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Basic Mathematics-II




                    Notes          Now  we  consider  the  following  cases  to  reduce  the  given  differential  equation  to  the
                                   homogeneous form.

                                   Case I

                                         a  b
                                   When     .
                                        A   B
                                   Putting x = X + h, y = Y + k; (h & k are constants)
                                   so that dx = dX, dy = dY.
                                   Equation (1) becomes:

                                                  dY    a X h   b  Y  k c
                                                          
                                                    
                                                          
                                                  dX  A X h   B  Y  k c
                                                                 
                                                  dY   aX   bY   ah bk c 
                                                                    
                                   or                                                                    …..(2)
                                                  dX  AX   BY   Ah Bk c 
                                                                  
                                                                     
                                   Choosing h & k such that (2) becomes homogeneous.
                                   Thus ah + bk + c = 0 and Ah + Bk + c = 0
                                           h       k       1
                                   so that            
                                         bC   Bc  cA  Ca  aB   Ab
                                             bC   Bc  cA  Ca
                                         h       ,k  
                                             aB  Ab   aB   Ab

                                             a    b  i . .,aB  bA   0 
                                                  e
                                                            
                                           A   B            
                                    Equation (2) becomes
                                          dY   aX   bY
                                             
                                          dX   AX   BY
                                   which is homogeneous in X & Y and can be solved by putting Y = vX.

                                   Case II

                                         a  b
                                   When     , i.e., aB – bA = 0.
                                        A   B
                                   The case (1) fails.
                                        a  b   1
                                   Now          (say)
                                        A  B  M
                                   so that  A = ma, B = mb.

                                                  
                                          dy   ax by   c
                                                               
                                                           f ax by 
                                                   
                                          dx  m ax by   c
                                    Equation (1) reduces to
                                   which can be solved by putting ax + by = t.



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