x2y′′+(2−x)y′=0⋯⋯⋯(eqn1)Comparing eqn1 to the general equation below:P(x)y′′+Q(x)y′+R(x)y=0, yieldsP=x2,Q(x)=(2−x),and R(x)=0Now, by definition we have that a point xo is a singular point if P(x0)=0.Thus equating our P(x)=0,yieldsx2=0⇒x=0.The singular points of the given differential equation on the x-axis is x = 0.To classify the singular point xo=0, we need to compute,limx→x0(x−xo)P(x)Q(x),and limx→x0(x−xo)2P(x)R(x)Now, since xo=0, we have:limx→x0(x−xo)P(x)Q(x)=limx→0(x−0)(x22−x)=limx→0x2−x=∞,andlimx→x0(x−xo)2P(x)R(x)=0<∞. Since, limx→x0(x−xo)2P(x)R(x)<∞ but limx→x0(x−xo)P(x)Q(x)=∞,We conclude that the singular point xo=0 is an irregular singular point.
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