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4) The functions y1(t) = t and y2(t) t are solutions of the homogeneous differential equation t'y 4ty +6y = 0 on (0, 00).

4) The functions y1(t) = t and y2(t) t are solutions of the homogeneous differential equation t'y" 4ty +6y = 0 on (0, 00). When using variation of parameters with y = u1(t)t+u2(t)t3 to find a solution of the nonhomogeneous differential equation t?y" 4ty' + 6y = 4t, what is the function u, (t) ? (1) 4 (2) -4 4 (3) 4 (5) -4t (6) -4t2

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