Riley hobson bence · Problem 15.37

Higher-order ordinary differential equations

Mathematics — Methods · higher-order-odes-isobaric

Problem

Consider the equation xpy′′+n+3−2pn−1xp−1y′+(p−2n−1)2xp−2y=yn,x^p y'' + \frac{n+3-2p}{n-1}x^{p-1}y' + \left(\frac{p-2}{n-1}\right)^2 x^{p-2}y = y^n, in which p≠2p \neq 2 and n>−1n > -1 but n≠1n \neq 1. For the boundary conditions y(1)=0y(1) = 0 and y′(1)=λy'(1) = \lambda, show that the solution is y(x)=v(x)x(p−2)/(n−1)y(x) = v(x)x^{(p-2)/(n-1)}, where v(x)v(x) is given by ∫0v(x)dz[λ2+2zn+1/(n+1)]1/2=ln⁡x.\int_0^{v(x)} \frac{dz}{\left[\lambda^2 + 2z^{n+1}/(n+1)\right]^{1/2}} = \ln x.

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