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A particular solution is given by: Y (t): ds y1(t) * Y2(t) y3(t) Therefore the solution of the initial-value problem is: y(t) +Y(t)

A particular solution is given by: Y (t): ds y1(t) * Y2(t) y3(t) Therefore the solution of the initial-value problem is: y(t) +Y(t)

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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Both images shown are the same problem,  I have gotten that a= 9 and y2 is cos4t and y3 is sin4t

Y(t) = | A particular solution is given by: nt Y(t) = | ds · y1(t) %3D to +([ · Y2(t) +([ Y3(t) Therefore the solution of the initial-value problem is: y(t) = +Y(t)
Transcribed Image Text:Y(t) = | A particular solution is given by: nt Y(t) = | ds · y1(t) %3D to +([ · Y2(t) +([ Y3(t) Therefore the solution of the initial-value problem is: y(t) = +Y(t)
Find the solution of the initial-value problem 7 227 y" – 9y" + 16y' - 144y = sec 4t, y(0) = 2, y'(0) = -, y"(0) = A fundamental set of solutions of the homogeneous equation is given by the functions: Y1(t) = eat where a = Y2(t) Y3(t) = %3D A particular solution is given by: Y(t) = ds y1(t) to
Transcribed Image Text:Find the solution of the initial-value problem 7 227 y" – 9y" + 16y' - 144y = sec 4t, y(0) = 2, y'(0) = -, y"(0) = A fundamental set of solutions of the homogeneous equation is given by the functions: Y1(t) = eat where a = Y2(t) Y3(t) = %3D A particular solution is given by: Y(t) = ds y1(t) to
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