求下列微分方程的特解(1)xy'+y-e^x=0,y|(x=a ) =b(2)y'-(2/x)y=(1/2)x,y|(x=1) =2

来源:学生作业帮助网 编辑:作业帮 时间:2024/08/15 14:06:03
求下列微分方程的特解(1)xy'+y-e^x=0,y|(x=a ) =b(2)y'-(2/x)y=(1/2)x,y|(x=1) =2
x){Ɏӟ[ٴWt?s0|R]R75@(X~D M ٕ ~m Ҫ@h10I*'`5X qÔ; L@NE%v֩HԴMIL*KuVJ@/f}u=̪u3Ү/.H̳ Cuv0=PGF<Ե59+'BirQ

求下列微分方程的特解(1)xy'+y-e^x=0,y|(x=a ) =b(2)y'-(2/x)y=(1/2)x,y|(x=1) =2
求下列微分方程的特解
(1)xy'+y-e^x=0,y|(x=a ) =b
(2)y'-(2/x)y=(1/2)x,y|(x=1) =2

求下列微分方程的特解(1)xy'+y-e^x=0,y|(x=a ) =b(2)y'-(2/x)y=(1/2)x,y|(x=1) =2
(1)xy'+y-e^x=0
(xy)'=e^x,xy=e^x+C,y(a)=b,ab-e^a=C xy=e^x+ab-e^a
(2)y'-(2/x)y=(1/2)x
由通解公式:y=Cx^2+x

(1)xy'+y-e^x=0
(xy)'=e^x,->xy=e^x+C又y(a)=b,, ab-e^a=C ->xy=e^x+ab-e^a
(2)y'-(2/x)y=(1/2)x
->y=2x^2+lnx/2