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dy

d^{N}y

 (2.1) 
 (2.2) 
 (2.3) 
 (2.4) 
 (2.5) 
 (2.6) 
B=B 
^
z

 (2.7) 
 (2.8) 
 (2.9) 
 (2.10) 
df

 (2.11) 
 (2.12) 
 (2.13) 
 (2.14) 
 (2.15) 

∆

 (2.17) 
 (2.18) 
 (2.19) 
 (2.20) 
 (2.21) 
 (2.22) 
 (2.23) 
 (2.24) 
 (2.25) 
 (2.26) 
 (2.27) 
d
 z=Az 
d
 z=λz 
 (2.28) 
 (2.29) 
 (2.30) 
d^{2}y
 = −1 
A y + B 
dy
 + C 
d^{2}y
 + D 
d^{3}y
 = E 
d^{2} y
 = 2  ⎛ ⎝ 
dy
 ⎞ ⎠ 
2  − y^{3} 
 (2.31) 
(a)  Integrate
term by term to find the solution for y to thirdorder in x.  
(b)  Suppose y_{1} = f_{0} x. Find y_{1}−y(x) to secondorder in x.  
(c)  Now consider y_{2} = f(y_{1},x) x, show that it is equal to f(y,x)x plus a term that is thirdorder in x.  
(d)  Hence find y_{2}−y to secondorder in x.  
(e)  Finally show that y_{3} = ^{1}/_{2} ( y_{1}+ y_{2}) is equal to y accurate to secondorder in x. 



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