tany(dy/dx=tany,求y的解析式)
dy/dx=tany,求y的解析式?
网友:川水往事 提问
dy/dx=tany,求y的解析式
五星知识达人网友:独行浪子会拥风 解答于 2022-01-03 10:09
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的方程
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.
1楼网友:鱼芗 解答于 2022-06-03 10:29
????? 你已经得出答案了啊 y‘就是dy/dx啊看来你还分不太清 你看看书就明白了。
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