Documents

2.First Order and First Degree

Description
ode
Categories
Published
of 13
All materials on our website are shared by users. If you have any questions about copyright issues, please report us to resolve them. We are always happy to assist you.
Related Documents
Share
Transcript
   O. D. E. Chapter2_4 Chapter2 Ordinary Differential Equations of the First Order and First Degree   General form: 1.  M  (  x ,  y ) dx   +    N  (  x ,  y ) dy      0 (2.1 a ) 2.  y        f  (  x ,  y ) (2.1 b ) I. Separable Differential equations  Form:  M  1 (  x )  N  1 (  y ) dx   +    M  2 (  x )  N  2 (  y ) dy      0     dy y N  y N dx x M  x M  )()()()( 1221    C   Ex1 9  yy    +   4  x      0 Solution: 9  ydy   +   4  xdx      0   9  y 2  +   4  x 2     C   Ex2  y       1   +    y 2  Solution: 2 1  ydy     dx     tan  1  y       x   +   C        y      tan(  x   +   C  ) Ex3  y    +   5  x 4  y 2     0,  y (0)      1 Solution:   C  x ydx x ydy   542 15    y (0)      1     1      C      The solution is  y 1     x 5  +   1 Ex4  y        2  xy ,  y (0)      1 Solution:  ydy     2  xdx     ln  y        x 2  +   C         y      C 2   x e       y      C  e  x   2    y (0)      1   1      Ce 0      C       1   The solution is  y      e  x   2     O. D. E. Chapter2_5 II. Reducible to separable differential equations  1. Homogeneous equation Definition: If  f  ( tx , ty )      t  r   f  (  x ,  y ), then  f  (  x ,  y ) is a homogeneous function of degree r  . Substituting t       1/  x  into  f  ( tx , ty )      t  r   f  (  x ,  y ), we have )(),1(),(),( 1),1(  x y F  x x y f   x y x f   y x f    x x y f    r r r    . In particular r       0, then  f  (  x ,  y )       F  (  x y ) If  M  (  x ,  y ) dx   +    N  (  x ,  y ) dy      0, when  M  (  x ,  y ) and  N  (  x ,  y ) are of the same degree in  x  and  y ,  M  /  N   is a homogeneous function of degree zero, and the differential equation can be written as )(  x y F  N  M dxdy  . (2.2) Let  y      ux , equation (2.2) becomes uu F du xdxu F dxdu xu  )()(  is separable. Ex5 2  xyy        y 2  +    x 2     0 Solution: Let  y      ux , the equation becomes 2  xux ( u   +   u   x )      u 2  x 2  +    x 2     0   2 u ( u   +    xu  )      u 2  +   1      0   2  xuu    +   u 2  +   1      0 21  2 uduudx x        ln(1   +   u 2 )       ln  x   +   C       1   +   u 2     C  /  x  1+  xC  x y       2       x 2  +    y 2     Cx  Ex6 (  x   +    y cos  y x ) dx       x cos  y x dy      0 Solution: Let  y      ux     (  x   +   ux cos u ) dx       x cos u ( udx   +    xdu )      0   cos udu     dx x  sin u      ln  x   +   C      sin  y x    ln  x      C      O. D. E. Chapter2_6 2.  M  (  x ,  y ) and  N  (  x ,  y ) are linear in  x  and  y  Form: ( a 1  x   +   b 1  y   +   c 1 ) dx   +   ( a 2  x   +   b 2  y   +   c 2 ) dy      0 (2.3) (1) If aabb 1212  , let  x       X    +    ,  y      Y    +    , equation (2.3) becomes [( a 1  X    +   b 1 Y  )   +   ( a 1    +   b 1    +   c 1 )] dX    +   [( a 2  X    +   b 2 Y  )   +   ( a 2    +   b 2    +   c 2 )] dY       0. (2.4) We choose   00 222111 cbacba  Then equation (2.4) reduces to ( a 1  X    +   b 1 Y  )   dX    +   ( a 2  X    +   b 2 Y  )   dY       0 is a homogeneous equation. (2) If aabbk  cc 121212    , equation (2.3) becomes [ k    ( a 2    x   +   b 2    y )   +   c 1 ]   dx   +   ( a 2    x   +   b 2    y   +   c 2 )   dy      0. (2.5) Let v      a 2  x   +   b 2  y      dy     dv a dxb   22 , equation (2.5) becomes ( kv   +   c 1 ) dx   +   ( v   +   c 2 )  dv a dxb   22    0 0)]([ 222221    dvbcvdxcvbackv  is a separable one. (3) If aabbcc 121212       k  , equation (2.3) becomes k  ( a 2  x   +   b 2  y   +   c 2 ) dx   +   ( a 2  x   +   b 2  y   +   c 2 ) dy      0. (2.6) a. If a 2  x   +   b 2  y   +   c 2     0, we get only a trivial solution.  b. If a 2  x   +   b 2  y   +   c 2     0, equation (2.6) reduces to kdx   +   dy      0     y       kx   +   C  .   O. D. E. Chapter2_7 Ex7 (4  x   +   3  y   +   1) dx   +   (  x   +    y   +   1) dy    0 Solution:   4 3 1 01 0  x y x y            x      2,  y       3 Let  X        x      2, Y      y   +   3, we have (4  X    +   3 Y  ) dX    +   (  X    +   Y  ) dY       0 Let Y       uX      (4  X    +   3 uX  ) dX    +   (  X    +   uX  )( udX    +    Xdu )      0 C  y x x y xC uu X C uu X duuu X dX duuu X dX duu X dX uu    122)12ln( 21)]2(ln[ 21)2ln(ln0 )2(1210)2(10)1()44( 222  Ex8 (2  x      4  y   +   5)  y    +    x      2  y   +   3      0 C  y x y xC  y x y x x C uu xC u u xC duudxduudxduuudxduudxu uuuuuu uuuu y yu y xu        )1184ln(848]11)2(4ln[)2(48 8)114ln(48)114ln( 812)1142/121(0)1142/121(0114520)52()114( 0)52(114062)1)(52( 0321)52( becomesequationThe 21212Let:Solution 1111  Ex9 (  x       y      1) dx   +   (2  x      2  y      2) dy      0 Solution: Dividing by  x       y      1    dx   +   2 dy      0     x   +   2  y      C   
Search
Tags
Related Search
We Need Your Support
Thank you for visiting our website and your interest in our free products and services. We are nonprofit website to share and download documents. To the running of this website, we need your help to support us.

Thanks to everyone for your continued support.

No, Thanks