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^2y/dt^2+y=0,想看看解法

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• 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?
• 令x=cost,变换方程d^2y/dx^2-x/(1-x^2)*dy/dx+y/(1-x^2)=0
• 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^2y/dt^2+y=0,想看看解法
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