求下列微分方程的通解(1)dx+xydy=y平方dx+ydy (2)xy'-ylny=0 (3)xdy+dx=e的y次方

问题描述:

求下列微分方程的通解
(1)dx+xydy=y平方dx+ydy (2)xy'-ylny=0 (3)xdy+dx=e的y次方dx
1个回答 分类:数学 2014-10-28

问题解答:

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(1)
dx+xydy=y=y^2dx+ydy
==>(xy-y)dy=(y^2-1)dx
==>(x-1)ydy=(y^2-1)dx
==>ydy/(y^2-1)=dx/(x-1)
两边积分,得:
ln(y^2-1)/2=ln(x-1)+lnC
==>y^2-1=e^[C(x-1)^2]
==>y=±(e^[C(x-1)^2]+1)^(1/2).
(2)
(3)
xdy+dx=e^ydx
==>xdy=(e^y-1)dx
==>dy/(e^y-1)=dx/x
对左边积分,可设t=e^y,
所以y=lnt,dy=1/tdt,
dy/(e^y-1)=dt/t(t-1)=(1/(t-1)-1/t)dt
对t积分,得:ln(t-1)-lnt,
变成y,即:ln(e^y-1)-y
两边积分,得:
ln(e^y-1)-y=lnx+lnC,
==>1-1/e^y=Cx.
==>y=ln(1/(1-Cx)).
 
 
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