设x=e^(-t) 试变换方程x^2 d^2y/dx^2 +xdy/dx+y=0网上有种解法如下（网友franciscococo提供）：x=e^(-t),即dx/dt= -e^(-t)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这种解法是否正确?d^2y/dx^2= [d(dy/dx) /dt] * dt/dx= [-e^t *d^2y/dt^2 -e^t *dy/dt] * (-e^t)这步又是怎么得到的?
问题描述：
设x=e^(-t) 试变换方程x^2 d^2y/dx^2 +xdy/dx+y=0
网上有种解法如下（网友franciscococo提供）：
x=e^(-t),即dx/dt= -e^(-t)
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
这种解法是否正确?
d^2y/dx^2
= [d(dy/dx) /dt] * dt/dx
= [-e^t *d^2y/dt^2 -e^t *dy/dt] * (-e^t)
这步又是怎么得到的?
答