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.(1) t is straightforward to show that y = k-1 is a solution to the homog quation

.(1) t is straightforward to show that y = k-1 is a solution to the homog quation

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter9: Systems Of Equations And Inequalities
Section9.7: The Inverse Of A Matrix
Problem 32E
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3.5.2 Example B (1) It is straightforward to show that y equation = k-1 is a solution to the homogeneous 2k + 1 k Yk = 0. - 1 (3.151) Yk+2 Yk+1+ We will now use this to determine the general solution to the inhomogeneous equation 2k + 1 k Yk+1+ Yk k 1 k(k + 1). Yk+2 (3.152) k With the identification k Yk Yk Uk, R: = k(k + 1), (3.153) k equation (3.120) becomes (k + 1)Auk+1 – kAuz = k(k + 1). (3.154) This equation has the solution Auk = A/k + /3(k² – 1), (3.155) where A is an arbitrary constant. If we define k-1 $(k) (3.156) IWI
Transcribed Image Text:3.5.2 Example B (1) It is straightforward to show that y equation = k-1 is a solution to the homogeneous 2k + 1 k Yk = 0. - 1 (3.151) Yk+2 Yk+1+ We will now use this to determine the general solution to the inhomogeneous equation 2k + 1 k Yk+1+ Yk k 1 k(k + 1). Yk+2 (3.152) k With the identification k Yk Yk Uk, R: = k(k + 1), (3.153) k equation (3.120) becomes (k + 1)Auk+1 – kAuz = k(k + 1). (3.154) This equation has the solution Auk = A/k + /3(k² – 1), (3.155) where A is an arbitrary constant. If we define k-1 $(k) (3.156) IWI
(1) (1) Y+2(uk+2 – Uk+1) – qkY" (Uk+1 – Uk) = Rk, (3.119) or (1) .(1) Yk+2Auk+1 – qkY'Auk R. (3.120) Lot r. Au: thoroforo
Transcribed Image Text:(1) (1) Y+2(uk+2 – Uk+1) – qkY" (Uk+1 – Uk) = Rk, (3.119) or (1) .(1) Yk+2Auk+1 – qkY'Auk R. (3.120) Lot r. Au: thoroforo
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