Finding Particular Solutions using Variation of Parameters  (continued)

 3 Finally substitute  yp(x), dyp(x)/dx ,  d2yp(x)/dx2   into d2y/dx2  +  b dy/dx  +  cy  =  f(x)   The result is a second equation to determine  du1/dx  and  du2/dx .          This second equation has the form:                        du1(x)/dx  dy1(x)/dx  +  du2(x)/dx dy2(x)/dx  =  f(x)              (equation 2)   where we used that  both    y1(x)  and    y2(x)      satisfy the homogeneous d.e.                               a d2y/dx2  +  b dy/dx  +  cy  =  0 4 The next step is to solve equations (1) and (2), using algebra, for  du1(x)/dx  and du2(x)/dx  .Note, you have two equations in two unknowns.   The unknowns are the derivatives of u1(x)  and  u2(x) .        They equations are as follows:          (repeated for you here)                                        du1(x)/dx  y1(x)         +  du2(x)/dx y2(x)  =  0               (equation 1)                             du1(x)/dx  dy1(x)/dx  +  du2(x)/dx dy2(x)/dx  =  f(x)    (equation 2)    Integration then gives  u1(x) and u2(x).  It turns out that the constants of integration end up being absorbed in the complementary solution.   Note:  This integration may not be simple.   Finally, the particular solution is      yp (x)  =   u1(x) y1(x)  +  u2(x) y2(x)