The solution of y" = 6(t – 1) – 8(t – 3), y(0) = y'(0) = 0 is y(t) = %3D %3D (A) (t – 1)u(t – 1) - (t - 3)u(t – 3) (в) t(u(t — 1) — и (t — 3)] (C) (t – 1)[u(t – 1) – u(t – 3)] (D) (t — 3)[u(t — 1) — и(t — 3)] (E) (t – 1) u(t - 1) – (t – 3)u(t – 3)
The solution of y" = 6(t – 1) – 8(t – 3), y(0) = y'(0) = 0 is y(t) = %3D %3D (A) (t – 1)u(t – 1) - (t - 3)u(t – 3) (в) t(u(t — 1) — и (t — 3)] (C) (t – 1)[u(t – 1) – u(t – 3)] (D) (t — 3)[u(t — 1) — и(t — 3)] (E) (t – 1) u(t - 1) – (t – 3)u(t – 3)
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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A linear function can just be a constant, or it can be the constant multiplied with the variable like x or y. If the variables are of the form, x2, x1/2 or y2 it is not linear. The exponent over the variables should always be 1.
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