Variation of Parameters

Date: 11/10/2000 at 12:06:02
From: Megan Hall
Subject: Diff. Eq. and Variation of Parameters

I know that 

     y" + 4y = sec(x) 

is a variation of parameters problem, but I can't seem to solve it. 
I'm not using the Cramer's Rule form, but the other standard form that 
involves solving for u and v as a system of equations. Help! Please 
show me how to solve this one. I think I'm getting lost in solving for 
u and v and integrating.

Date: 11/11/2000 at 11:08:49
From: Doctor Anthony
Subject: Re: Diff. Eq. and Variation of Parameters

The equation is
     
     (D^2 + 4)y = sec(x)

The complementary function (CF) is found from the auxiliary equation

     m^2 + 4 = 0    
           m = +-2i

The CF gives  

     y = a*cos(2x) + b*sin(2x) 

where a and b are arbitrary constants.

For the general solution we use the method of variation of parameters.

Since the two linearly independent solutions of the CF are known, we 
can find a particular solution as follows:

     y = u1*cos(2x) + u2*sin(2x)

where u1 and u2 are unknown functions of x which are to be determined.

Then differentiating

     y' = 2u1(-sin(2x)) + u1'.cos(2x) + 2u2(cos(2x)) + u2'.sin(2x)
        = -2u1.sin(2x) + 2u2.cos(2x) + u1'.cos(2x) + u2'.sin(2x)
     y" = etc.

Having then substituted into the original differential equation we end 
up with

     u1'*cos(2x) + u2'*sin(2x) + 0           = 0
     u1'*(-sin(2x)) + u2'*cos(2x) - sec(x)/2 = 0 

Solving for u1' and u2' by determinants we get

           u1'                  -u2'                     1
    -----------------  =  ------------------  =  -----------------
   |sin(2x)     0    |   | cos(2x)     0    |   | cos(2x)  sin(2x)|   
   |cos(2x) -sec(x)/2|   |-sin(2x) -sec(x)/2|   |-sin(2x)  cos(2x)|

           -sin(2x)sec(x)/2    -sin(x)
     u1' = ---------------- =  ------- = -sin(x) 
                   1              1

           +cos(2x)sec(x)/2       
     u2' = ---------------- = cos(x) - (1/2)sec(x)  
                  1                            

and integrating 

     u1 = cos(x)
     u2 = sin(x) - (1/2)ln[sec(x) + tan(x)] 

P.I. is 

     y = u1*cos(2x) + u2*sin(2x)
       = cos(x)cos(2x) + sin(2x)*[sin(x)-(1/2)ln[sec(x) + tan(x)]]

And so the general solution is

     y = a*cos(2x) + b*sin(2x) 
         + cos(x)cos(2x) + sin(2x)*[sin(x)-(1/2)ln[sec(x) + tan(x)]] 

- Doctor Anthony, The Math Forum
  http://mathforum.org/dr.math/