x^2y"-2xy'+2y=x^3lnx
设x=e^t,t=lnx
y'=(y't)*(1/x)y''=((y''t)-y't)/x^2,代入得:
(y''t-y't)-2y't+2y=te^3t
y''t-3y't+2y=te^3t
特征方程的根为:1和2
设特解y*=(At+B)e^3t,代入求得:A=1/2,B=-1/4
y=C1e^t+C2e^2t+(1/4)(2t-1)e^3t
=C1x+C2x^2+(1/4)(2lnx-1)x^3
y'=(y't)*(1/x)y''=((y''t)-y't)/x^2什么意思没看懂,还有特解y*=(At+B)e^3t中AB怎么求出来的,是代入e^2t*y"-2e^ty’+2y=te^3t中么
这是典型的欧拉方程现在把y看成t的函数,t是x的函数y'(x)=y'(t)t'(x)=y'(t)/x推出:xy'(x)=y'(t)y''(x)=(y''(t)t'(x)x-y'(t))/x^2=((y''(t)-y'(t))/x^2,推出:x^2y''(x)=y''(t)-y'(t)这样,原方程变为:y''(t)-3y'(t)+2y=te^3t特解是代入:y''(t)-3y'(t)+2y=te^3t