QUESTION 5The solution of associated homogeneous equation ofa =7a +6a ,+n²+2" with a,=9, a,=10 and a,=32 is0(-3)"+c,(-2)*+c,(-1)"0,3"+c,(-2)" +c, 1"+C.o,3"+c,(-2)" +c,(-1)*,3" +C, 2" c,(-1)"1.5 pc

Answered: Image /qna-images/answer/f05ae337-c588-4eaa-ae97-cd172cf83572.jpg

QUESTION 5 The solution of associated homogeneous equation of a =7a+ +6 a+n2+2" with a,=9, a,=10 and a,=32 is o(-3)"+c,(-2)"+c,(-1)* o,3"+c,(-2)" +c, 1" +C. o,3" +c,(-2)" +c,(-1)* ,3" +c, 2" c,(-1)" C. 1.5 pc

QUESTION 5 The solution of associated homogeneous equation of a =7a+ +6 a+n2+2" with a,=9, a,=10 and a,=32 is o(-3)"+c,(-2)"+c,(-1)* o,3"+c,(-2)" +c, 1" +C. o,3" +c,(-2)" +c,(-1)* ,3" +c, 2" c,(-1)" C. 1.5 pc

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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Differential Equations

Have you ever observed the movement of the satellites and rockets or how the planets go around the sun in their orbits? How will you determine their motion? Definitely they follow some scientific laws and these kinds of problems are framed as differential equations.

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Match column B with the correct answer on column A, write only the capital letter of the answer on the blank provided at the left side/right side of the test paper. Match from column A the solution of the differential equations on column B. Column A A. xlnx + ylny = c B. 2y = x4 + cx2 C. e(x^2) + y2 = c D. x2 = cy3 E. x2 y = 2x4 + c F. ylnx + xlny = c G. x2 (x2 + 2y2) = c H. x (y + x2) = c (y – 2x) I. xy2 – x2y + 3x2 – 2y = c J. xy3 = 2y2 + 4y + 4 + cey K. y (x2 + c) + 2 = 0 L. (y – x) (y + 3x)3 = cx3 M. x2 y = 2x2 + c N. x (y + 1) = (1 + cx) ey O. e(-x^2) + y-2 = c P. (y + x2) = c (y – 2x) Q. x2y2 + 2xy – x2 = c R. x4 = c2 (4x2 + y2) S. x2 (x+ 2y) = c T. x2 + y2 – 3xy = c Column B (x – y) (4x + y) dx + x (5x – y) dx = 0 xy3 dx + e(x^2) dy = 0 y dx + (3x – xy + 2) dy = 0 2 (2x2 + y2) dx – xy dy = 0 (x2 + y2) – xy dy = 0 (2x – 3y) dx + (2x – 3x) dy = 0 x2 yy’ = ey (x4 + 2y) dx – x dy = 0 (3x2 – 2xy + 3y2) dx = 4xydy
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