Consider the equation
xpy′′+n−1n+3−2pxp−1y′+(n−1p−2)2xp−2y=yn,
in which p=2 and n>−1 but n=1. For the boundary conditions y(1)=0 and y′(1)=λ, show that the solution is y(x)=v(x)x(p−2)/(n−1), where v(x) is given by
∫0v(x)[λ2+2zn+1/(n+1)]1/2dz=lnx.