Solution of Nonhomogeneous D.E. using Variation of Parameters (continued)                                   

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2.

 

 

Use algebra to solve (1) and (2) namely

 

                        u1’(x) sin 3x  +  u2’(x) cos 3x  =   0                 - - - - -    (1)

 

                       3u1’(x) cos 3x  -  3u2’(x) sin 3x   =  2 sec  3x  - - - - - -    (2)

 

 

for   u1’(x)   and   u2’(x) .     The result is:

 

                             u1’(x)   =  2/3 ,            u2’(x)   =   -2/3  tan 3x

 

 

     Now  continuing            u1’(x)   =  2/3 ,            u2’(x)   =   -2/3  tan 3x

 

 

Integrate these expressions to find     u1(x)     and    u2(x) .  The result is:

 

 

  u1(x)    =  2x/3  +  B1      and      u2(x)    =   (2/9) ln (cos 3x)  +  B2

 

 

Therefore the particular solution

 

                           yp(x)   =   u1(x) y1(x)  +  u2(x) y2(x)     is

 

 

                 yp(x)  =   (2x/3  + B1 ) sin 3x   +  (2/9) ln (cos 3x  + B2  ) cos 3x

 

 

and the general solution for y(x) is the sum of the complementary and particular

solutions.  This is the final result (below).

 

 

 y(x)  =  C1  sin 3x   +  C2  cos 3x   +  (2x/3) sin 3x   +   (2/9) ln (cos 3x) cos 3x\

 

(here C1  and  C2  are constants of integration)

 

 

where we have absorbed constants of integration  B1    and  B2   into   C1    and    C2   

 


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