So Solution of tion of Non Non-Ho Homogeno mogenous us Line - PowerPoint PPT Presentation

So Solution of tion of Non Non-Ho Homogeno mogenous us Line inear ar DE DE wit with h Co Consta nstant nt Co Coef efficien icients ts

Consider the DE:      n n 1       ... ( ) a y a y a y a y f x  n n 1 1 0   y y y G c p y The complementary (or homogenous) solution c y p The particular solution y The general solution G

y We have two methods for obtaining p The first is called: undetermined coefficient method The second called: variation of parameter method

1- The Undetermined Coefficient Method: The form of the particular solution depends on the form of the inhomogeneous function f(x). This form contains arbitrary constant which can be computed by substituting in the given DE and we summarize this in the following table.

Notes:

Example Find the general solution of the DE:        2  y y y y x x  2   y Ax Bx C Solution p        y 2 Ax B , y 2 , A y 0 p p p  3   2     1 0                 2 2 0 2 A 2 Ax B Ax Bx C x x      1, i 1 2,3    2 coeff . of x A 1   x    y c e coeff . of x 2 A B 1 c 1      0 coeff . of x B 2 A C 0 c cos x c sin x 2 3       1, 3, 1 A B C     2 3 1 y x x p

Example Find the general solution of the DE:      2 12 6 y y x x   Solution  2 2   y x Ax Bx C p         3 2   2    2  0 24 Ax 6 B 12 Ax 6 Bx 2 C 12 x 6 x 2    . 12 12 coeff of x A     1, 0    1 2,3 coeff . of x 24 A 6 B 6    0    x coeff . of x 6 B 2 C 0 y c c x c e c 1 2 3       A 1, B 5, C 15     4 3 2 y x 5 x 15 x p

Example Find the general solution of the DE: e       5 x y 10 y 25 y 4 Solution   e   2 5 x y Ax      2 p 10 25 0   e     2 5 x y A 2 x 5 x      p 5   1 2 e      2 5 x y A 2 20 x 25 x p  c x e     5 x A  y c 2 c 1 2   e   2 5 x y 2 x p

Example Find the general solution of the DE:      y 3 y 2 y x sin x Solution         y A Bx cos x C Dx sin x p      2 3 2 0            y A D Cx cos x C B Ax sin x p       1, 2 1 2              y 2 C B Ax cos x 2 A D Cx sin x   p   x 2 x y c e c e c 1 2  3 17 1 3     A , B , C , D 10 50 10 25      3 17 1 3         y x cos x x sin x p     10 50 10 25

2- The Variation of Parameter Method: The method of variation of parameter is more general than the undetermined coefficient method. Steps of Solution: (1) Find the fundamental set of solution of the homogeneous D.E:     y c y c y ... c y c 1 1 2 2 n n (2) The particular solution will take the form:     y u ( x ) y ( x ) u ( x ) y ( x ) ... u ( x ) y ( x ) p 1 1 2 2 n n

  u x (3) Evaluate the value of by the form: i ... y y y 1 2 n    y y ... y   1 2 n  W y y , , y ,..., y 1 2 3 n ... ... ... ...          n 1 n 1 n 1 y y ... y n 1 2 i th column y y ... 0 y 1 2 n    ... 0 y y y 1 2 n    W y y , , y ,..., y i 1 2 3 n ... ... ... 0 ...            n 1 n 1 n 1 ... y y f x y n 1 2 W x ( )    i u x ( ) , i 1,2,3,..., n i ( ) W x

Example Find the general solution of the DE: x e      y 2 y y x Solution      y 2 y y 0 𝝁 2 − 𝟑𝝁 + 𝟐 = 0 (𝝁 − 𝟐) 𝟑 = 𝟏 𝐳 = (𝒅 𝟐 + 𝒅 𝟑 𝒚)𝑓 𝑦 𝝁 =1,1

  x x y u e u xe 1 2 p x x e xe   2 x W e   x  x e x 1 e x x e 0 0 xe 2 x e      2 x W W e x x 1 2 e e    x x x x 1 e e x x  2 x e W       1 dx x u dx 1 2 x W e 2 x e  W   x   2 dx ln x u dx 2 2 x W e

Example Find the general solution of the DE: 1   y 9 y 4sin3 x Solution   y 9 y 0    2 9 0      3 , i 3 i 1 2   y c cos3 x c sin3 x 1 2 c

  y u cos3 x u sin3 x p 1 2 cos3 x sin3 x   3 W  3sin3 x 3cos3 x cos3 x 0 0 sin3 x cos3 x 1      W W 1 1 2  1 3sin3 4sin3 x x 3cos3 x 4 4sin3 x 4sin3 x 1 1 W        1 dx x u dx 1 12 12 W W 1 cos3 xdx 1 ln sin3      2 u dx x 2 W 12 sin3 x 36