求解带初值的微分方程xy''-xy'-y=0y(0)=0,y'(0)=1

Question

求解带初值的微分方程
xy''-xy'-y=0
y(0)=0,y'(0)=1

Answer 1

(xy)'=y+xy'
(xy)''=xy''+2y'
xy''-xy'-y=0
xy''=xy'+y
(xy)''-2y'=(xy)'
u=xy,v=y
u''-2v'=u'
u''-u'=2v'
两边积分得
u'-u=2v
u=C1e^v+e^(-v)+C
即xy=C1e^y+e(-y)+C
y(0)=0,即0=C1+1+C