那么dy/dx=(dy/dt) / (dx/dt)= -e^t *dy/dt，
而
d^2y/dx^2
= [d(dy/dx) /dt] * dt/dx
= [-e^t *d^2y/dt^2 -e^t *dy/dt] * (-e^t)
=e^(2t) *d^2y/dt^2 +e^(2t) *dy/dt
所以x^2 d^2y/dx^2= d^2y/dt^2 +dy/dt，
而xdy/dx= -dy/dt，
于是原方程可以变换为：
d^2y/dt^2 +y=0
设x=e^(-t) 试变换方程x^2 d^2y\/dx^2 +xdy\/dx+y=0
=e^(2t) *d^2y\/dt^2 +e^(2t) *dy\/dt 所以x^2 d^2y\/dx^2= d^2y\/dt^2 +dy\/dt，而xdy\/dx= -dy\/dt，于是原方程可以变换为：d^2y\/dt^2 +y=0
设x=e^(-t) 试变换方程x^2 d^2y\/dx^2 +xdy\/dx+y=0
-e^t dy\/dt](-e^t)=e^(2t)d^2y\/dt^2 +e^(2t)dy\/dt 所以x^2 d^2y\/dx^2= d^2y\/dt^2 +dy\/dt，而xdy\/dx= -dy\/dt，于是原方程可以变换为:d^2y\/dt^2 +y=0
设x=e^(-t),变换方程x^2*d^2y\/dx^2+x*dy\/dx+y=0
x * dy\/dx = - dy\/dt, x² * d²y\/dx² = dy\/dt + d²y\/dt²代入原方程，得：d²y\/dt² + y= 0
d²y\/dx²+2dy\/dx+y=e∧x
x=e^(-t),dx\/dt = -e^(-t) = -x dy\/dx = (dy\/dt) \/ (dx\/dt) = (-1\/x) * dy\/dt d²y\/dx² = (1\/x²) * dy\/dt + (-1\/x) * d²y\/dt² \/ (dx\/dt) = (1\/x²) * dy\/dt + (1\/x²) * d²y\/dt²= ...
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...
\/[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.
dy\/dx=tany,求y的解析式
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)\/(...
令y=tanz,试变换方程 d2y\/dx^2=2+((dy\/dx)^2) *(2(1+y)\/(1+
只能求出y '与y 的关系
令x=cost,变换方程d^2y\/dx^2-x\/(1-x^2)*dy\/dx+y\/(1-x^2)=0 答案是d^...
d^2y\/dx^2=d(dy\/dx)\/dx=d(-dy\/(sintdt))\/(-sintdt)=(-(d^2y\/dt*sint-dy\/dt*cost)\/(sint)^2)dt\/(-sintdt)=d^2y\/dt^2\/(sint)^2-dy\/dt*cost\/(sint)^3原方程可化为1\/(sint)^2*d^2y\/dt^2-cost\/(sint^3)*dy\/dt+cost\/(sint)...
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二元一次方程的最值怎么求
例如，考虑二元一次方程 f(x, y) = ax^2 + bxy + cy^2 + dx + ey + f = 0，其中 a、b、c、d、e、f 为常数，需求解最值。利用配方法将二次项的系数变为完全平方形式，即利用常数项 f 的符号确定配方方式；将配方后的方程表示成二元一次方程 g(x, y) = p(x - h)^2 + q...
