Consider the differential equation zy" – v = (1) Making the changes of variable * = e', y = u(t)e 2* in the differential equation (1), we obtain the differential equation, u depending on t, u" – 6u' – bu = 0. The solution of the differential equation (1), depending on the parameter t, with constant c, and C2, is given by: a) { 2(t) y(t) ´z(t) et et (c1 cos(tv/15) + c2 sen(t/15)) %3D S¤(t) b) et y(t) ´z(t) c) y(t) et e3 d) { 2(t) l y(t) S¤(t) et et (c1 cos(t/15) + c2 sen(t/15)) || ||

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