Question

求微分方程第4题

Answer 1

4.求微分方程 dy/dx=2xy/(x²-y²)的通解
解：dy/dx=(2y/x)/[1-(y/x)²]...........(1)
令y/x=u，即y=ux;
代入(1)式得：

u+x(du/dx)=2u/(1-u²)
x(du/dx)=(u+u³)/(1-u²)
分离变量得 [(1-u²)/(u+u³)]du=dx/x
取积分 ∫[(1-u²)/(u+u³)]du=∫dx/x
∫[1/u-2u/(1+u²)]du=lnx+lnc=lncx
lnu-ln(1+u²)=lncx
即ln[u/(1+u²)]=lncx
故u/(1+u²)=cx,将u=y/x代入得：
(y/x)/[1+(y/x)²]=xy/(x²+y²)=cx
故得隐性通解为：y/(x²+y²)=c