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146_pdfsam_math 54 differential equation solutions odd

146_pdfsam_math 54 differential equation solutions odd -...

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Chapter 3 y 2 = ( 1 + 4 h + 8 h 2 ) y 1 = ( 1 + 4 h + 8 h 2 ) 1 3 ( 1 + 4 h + 8 h 2 ) = 1 3 ( 1 + 4 h + 8 h 2 ) 2 , y 3 = ( 1 + 4 h + 8 h 2 ) y 2 = ( 1 + 4 h + 8 h 2 ) 1 3 ( 1 + 4 h + 8 h 2 ) 2 = 1 3 ( 1 + 4 h + 8 h 2 ) 3 . Continuing this way we see that y n = 1 3 ( 1 + 4 h + 8 h 2 ) n . (3.16) (This can be proved by induction using equation (3.15) above.) We are looking for an ap- proximation to our solution at the point x = 1 / 2. Therefore, we have h = 1 / 2 x 0 n = 1 / 2 0 n = 1 2 n n = 1 2 h . Substituting this value for n into equation (3.16) yields y n = 1 3 ( 1 + 4 h + 8 h 2 ) 1 / (2 h ) . 7. For this problem, f ( x, y ) = x y 2 . We need to approximate the solution on the interval [1 , 1 . 5] using a step size of h = 0 . 1. Thus the number of steps needed is N = 5. The inputs to the subroutine on page 129 are
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