Given the initial value problem x – 6x1 + x2 = 0 x, – 5x1 – 4x2 = 0 T, (0) — - 2, г2(0) — 0 %3D find 12(t). O r2(t) = 2e5t (cos (2t) + sin (2t)) O r2(t) = e5t (cos (2t) + sin (2t)) O r2(t) = e5t sin (2t) O 12(t) = -5ešt sin (2t) O r2(t) = 5e5t sin (2t)

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