Answer to Quiz 11 Fri 19 Apr 2002

Consider y"(x) + x^2 y(x) = 2x, with y(0) = 1 and y(1) = -1.

1. With our first guess y'(0) = 1 we find 1.37986 at x = 1. Our second guess y'(0) = -1 yields -0.132521 at x = 1. What is your next guess for y'(0) in the shooting method?

      p(1) = 1.37986    p(-1) = -0.132521
   
              z - 1                 z + 1
      p(z) = ------- (-0.132521) + ------- 1.37986
             -1 - 1                 z + 1
   
              1                              1
           = --- ( 0.132521 + 1.37986 ) z + --- ( -0.132521 + 1.37986)
              2                              2
   
           = 0.756191 z + 0.623670
   
      p(z) = -1 = 0.756191 z + 0.623670
   
                -1 - 0.623670
           z = --------------- = -2.14717
                   0.756191
   
   

2. Suppose we modify the differential equation above into y"(x) + y^2 = 2x. Is the modified problem easier or harder than the original one? Justify your answer.

   The modified differential equation is nonlinear. Therefore, the linear interpolation will no longer yield the correct value for y'(0). Thus the modified problem is harder than the original one.