求y''+2y'+y=e^-x/x的通解y''+2y'+y=(e^-x)/x

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求y''+2y'+y=e^-x/x的通解y''+2y'+y=(e^-x)/x

求y''+2y'+y=e^-x/x的通解y''+2y'+y=(e^-x)/x
求y''+2y'+y=e^-x/x的通解
y''+2y'+y=(e^-x)/x

求y''+2y'+y=e^-x/x的通解y''+2y'+y=(e^-x)/x
p=dy/dx,y''=dp/dx=pdp/dy
pdp/dy+2p+y=0
pdp/dy=-2p
dp=-2dy
p=-2y+c
p=-2y+u
dp/dy=-2+u'
(-2y+u)(-2+u')+2(-2y+u)+y=0
4y-2yu'-2u+uu'-4y+2u+y=0
2yu'=y
2du=dy
u=y/2
p=-2y+y/2=-3y/2
-3y/2*(-3/2)-3y+y=e^(-x)/x
y=4e^(-x)/x