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or De+1 Ck+1 Ck = a = arbitrary constant. De (5.82) These latter equations have the solutions Ck = Aa and De = Ba, (5.83)

or De+1 Ck+1 Ck = a = arbitrary constant. De (5.82) These latter equations have the solutions Ck = Aa and De = Ba, (5.83)

Advanced Engineering Mathematics
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 blue

5.3.1 Example A The substitution of equation (5.65) into the equation 2(k + 1, l) = z(k, l+ 1) (5.76) gives Ak+1u = x*u'+1 or X= µ. (5.77) Therefore, a special solution of equation (5.76) is z(k, l) = µk+e. (5.78) Summing solutions of this form gives k+l 2(k, l) = (5.79) where C(u) is chosen so that the integral is defined. From equation (5.79) we conclude that a general solution of equation (5.76) is z(k, l) = f(k+ l), (5.80) where f is an arbitrary function of k+l. This result has already been obtained by the method of Section 5.2. Using the separation-of-variables method, where z(k, l) rewrite equation (5.76) as Ck De, we Ck+1De = CkDe+1, (5.81) or Ck+1 De+1 arbitrary constant. (5.82) Ck De These latter equations have the solutions Ck = Aak and De = Ba, (5.83) where A and B are arbitrary constants. If we let a = µ, then we obtain the same situation as in the above calculation using Lagrange's method. Thus, the solution is given by equation (5.80).
Transcribed Image Text:5.3.1 Example A The substitution of equation (5.65) into the equation 2(k + 1, l) = z(k, l+ 1) (5.76) gives Ak+1u = x*u'+1 or X= µ. (5.77) Therefore, a special solution of equation (5.76) is z(k, l) = µk+e. (5.78) Summing solutions of this form gives k+l 2(k, l) = (5.79) where C(u) is chosen so that the integral is defined. From equation (5.79) we conclude that a general solution of equation (5.76) is z(k, l) = f(k+ l), (5.80) where f is an arbitrary function of k+l. This result has already been obtained by the method of Section 5.2. Using the separation-of-variables method, where z(k, l) rewrite equation (5.76) as Ck De, we Ck+1De = CkDe+1, (5.81) or Ck+1 De+1 arbitrary constant. (5.82) Ck De These latter equations have the solutions Ck = Aak and De = Ba, (5.83) where A and B are arbitrary constants. If we let a = µ, then we obtain the same situation as in the above calculation using Lagrange's method. Thus, the solution is given by equation (5.80).
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