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