设y=e^x是微分方程xy'+p(x)y=x的一个解,求此微分方程的

Question

设y=e^x是微分方程xy'+p(x)y=x的一个解,求此微分方程的

Answer 1

∵y=e^x
∴y'=e^x
∵y=e^x微分方程xy'+p(x)y=x解
∴x*(e^x)+p(x)*(e^x)=x
=>p(x)=x*[(1-e^x)/(e^x)]
∴微分方程xy'+p(x)y=x微分方程xy'+x*[(1-e^x)/(e^x)]*y=x即y'+[(1-e^x)/(e^x)]*y=1
设微分方程y'+[(1-e^x)/(e^x)]*y=1相应齐次微分方程
y'+[(1-e^x)/(e^x)]*y=0
=>dy/dx=-[(1-e^x)/(e^x)]*y
=>dy/y=-[(1-e^x)/(e^x)]*dx
=>∫dy/y=∫-[(1-e^x)/(e^x)]*dx
=>lnlyl=∫-[e^(-x)-1]*dx
=>lnlyl=e^(-x)+x+C
=>y=C*[e^(e^(-x))]*(e^x)
设微分方程y'+[(1-e^x)/(e^x)]*y=1通解y=C(x)*[e^(e^(-x))]*(e^x)
则y'=C'(x)*[e^(e^(-x))]*(e^x)+C(x)*[(e^(e^(-x)))*(-e^(-x))*(e^x)+ (e^(e^(-x)))*(e^x)]
=C'(x)*[e^(e^(-x))]*(e^x)+C(x)*[e^(e^(-x))]*[(e^x)-1]
代入微分方程y'+[(1-e^x)/(e^x)]*y=1得
C'(x)*[e^(e^(-x))]*(e^x)+C(x)*[e^(e^(-x))]*[(e^x)-1]+[(1-e^x)/(e^x)]*C(x)*[e^(e^(-x))]*(e^x)=1
=>C'(x)*[e^(e^(-x))]*(e^x)=1
=>C'(x)=e^[-e^(-x)]*e^(-x)
=>C(x)=∫e^[-e^(-x)]*e^(-x)dx
=>C(x)=-∫e^[-e^(-x)]d(-e^(-x))
=>C(x)=-e^[-e^(-x)]+C
∴微分方程y'+[(1-e^x)/(e^x)]*y=1通解y=[-e^[-e^(-x)]+C]*[e^(e^(-x))]*(e^x)
即y=-e^x+C*[e^(e^(-x))]*(e^x)
当x=ln2,y=0时
0=-2+C*(e^(1/2))*2
=>C=e^(-1/2)
∴满足条件y(ln2)=0特解y=-e^x+[e^(-1/2)]*[e^(e^(-x))]*(e^x)