case_1:λ=−p2<0x2y′′+xy′−p2y=0⇒y1=xp,y2=x−p⇒y=c1⋅xp+c2⋅x−pandy′(1)=y′(e2π)=0⇒c1=c2=0.case_2:λ=0x2y′′+xy′=0⇒y=c1⋅ln(x)+c2andy′(1)=y′(e2π)=0⇒c1=0.case_3:λ=p2>0x2y′′+xy′+p2y=0⇒y=c1sin(pln(x))+c2cos(pln(x))andy′(1)=y′(e2π)=0⇒y′=xp(c1cos(pln(x))−c2sin(pln(x)))⇒c1=0,p=2k,k∈Z.λ=(2k)2,y=c2⋅cos(2k⋅ln(x));
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