Question

作变换u=tany,x=e的t次幂 试将方程 x^2d^2y/dx^2+2x^2(tany)(dy/dx)^2+xdy/

Answer 1

u=tany,x=e^t.

du=(secy)^2dy=[1+(tany)^2]dy=(1+u^2)dy,

dy=du/(1+u^2), dx=e^tdt.

dy/dx=1/[e^t(1+u^2)]du/dt,

d^2y/dx^2=d(dy/dx)/(e^tdt)

=(d^2u/dt^2-du/dt)/[e^(2t)(1+u^2)]-2u(du/dt)^2/[e^(2t)(1+u^2)^2].

sinycosy=sin(2y)/2=tany/[1+(tany)^2]=u/(1+u^2).

于是,

x^2d^2y/dx^2+2x^2(tany)(dy/dx)^2+xdy/dx-sinycosy

=(d^2u/dt^2-u)/(1+u^2).

从而化为u关于t的方程 d^2u/dt^2-u=0.