Question

d²y/dx²+2dy/dx+y=e∧x

设x=e^(-t),变换方程x^2*d^2y/dx^2+x*dy/dx+y=0
设x=e^(-t),变换方程(x^2)*d^2y/dx^2+x*dy/dx+y=0
答案是d^x/dt^2+y=0

Answer 1

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
= (1/x²) * [ dy/dt + d²y/dt² ]
x * dy/dx = - dy/dt,x² * d²y/dx² = dy/dt + d²y/dt²
代入原方程,得：d²y/dt² + y= 0