Solution of Nonhomogeneous D.E. using Variation of Parameters                                     

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1.

 

Find a particular solution for the d.e.         y’’   +   9 y   =   2  sec 3x

 

Start with the complementary solution of the homogeneous d.e.

 

                                    y’’  +  9 y   =   0      ( Assume  y(x)  =  erx )

 

The characteristic equation is:  r2  +  9  =  0

 

So      yc   =  C1  sin 3x   +  C2  cos 3x       is the complementary solution of the d.e.

 

 

Therefore,  y1(x) = sin 3x  and  y2(x) = cos 3x    and recall

 

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

 

                     yp (x)  =  u1 (x) sin 3x   +  u2 (x) cos 3x

 

For the particular solution where u1 (x) and  u2 (x)  are yet to be determined.

 

Next differentiate  yp (x)  which gives

 

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

 

Key Step:

 

 With the method of parameters we eliminate the possibility of second derivatives

 

of  u1(x)  and u2(x) by setting the following:  

 

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

 

So the derivative of  yp  becomes       yp’   =   3u1(x) cos 3x  -  3u2(x) sin 3x

 

Next calculate the second derivative of  yp  as follows:

 

                  yp’’   =   3u1’(x) cos 3x  -  3u2’(x) sin 3x  -  9u1(x) sin 3x  -  9u2(x) cos 3x

 

and           9 yp (x)  =  9 u1 (x) sin 3x   +  9 u2 (x) cos 3x

 

Substitute  yp’’    and  9 yp    into the original d.e. to obtain:

 

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

 

Click here to continue with this example.

 



Copyright © 2011 Richard C. Coddington

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