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Question 2 Question 1 Find the solution. Find the solution. 6y'-2y = xy4 dy x-+2y = –, y (2) = 1 dx sin x A - (x- 1) + се "Х A x = (2 + cot2 - cosx) /y2 B y -3 - (х-1) + се "Х B су %3D (3 — сos 2x tanx) /x2 (c) y-3= - - (х— 1) — се* C) y = (4 + cos 2 cosx) /x? COSX 1 y = (4 + sin 2 + cosx?)/x2 D y³= - (х+ 1) + се "Х

Question 2 Question 1 Find the solution. Find the solution. 6y'-2y = xy4 dy x-+2y = –, y (2) = 1 dx sin x A - (x- 1) + се "Х A x = (2 + cot2 - cosx) /y2 B y -3 - (х-1) + се "Х B су %3D (3 — сos 2x tanx) /x2 (c) y-3= - - (х— 1) — се* C) y = (4 + cos 2 cosx) /x? COSX 1 y = (4 + sin 2 + cosx?)/x2 D y³= - (х+ 1) + се "Х

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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Question 2 Question 1 Find the solution. Find the solution. 6y'-2y = xy4 dy x-+2y = –, y (2) = 1 dx sin x A - (x- 1) + се "Х A x = (2 + cot2 - cosx) /y2 B y -3 - (х-1) + се "Х B су %3D (3 — сos 2x tanx) /x2 (c) y-3= - - (х— 1) — се* C) y = (4 + cos 2 cosx) /x? COSX 1 y = (4 + sin 2 + cosx?)/x2 D y³= - (х+ 1) + се "Х
Transcribed Image Text:Question 2 Question 1 Find the solution. Find the solution. 6y'-2y = xy4 dy x-+2y = –, y (2) = 1 dx sin x A - (x- 1) + се "Х A x = (2 + cot2 - cosx) /y2 B y -3 - (х-1) + се "Х B су %3D (3 — сos 2x tanx) /x2 (c) y-3= - - (х— 1) — се* C) y = (4 + cos 2 cosx) /x? COSX 1 y = (4 + sin 2 + cosx?)/x2 D y³= - (х+ 1) + се "Х
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