2. given that y1 (t) = t3 is a solution of the equation 3t2y"-5ty3y = 0 for t > 0 (1) find a second linearly independent solution y2(t) of (1) solution: we proceed by using the reduction of order method. thus, we need to find a function v such that y2(t) v(t)y1(t) is solution of equation (1). if v not a constant function, then y2 is linearly independent (cf lecture notes). then, 2' vuk and inserting this into equation (1) yields 0 3t("y20'y vy{) 5t (v'yn + vy{) -3vyi 3y" + (6y- 5ty1) (3t°y{ - 5tyh -3y)u 3t°y" 13t', where in the third equality, we used that yi is a solution of (1) and that y1(t) = t3. the last equation is equivalent to 13 0 " for t 0 thus w= v is a solution of the 1st order linear ode 13 for t > 0 3 ct13/3 for t 0 and which (by seperation of variables) has the solution w (t) = any constant c e r. thus, v(t) = u cer and so w(s)ds ct10/3 for t > 0 and any constant y2(t) = v(t)y1(t) = t 10/3 f3 since v is a non-constant function, {yı, y2} is linearly independent for t0