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dt=(tfin-tin)/(N-1);
u=sin(t);
a=(1/(2*pi))*ones(1,N);
b=ones(1,N);
y(1)=0;
D(1)=(1/a(1))*(u(1)-b(1)*y(1));

for k=2:N
y(k)=((2*a(k)/dt+b(k))^(-1))*...
(2*a(k)*y(k-1)/dt+a(k)D(k-1)+u(k));
D(k)=(2/dt)*(y(k)-y(k-1))-D(k-1);
end

plot(t,y,t,u,'--')

In-Class Exercise

Pb. 4.37 Plot the amplitude of y, and its dephasing from u, as a function of
a for large t.

Example 4.9
Find the first-order iterative scheme to solve the second-order differential
equation given by:

2
dy dy
at() 2 + bt() + ct y() = u t() (4.34)
dt dt
dy
with initial conditions yt( = 0 ) and given.
dt t=0
Solution: Substituting the above first-order expression of the iterators for the
first-order and second-order numerical differentiators [respectively Eqs.
(4.16) and (4.30), into Eq. (4.34)], we deduce the following iterative equation
for y(k):

 ak () bk ()  −1
yk () =  4 + 2 + ck ()  ×
 (∆ t) 2 t ∆ 
(4.35)
  ak () bk ()  ak ()  
 yk ( − 14 2 + 2  + Dk ( − 1 4 + bk() + a k D k() 2 ( − 1 ) + u k() 
)
)



  (∆ t) ∆ t   t ∆  
© 2001 by CRC Press LLC