(d^2)y/d(x^2)这个式子怎样化简啊?x=2t-1,te^y+y+1=0求(d^2)y/d(x^2)丨t=0求高手指导.

Question

(d^2)y/d(x^2) 这个式子怎样化简啊?
x=2t-1 ,te^y+y+1=0 求(d^2)y/d(x^2)丨t=0 求高手指导.

Answer 1

(d^2)y/d(x^2)不就是y关于x的二阶导数嘛,还化什么简?!
由条件x=2t-1得t=(x+1)/2.带入te^y+y+1=0得：(x+1)e^y+2y+2=0
所以,上式两边对x求导化简得：e^y+[(x+1)e^y+2]dy/dx=0.再次在等式两边对x求导化简得：[(x+1)e^y+2]*(d^2)y/d(x^2)+(x+1)e^(2y)(dy/dx)^2+2e^y(dy/dx)=0.
当t=0时,带回条件知x=-1,y=-1.
将所得带入e^y+[(x+1)e^y+2]dy/dx=0求得：dy/dx=-1/(2e).再将所得全部带入[(x+1)e^y+2]*(d^2)y/d(x^2)+(x+1)e^(2y)(dy/dx)^2+2e^y(dy/dx)=0中就有：
(d^2)y/d(x^2)=e^(y-1)/2
即：当t=0时,(d^2)y/d(x^2)=e^(y-1)/2.