Explain this step please Math / Advanced Math / Q&A Library / 3.5.2 Example B (1) It i.3.5.2 Example B (1) It is straightforward to...Step 21→ Uk = uj + Ap(k) + (k – 1) [k(2k – 1) – 6)],where φ (k) Σ18k-1 1i=1 _i→ uk = u1 + Ad (k) + (k – 1) [2k² – k – 6]→ Uk = u1 + A$(k) +(k – 1)(k – 2)(2k + 3)— ик 3D АФ (k) + k (k + 1) (2k — 5) + в, wherek(k+1)(2k5) + B,> Uk =uj = -B – , and B is a arbitrary constant.1So, general solution of (3.152) is(k – 1)u:-(x) (k – 1)+ (k – 1)k(k + 1) (2k – 5) + B(k – 1).

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Description: Explain this step please Math / Advanced Math / Q&A Library / 3.5.2 Example B (1) It i.3.5.2 Example B (1) It is straightforward to...Step 21→ Uk = uj + Ap(k) + (k – 1) [k(2k – 1) – 6)],where φ (k) Σ18k-1 1i=1 _i→ uk = u1 + Ad (k) + (k – 1) [2k² – k

Answered: Image /qna-images/answer/02da7c16-4e77-4dbb-8c6e-9248a9e1c1ba.jpg

Answered: → uk = u1 + A¢(k) + (k – 1) (k – 2)(2k…

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→ uk = u1 + A¢(k) + (k – 1) (k – 2)(2k + 3) → Uk = A¢(k) + where k(k+ 1) (2k – 5) + B, uj = -B – , and B is a arbitrary constant. |

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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