变换?要怎么变换?
再问： 我看不懂题、书后又没答案才提问的呀
再答： 哦，大概是让你用x=e^(-t)带入进行计算 x=e^-t dx/dt = -e^(-t)=-x dy/dt= dy/dx *dx/dt = -xdy/dx ==> dy/dx = (-1/x) dy/dt = - e^t dy/dt d^2y/dt^2= d(dy/dt)/dt = d(-xdy/dx)/dt = -dx/dt dy/dx + -xd(dy/dx)/dt =xdy/dx - xd(dy/dx)/dx *dx/dt = xdy/dx +x^2d^2y/dx^2 = e^(-t) * (- e^t dy/dt) + e^(-2t) d^2y/dx^2 = d^2y/dt^2 ==> d^2y/dx^2 = e^(2t)[d^2y/dt^2 + dy/dt) 记y'=dy/dt, y'' = d^2y/dt^2，则得到 dy/dx = -e^ty' d^2y/dx^2 = e^(2t)y'' +e^(2t)y' 带入x^2*(d^2y/dx^2)+x*(dy/dx)+y=0得到 e^(-2t) ( e^(2t)y'' +e^(2t)y') + e^(-t)(-e^ty') +y =0 就是y'' +y' -y' +y =0; y'' +y=0 这是一个非常常见的方程，其解为y=c1 sint + c2 cost 再把x=e^(-t) ==> t= -lnx待人就是解了
