12y Consider the differential equation xy" – y' = (1) Making the changes of variable x=e', y = u(t)e 2" in the differential equation (1), we obtain the differential equation, u depending on t, given by u"-6u'-4u = 0 The solution of the differential equation (1), depending on the parameter t, with constant c1 and C2, is given by: a) { z(1) y(t) et e* (cı cos(tv/13) + c2 sen(t/13)) S ¤(t) et b) | y(t) et (cietvi3 + cze-tV13 S 2(t) = et c) y(t) e3t -t/13 + c2e Į æ(t) et d) l y(t) e' (c1 cos(tv/13) + C2 sen(tv/13))

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