Me Metho thods ds of of Sol Solutio ution n of of fi first - PowerPoint PPT Presentation

Me Metho thods ds of of Sol Solutio ution n of of fi first rst or order der o ode de

Let M (x ,y) dx + N(x ,y) dy = 0 Exact differential equation   M N  Then   y x Let u = u(x ,y ) =c be a solution Then   u u 𝝐𝒗 𝝐𝒗   𝝐𝒚 𝒆𝒚 + 𝝐𝒛 𝒆𝒛 = 𝟏 M , N du=   x y Then   2 2     M N M u N u    ,         y y x x x y y x

(4) Exact Differential Equations:   This type takes the form: M x y dx ( , ) N x y dy ( , ) 0   M N  Such that:   y x  The solution function will take the form: f ( x , y ) c     f ( x , y ) M ( x , y ) dx h ( y ) c f y  N ( x , y )

Example     2   2   Solve the DE 2 y x 3 dx 2 yx 7 dy 0 Solution   M N    2   2  4 yx N 2 yx 7 M 2 y x 3   y x     2 ( , ) (2 3) ( ) f x y y x dx h y dh y ( )      2 2 2 f 2 yx f x y ( , ) y x 3 x h y ( ) y dy dh y ( )   h y ( ) 7 y 7 dy     2 2 f ( x , y ) y x 3 x 7 y c

Example     3  2   2   Solve the DE y y sin x x dx 3 xy 2 y cos x dy 0 Solution   M N    2 3 y 2 sin y x   y x   2   f x y ( , ) (3 xy 2 cos ) y x dy g x ( ) dg x ( )     3  2  3 2 f y y sin x f x y ( , ) xy y cos x g x ( ) x dx dg x ( )     2 g x ( ) 1/ 2 x x dx 1     3 2 2 f x y ( , ) xy y cos x x c 2

Solve the differential eqns  1  ln ydx xy dy =0 2 3   (3x sin y) dx (x cosy)dy 0 y y   e dx xe dy 0  x x 1   e ln ydx e y dy 0 2 2    2xydx (x 3y )dy 0 2    2xy dx (x cosy)dy 0 3 2 2    2xy dx (3x y cosy)dy 0

(5) DE ’ s Reducible to Exact:   M x y dx ( , ) N x y dy ( , ) 0 This type takes the form:   M N  Such that:   y x Sometimes there exist a suitable function called an integrating factor such that,     M x y dx ( , ) N x y dy ( , ) 0 which is exact DE.

 ( , ) x y The problem now is: how to obtain Case (I):     1  M N   A x dx ( )   ( ) x e   A x ( )     N y x Case (II):         B y dy ( ) 1 M N   ( ) y e   B y ( )     M y x

Example     2    2    Solve the DE 3 2 0 y xy y dx x xy x dy Solution       2 2 N x 3 xy 2 x M y xy y   N M       2 x 3 y 2 2 y x 1   x y          x y 1 1 M N 1        B y ( )         M y x y x y 1 y 1  dy      B y dy ( )   ln y  y ( ) y e e e y

          2 2 y xy y dx x 3 xy 2 x dy 0     3  2  2  2  2   y xy y dx yx 3 xy 2 xy dy 0   M N      3 2 2 f x y ( , ) ( y xy y ) dx h y ( ) 1  3  2 2  2  f x y ( , ) xy x y xy h y ( ) 2 dh y ( )  2  2   f 3 xy x y 2 xy y dy dh y ( )     0 ( ) h y c dy 1     3 2 2 2 f x y ( , ) xy x y xy c 2

Example     2  3   Solve the DE cos sin cos 1 0 x x dy y x dx Solution    2 N cos x sin x 3 M y cos x 1   M N  3  3  2 cos x cos 2cos sin x x x   y x     2 1 M N 2cos sin x x 2sin x       A x ( )   2   N y x cos x cos x sin x 2   1 sin x dx   ln  2  1   2ln cos x        2 cos x cos x ( ) x e e e sec x 2 cos x

       2 3 cos x sin x dy y cos x 1 dx 0        2 sin x dy y cos x sec x dx 0    f x y ( , ) (sin ) x dy g x ( ) dg x ( )     f x y ( , ) y sin x g x ( ) f y cos x x dx dg x ( )        2 sec ( ) tan x g x x c dx    f x y ( , ) y sin x tan x c

(6) Linear DE ’ s: The general first order linear DE takes the form: dy dx   p x y ( ) q x ( ) Solution steps:    p x dx ( ) ( ) x e   1     y x ( ) ( ) ( ) x q x dx c  ( ) x

Example dy dx   Solve the DE cos x y sin x 1 Solution dy sin x 1   y cos cos dx x x 1 sin x   q x ( ) p x ( ) cos x cos x   1 sin x dx   ln  1           lncos x cos x cos x ( ) x e e e sec x cos x   1 1  tan       y sec x x c y ( x ) sec x dx c   sec x cos x

(7) DE ’ s Reducible to Linear (Bernoulli DE): The general form of Bernoulli DE is: dy dx   ( ) n p x y ( ) q x y y   1 n u du dx     (1 n p x u ) ( ) (1 n q x ) ( )

Example dy dx   2 Solve the DE x 4 y x y Solution 1 4 dy       2 y x y 1 1 n         2 1/2 u ( x ) x x dx c dx x      2  x 2 q x   p x 1   2 1 du 2 1        u x ( ) x ln x c 2 u x le t u y   2 dx x 2 2    1 2   ln  dx 1        2ln   x x x ( ) x e e e 2 x

Solve    2 x ( y 1 ) dx ydy 0 1. 2. sin x. sin y dx + cos x cos y dy 2 x 2 x   ye dx ( 4 e ) dy 3. 4. x cos y dx + tan y dy = 0 2 3    tan y . y sin x 0 5. 2    y 1 y 6.

Solve