用variationofparameters的方法解x^2y''+xy'-y=1/(x+1)先设左边式子=0解得y=c1x+c2xlnx设y=u1x+u2xlnx

Question

用variation of parameters的方法解x^2y''+xy'-y = 1/(x+1)
先设左边式子=0 解得y=c1x+c2xlnx
设y=u1x+u2xlnx

Answer 1

令 y=u(x)x+v(x)xlnx
则 y'=u'x+u+v'xlnx+vlnx+v
不妨设 u'x+v'xlnx=0
则 y''=u'+v'lnx+v/x+v'
因为 y''+(1/x)y'-(1/x^2)y=1/(x^2(x+1))
将y'',y'和y关于u,v,x的表达式代入上式左侧
可得v'+2v/x=1/(x^3+x^2)
因此v=1/x^2(ln|x+1|+c)
然后解出u
Y=ux+vxlnx为所求的particular solution
General solution为y=c1x+c2xlnx+Y