Final Exam Solution - King Fahd University of Petroleum and Minerals

Exercise 2 (12 points)Consider the following differential equationa) Find the indicial equation roots.x 2 y ′′ − xy ′ + 5 (x − 1)y = 0.4b) Let y 1 = ∑ n≥0 C nX n+r 1, with C 0 = 1 be the power series solution corresponding tothe largest root of the indicial equation. Find the value of C 1 + C 2 .Solutiona) The indicial equation is of the form r(r − 1) + a 0 r + b 0 = 0 with: (1 point)a 0 = lim x→0 x(− x ) = −1 and bx 2 0 = lim x→0 x 2 [ 5( x−1 )] = − 5 . (1 point)4 x 2 4The indicial equation is: r(r − 1) − r − 5 = 4 r2 − 2r − 5 = (r − 5)(r + 1 ) = 0. (1 point)4 2 2Hence r 1 = 5 > r 2 2 = − 1 . (1 point)2Total: 4 pointsb) We have r 1 = 5 2the largest root, theny 1 = ∑ ∞n=0 C nx n+ 5 2 , y ′ 1 = ∑ ∞n=0 (n + 5 2 )C nx n+ 3 2 and y ′′1 = ∑ ∞n=0 (n + 5 2 )(n + 3 2 )C nx n+ 1 2 .⇒ (1 point)Substituting in the DE gives:∑ ∞n=0 (n+ 3 2 )(n+ 5 2 )C nx n+ 5 2 − ∑ ∞n=0 (n+ 5 2 )C nx n+ 5 2 + 5 4∑ ∞n=0 C nx n+ 7 2 − 5 4∑ ∞n=0 C nx n+ 5 2 = 0⇒ (1 point)Thus ∑ ∞[n=0 (n +3)(n + 5) − (n + 5) − ∑ 52 2 2 4]Cn x n+ 5 2 + 5 ∞4 m=1 C m−1x m+ 5 2 = 0.⇒ (1 point)Hence, 0 + ∑ ∞n=1 [n(n + 3)] C nx n+ 5 2 + 5 4∑ ∞n=1 C n−1x n+ 5 2 = 0. ⇒ (1 point)Finally we have ∑ ∞[n=1 n(n + 3)Cn + 5C ]4 n−1 xn+ 5 2 = 0. ⇒ (2 points)The recurrence relation is: C n = − 54n(n+3) C n−1, for n = 1, 2, ... ⇒ (1 point)With C 0 = 1 we get C 1 = − 516 and C 2 = 5128 .The value of C 1 + C 2 is − 35 . ⇒ (1 point)128Total: 8 points3