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/dx-si_数学解答

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/dx-sinycosy=0 化为u关于t的方

人气：478 ℃　时间：2020-10-01 14:07:30

解答

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].sinycos...