4 x W e = = + 4 x ( 1) W x xe + 2 2 2 x x x 2 2 e xe e 1 + + 2 2 2 x x x ( 2 x e xe e 2 x 0 1) e = = + 4 x ( 1) W x e 2 + 2 2 x x 2 ( e x e W W W W 1 = 2 = Step 2-3: = + = 2 1 u x u x x 2 1

  8. 238 1 3 + + 1 2 c 1 Step 2-4: = = = + 2 3 2 ( ) u u dx x x dx x x c 1 1 1 2 2 = = + = 2 ( 1) u u dx x dx x x 2 2 1 3 1 2 1 2 1 6 1 2 Step 2-5: = + + = + 3 2 2 2 2 3 2 2 x x x ( ) ( ) ( ) y x x e x x xe x x e p 1 6 1 2 = + + + 2 2 3 2 2 x x x ( ) y ce c xe x x e Step 3: 1 2

  9. 239 + = 4 36 csc3 y y x Example 2 (text page 163) + = x c + = cos3 sin3 4 36 0: cy c x y y Step 1: solution of 1 2 + = = ( ) f x csc3 / 4 x 9 csc3 / 4 Step 2-1: standard form: y y x Step 2-2: 0 sin3 x cos3 3sin3 sin3 3cos3 x x = = 3 W = = 1/ 4 W 1csc3 4 x x 1 3cos3 x x cos3 0 x cos3 4 sin3 x x 1 = = W 1 4 2 sin3 csc3 x x W W W W cos3 x x 1 1 1 = 2 = = = u u 1 2 Step 2-3: 12 12 sin3 x u = 1ln sin3 36 Step 2-4: = u x 12 1 2 1 cos3 xdx x ( ) 12 sin3

  10. 240 x 1 = + cos3 sin3 ln sin3 x y x x Step 2-5: 12 36 p x 1 Step 3: = + = + + cos3 sin3 cos3 sin3 ln sin3 x y y y c x c x x x 12 36 1 2 c p Note: Interval (0, /6) (0, /3)

  11. 241 = 1/ y y x Example 3 (text page 164) = c e + x x cy ce = ( ) 1/ f x x 1 2 x x e e e = = 2 W x x e xedx Note: analytic x t edt t x ( page 49) x 0

  12. 242 4-6-5 Case of the Higher Order Linear DE ( ) ( ) ( ) 1 n n a x y x a x y + ( ) x ( ) ( ) x y ( ) ( ) + + + = ( ) n ( 1) n a x y x a g x 1 0 Solution of the associated homogeneous equation: = + + + + ( ) ( ) ( ) ( ) y 1 1 c y x c y x c y x c y x 2 2 3 3 c n n The particular solution is assumed as: = + + + + ( ) ( ) u x y x ( ) ( ) ( ) ( ) ( ) ( ) y u x y x u x y x u x y x 1 1 2 2 3 3 p n n W W = k ( ) ( ) k u x u x dx = ( ) u x k k

  13. 243 Process of the Higher Order Case Step 2-1 standard form ( ) ( ) x ( ) ( ) x ( ) ( ) x ( ) ( ) x a x a x a a a x g x a ( ) x ( ) x ( ) + + + + = ( ) n ( 1) n 1 1 0 n y y y x y a n n n n Step 2-2 Calculate W, W1, W2, ., Wn(see page 244) W W W W W W 1 = 2 = n = Step 2-3 u u u 1 2 n Step 2-4 . ( ) 1 1 u u x dx ( ) x dx ( ) x dx = = = 2 n u u u u 2 n ( ) x ( ) ( ) ( ) x y ( ) x ( ) x y ( ) x = + + + y u x y x u u Step 2-5 1 1 2 2 p n n

  14. 244 y y y y y y y y y y y y 1 2 3 n W W k = ( ) u x k 1 2 3 n = W 1 2 3 n ( 1) ( 2 1) ( 3 1) ( n 1) n n n n y y y y 1 0 0 y y y y y y y y y y + 1 2 1 1 k k n + 1 2 1 1 k k n = W k ( 2) ( 2 ( 2 ( ) f x 2) ( k 2) ( k 2) ( n 2) n n n n + n 0 ( ) f x y y y y y y y y y y 1 1 1 ( 1) 1) ( k 1) ( k 1) ( n 1) n n n n + n 1 1 1 ( ) ( ) x = / g x a n

  15. 245 0 0 ( ) ( ) x Wk: replace the kthcolumn of W by g x a ( ) f x = n 0 ( ) f x For example, when n = 3, y y y y y y y y y 1 2 3 = W 1 2 3 1 2 3 0 0 ( ) f x 0 0 ( ) f x 0 0 ( ) f x y y y y y y y y y y y y y y y y y y 2 3 1 3 1 2 = = = W W W 1 2 3 2 1 3 3 1 2 2 3 1 3 1 2

  16. 246 + = 4 sec2 y y x Exercise 30 = + x c + Complementary function: cos2 sin2 cy c c x 1 2 3 1 0 0 cos2 2sin2 4cos2 sin2 2cos2 4sin2 x x = = 8 W x x x x 0 0 cos2 2sin2 4cos2 sin2 2cos2 4sin2 x x = = 2sec2 W x x x x x 1 sec2 x 1 0 0 cos2 2sin2 4cos2 0 0 x 1 0 0 0 0 sin2 2cos2 4sin2 x = = = = 2tan2 W x x x 2 W x x 3 2 sec2 x sec2 x

  17. 247 1 W W sec2 tan2 4 W W W W x x 2 = = 1 = 3 = = = u 2 u u 3 1 4 4 1ln cos2 8 x 1ln sec2 8 = = u x u = + tan2 u x x 3 2 1 4 ( ) 1 ln sec2 8 = + + cos2 sin2 x y x c c x c x 1 2 3 1 8 ( ) + + + + tan2 cos2 ln cos2 sin2 x x x x x 4 for - /4 < x < /4 Note: - /4 , /4 are singular points

  18. 248 4-6-7 (1) associated homogeneous equation (2) (3) | | determinant (4) u1 (x) u2 (x) (5) f(x) = g(x)/an(x) ( 1storder standard form) (6) u1'(x) u2'(x) + c particular solution yp yp (7) an(x) = 0

  19. 249 4-7 Cauchy-Euler Equation 4-7-1 ( ) x ( ) x ( ) ( ) + + + + = ( ) n 1 ( 1) n n n a x y a x y a xy x a y g x 1 1 0 n n k not constant coefficients k ka x but the coefficients of y(k)(x) have the form of akis some constant associated homogeneous equation particular solution ( ) ( ) ( ) x + + ( ) n 1 ( 1) n n n a x y + x a x y 1 n n + = 0 a xy x a y 1 0

  20. 250 4-7-2 Associated homogeneous equation of the Cauchy-Euler equation ( ) ( ) 1 n n a x y x a x y x + ( ) + + + = ( ) n 1 ( 1) n n n 0 a xy x a y 1 0 Guess the solution as y(x) = xm, then ( ) m n m n + + n ( 1)( 2) 1 n a a x m m m x ( ) ) m n + m n + + + + 1 1 n ( 1)( 2) 2 x m m m x 1 n ( m n + 2 2 n ( 1)( 2) 3 a x m m m m n x 2 n + + 1 m a x m x = 1 m 0 a x 0

  21. 251 Delete xmon the previous page ( ) m n + ( 1)( 2) 1 a m m + + m n ( ( ) ) m n + + ( ( 1)( 1)( 2) 2) 2 3 a m m m 1 n auxiliary function a m m m m n 2 n : constant coefficient + + a m a 1 = 0 0 ! m ( ) ( ) k d dx = m k + 1 1 m m k x ( )! m k k

  22. 252 4-7-3 For the 2ndOrder Case ( ) x ( ) + + = 2 0 a x y a xy x a y 2 1 0 auxiliary function: a m m ( ) ( ) + + = 1 0 + + = a m a 2 0 a m a a m a 2 1 0 2 1 2 0 roots ( ) 2 + 4 a a a a a a ( ) 2 4 a a a a a a = 2 1 1 2 2 2 0 m = 2 1 1 2 2 2 0 m 1 a 2 a 2 2 [Case 1]: m1 m2and m1, m2are real two independent solution of the homogeneous part: m m and x x 1 2 = + m m cy c x c x 1 2 1 2

  23. 253 [Case 2]: m1= m2 Use the method of reduction of order m y x = 1 1 a 1 dx ( ) P x dx a x e e 2 ( ) x ( ) = = m y y x dx x dx 1 ( ) x 2 1 2 2 1 m y x 1 a a a ( ) x ( ) + + = ( ) 0, y y x y 0 1 Note 1: = P x 1 2 a x a x a x 2 2 2 a a = = m m 2 2 1 Note 2: 1 2 a 2

  24. 254 a a a a a 1 1 ln dx x 1 a a a x x x e e 2 2 ( ) x = 2 m = = = 2 2 1 m m m y x dx x dx x dx 1 1 1 1 2 a 2 2 2 m m m x x 1 1 1 2 a a a a 1 1 2 a a ( ) 1 = = = 1 m a m m 1 l n x x x dx x x dx x x 1 2 2 1 1 2 If y2(x) is a solution of a homogeneous DE then c y2(x) is also a solution of the homogeneous DE = m ln y x x If we constrain that x > 0, then 1 2 = + m m ln cy c x c x x 1 1 1 2

  25. 255 [Case 3]: m1 m2and m1, m2are the form of m j = + = m j 1 2 two independent solution of the homogeneous part: + j j and C x + x x x + = j j cy C x 1 2 j + + + = = = = ( )ln ln ln lnx j j x x j x ( ) e e e e ( ) + sin( ln ) cos( ln ) x x j x ( ) = cos( ln ) ) ln x sin( ln ) j x x ( ) x j x ( ) = + + ( ) [( )cos ( )sin ln ] cy x C C ( j C C x 1 2 1 2 = + [ cos x c ln sin ln ] cy x c x 1 2

  26. 256 Example 1 (text page 167) ( ) x ( ) 4 xy x = 2 2 0 x y y Example 2 (text page 168) ( ) x ( ) + + = 2 4 8 0 x y xy x y

  27. 257 Example 3 (text page 169) 1 2 ( ) 1 ( ) 1 y = = 1 y ( ) x + = 2 4 17 0 x y y

  28. 258 4-7-4 For the Higher Order Case Process: auxiliary function Step 1-1 roots n independent solutions Step 1-2 solution of the nthorder associated homogeneous equation Step 1-3

  29. 259 (1) auxiliary function m0 0 m x associated homogeneous equation (2) auxiliary function m0 k ln , , x x x x m m m m 2 1 k (ln ) , , (ln ) x x x 0 0 0 0 associated homogeneous equation

  30. 260 (3) auxiliary function + j j ( ) ( ) ( , cos ln sin x x x ) ln x associated homogeneous equation (4) auxiliary function + j j k ( ) ( ) ( ( ) sin ln (ln ) x x x ( ) ( ) ( ) 2 cos ln , cos ln ln , cos ln (ln ) , , x x x x x x x x 1 k cos ln (ln ) x x x ) ( ) 2 sin ln , sin ln ln , sin ln (ln ) , , x x x x x x x x 1 k associated homogeneous equation 2k

  31. 261 Example 4 (text page 169) ( ) x y ( ) x ( ) 8 xy x + + + = 3 2 5 7 0 x x y y auxiliary function ( )( ) ( ) + + + = 1 2 5 1 7 8 0 m m m m m m + + + + = 3 2 2 3 2 5 5 7 8 0 m m m m m m + + + = 3 2 2 4 8 0 m m m )( ) ( + + = 2 2 4 0 m m

  32. 262 4-7-5 Nonhomogeneous Case To solve the nonhomogeneous Cauchy-Euler equation: Method 1: (See Example 5) (1) Find the complementary function (general solutions of the associated homogeneous equation) from the rules on pages 252-255, 259-260. (2) Use the method in Sec. 4-6 (Variation of Parameters) to find the particular solution. (3) Solution = complementary function + particular solution Method 2: See Example 6 Set x = et, t = ln x

  33. 263 Example 5 (text page 169, illustration for method 1) ( ) 3 x y x xy x ( ) 3 + = 2 4 x 2 y x e Step 1 solution of the associated homogeneous equation auxiliary function ( ) 1 m m m = 1 + = 2 4 3 0 m m + = 3 3 0 m 1 m = 3 = c x c x + 3 2 cy 1 2 3 x x x = = Step 2-2 Particular solution 3 2 W x 2 1 3 0 x 3 0 x x = = 3 x = = 2 W x e 5 x 2 W x e 2 2 x 1 1 2 x e 2 2 x 2 3 x e W W W W 1 = 2 = Step 2-3 = = 2 x x u e u x e 2 1

  34. 264 1 = = + 2 x x x 2 2 u u dx x e xe e Step 2-4 1 2 = = x u u dx e 2 = + = 2 x x 2 2 p y 1 1 u y u y x e xe Step 2-5 2 2 Step 3 = c x c x + + 3 2 x x 2 2 y x e xe 1 2

  35. 265 Example 6 (text page 170, illustration for method 2) ( ) ( ) x y x xy x + = 2 ln y x Set x = et, t = ln x (Step 1) 1 x dt dy dx dt dy dx dt dy = = (chain rule) 2 1 x dt x dt = 1 d y dx d dx dx d y dy dt d dx dt dx d dy d dy = = = 2 2 2 1 x dt 1 x dt x 1 1 x dy dt d y dt dy dt = + 2 2 2 2 Therefore, the original equation is changed into 2 d dt d dt ( ) + = 2 ( ) y t ( ) y t y t t 2 (Step 2) This process can be simplified using the auxiliary function.

  36. 266 2 d dt d dt ( ) + = 2 ( ) y t ( ) y t y t t 2 (Step 3) = + + + t t ( ) y t 2 ce c te t 1 2 = c x c x + + + ( ) y x ln ln 2 ( t = ln x ) x x 1 2 (Step 4) Note 1: d dt k d y dx ( ) ( ) means D = + k 1 1 x D k D D y t t t t k (Step 2) (Step 2) Determine the auxiliary function, then replace m by Dt Note 2: : Cauchy-Euler equation auxiliary function

  37. 267 4-7-6 (1) Section 4-3 ex x x ln(x) auxiliary function mn (2) particular solution? Variation of Parameters ( ) m n + ( 1)( 2) 1 m m m (3) x = 0 (Why?)

  38. 268 Extra Problems: How do we solve ( ) ( ) + = 0 xy x y x (1) ( ) + = 2 ( 1) ( ) y x 0 x y x (2)

  39. 269 linear DE (1) numerical approach (Section 4-9-3) (2) using special function (Chap. 6) (3) Laplace transform and Fourier transform (Chaps. 7, 11, 14) (4) (table lookup)

  40. 270 (1) Section 4-7 DE (2) linear DE constant coefficient linear DE

  41. 271 Exercises for practicing Section 4-6 4, 5, 8, 13, 14, 17, 18, 21, 25, 28, 29, 34 Section 4-7 11, 17, 18, 20, 21, 24, 32, 35, 36, 37, 40, 42 Review 4 27, 28, 29, 30, 32, 42

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