(b) 2xy′′ (3−x)y′ − y = 0 Solution (a) Write the differential equation in standard form y′′ − 1 2x y′ 1 4x y = 0 We have P(x) = − 1 2x and Q(x) = 1 4x, so x = 0 is a singular point Taking limits we get lim x→0 xP(x) = −1/2 and lim
(b) 2xy′′ (3−x)y′ − y = 0 Solution (a) Write the differential equation in standard form y′′ − 1 2x y′ 1 4x y = 0 We have P(x) = − 1 2x and Q(x) = 1 4x, so x = 0 is a singular point Taking limits we get lim x→0 xP(x) = −1/2 and lim