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This is the first part of a two-part problem. Let P = -6 cos(6t) ý1(t) = - (sin(6t)) ] -6 sin(6t) y2(t) = -6 cos(6t) a. Show that y1 (t) is a solution to the system y = Py by evaluating derivatives and the matrix product (t) -6 0 Enter your answers in terms of the variable t b. Show that y2 (t) is a solution to the system y = Pj by evaluating derivatives and the matrix product Enter your answers in terms of the variable t.

This is the first part of a two-part problem. Let P = -6 cos(6t) ý1(t) = - (sin(6t)) ] -6 sin(6t) y2(t) = -6 cos(6t) a. Show that y1 (t) is a solution to the system y = Py by evaluating derivatives and the matrix product (t) -6 0 Enter your answers in terms of the variable t b. Show that y2 (t) is a solution to the system y = Pj by evaluating derivatives and the matrix product Enter your answers in terms of the variable 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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This is the first part of a two-part problem. Let P = cos(6t) y(t) = |- (sin(6t)) | -6 sin(6t) , Y2(t) = -6 cos(6t) a. Show that y1 (t) is a solution to the system y = Py by evaluating derivatives and the matrix product (t) -6 0 Enter your answers in terms of the variable t b. Show that y2 (t) is a solution to the system y = Pj by evaluating derivatives and the matrix product y2(t) Enter your answers in terms of the variable t.
Transcribed Image Text:This is the first part of a two-part problem. Let P = cos(6t) y(t) = |- (sin(6t)) | -6 sin(6t) , Y2(t) = -6 cos(6t) a. Show that y1 (t) is a solution to the system y = Py by evaluating derivatives and the matrix product (t) -6 0 Enter your answers in terms of the variable t b. Show that y2 (t) is a solution to the system y = Pj by evaluating derivatives and the matrix product y2(t) Enter your answers in terms of the variable t.
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