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MATH 2930 HW#14 Solution 2/25/2013 In all problems t>0. 35. t2y'' + ty' + y = 0, if we compare this to equation (ii) in problem 34, we see that α=1 and β=1. d2 y dy d2 y +(α−1) +β y= + y=0. If we let x = ln(t), we can obtain dx dx 2 dx 2 The two roots of the characteristic equation of this linear second-order differential equation with ±√ −4 =±i. The general solution is then constant coefficients are r = 2 y=c 1 sin ( x )+c 2 cos ( x)=c 1 sin (ln t)+c 2 cos (ln t). Remember to express the solution in terms of t! 36. t2y'' + 4ty' + 2y = 0, if we compare this to equation (ii) in problem 34, we see that α=4 and β=2. d2 y dy d2 y dy +(α−1) +β y= 2 +3 +2y=0. If we let x = ln(t), we can obtain 2 dx dx dx dx The two roots of the characteristic equation of this linear second-order differential equation with −3±√ 9−8 =−1,−2. The general solution is then constant coefficients are r = 2 c c y=c 1 e−x +c2 e−2x=c1 e−ln t+c 2 e−2 ln t = 1 + 22 . t t 37. t2y'' + 3ty' + 1.25y = 0, so we see that α=3 and β=1.25. d2 y dy d2 y dy +(α−1) +β y= +2 +1.25y=0. If we let x = ln(t), we can obtain 2 2 dx dx dx dx The two roots of the characteristic equation of this linear second-order differential equation with −2±√ 4−5 1 =−1± i. The general solution is then constant coefficients are r = 2 2 c1 1 1 1 1 ln t c 2 ln t −x −x −ln t −ln t y=c 1 e sin ( x)+c 2 e cos ( x)=c1 e sin( ln t)+c 2 e cos ( ln t )= sin ( )+ cos ( ). 2 2 2 2 t 2 t 2