Example A The second-order differential equation dy dy 3 + 2y = 0 dx (4.114) dæ? has the general solution y(x) c1e* + c2e2", (4.115) where c1 and c2 are arbitrary constants. The difference equation associated with this differential equation is Yk+2 – 3yk+1 +2yk = 0. (4.116) Its general solution is Yk = A+ B2k, (4.117) LINEAR DIFFERENCE EQUATIONS 131 since the characteristic equation r2 – 3r+2 = 0 has roots ri = 1 and r2 = 2; A and B are arbitrary constants. We now show how the result given by equation (4.117) can be obtained from equation (4.115). Let us calculate Dky(x); it is D*y(x) dn (c1e + cze2") = cie" +c22*e2. (4.118) dæn Therefore, D*y(x)\a=0 : C1 + c22k. (4.119) Yk which is the same as equation (4.117) except for the labeling of the arbitrary constants.
Example A The second-order differential equation dy dy 3 + 2y = 0 dx (4.114) dæ? has the general solution y(x) c1e* + c2e2", (4.115) where c1 and c2 are arbitrary constants. The difference equation associated with this differential equation is Yk+2 – 3yk+1 +2yk = 0. (4.116) Its general solution is Yk = A+ B2k, (4.117) LINEAR DIFFERENCE EQUATIONS 131 since the characteristic equation r2 – 3r+2 = 0 has roots ri = 1 and r2 = 2; A and B are arbitrary constants. We now show how the result given by equation (4.117) can be obtained from equation (4.115). Let us calculate Dky(x); it is D*y(x) dn (c1e + cze2") = cie" +c22*e2. (4.118) dæn Therefore, D*y(x)\a=0 : C1 + c22k. (4.119) Yk which is the same as equation (4.117) except for the labeling of the arbitrary constants.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Explain the determine
Transcribed Image Text:Example A
The second-order differential equation
dy
3
+ 2y = 0
dx
(4.114)
dx2
has the general solution
y(x) = c1e" + c2e2",
(4.115)
where c1 and c2 are arbitrary constants. The difference equation associated
with this differential equation is
Yk+2
3yk+1 + 2yk
0.
(4.116)
Its general solution is
Yk
A + B2k,
(4.117)
LINEAR DIFFERENCE EQUATIONS
131
since the characteristic equation r2 – 3r+2 = 0 has roots ri = 1 and r2 = 2; A
and B are arbitrary constants. We now show how the result given by equation
(4.117) can be obtained from equation (4.115).
Let us calculate Dky(x); it is
D*y(x)
dn
(cie + c2e?") = c1e" + c22*e2¤.
(4.118)
dxn
Therefore,
Yk = D*y(x)|x=0 = C1 + c22*,
(4.119)
which is the same as equation (4.117) except for the labeling of the arbitrary
constants.
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