设函数f(x)可导,满足(xex+f(x))ydx+f(x)dy=du(x,y),且f(0)=0,求f(x)及u(x,y
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设函数f(x)可导,满足(xex+f(x))ydx+f(x)dy=du(x,y),且f(0)=0,求f(x)及u(x,y))
ə【(xex+f(x))y】/əy=əf(x)/əx
xex+f(x)=f'(x)
f'-f-xe^x=0 .①
f'=f
f=c(x)e^x =>①
c'e^x=xe^x,c'=x
c(x)=x²/2+C
f(x)= (x²/2+C)e^x.
f(0)=0 ,C=0 =>f(x)= (x²/2)e^x..②
u(x,y)=∫【0,0;x,y】(x+x²/2)e^xydx+(x²/2)e^xdy
=∫[0,x]0dx+∫[0,y](x²/2)e^xdy
=(x²y/2)e^x.③
f(x)及u(x,y)):② ③
xex+f(x)=f'(x)
f'-f-xe^x=0 .①
f'=f
f=c(x)e^x =>①
c'e^x=xe^x,c'=x
c(x)=x²/2+C
f(x)= (x²/2+C)e^x.
f(0)=0 ,C=0 =>f(x)= (x²/2)e^x..②
u(x,y)=∫【0,0;x,y】(x+x²/2)e^xydx+(x²/2)e^xdy
=∫[0,x]0dx+∫[0,y](x²/2)e^xdy
=(x²y/2)e^x.③
f(x)及u(x,y)):② ③
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