æ±å¾®åæ¹ç¨ dy/dx=2y/(x-2y)满足åå§æ¡ä»¶y(0)=1çç¹è§£ï¼
解ï¼dy/dx=2(y/x)/[1-2(y/x)]...........â ï¼ ä»¤y/x=uï¼åy=uxï¼dy/dx=u'x+uï¼
代å
¥â å¼å¾ï¼x(du/dx)+u=2u/(1-2u)
å³æx(du/dx)=(u+2u²)/(1-2u);
å离åéå¾ï¼(1-2u)du/(u+2u²)=(1/x)dx
å积åï¼â«(1-2u)du/(u+2u²)=â«[(1/u)-4/(2u+1)]du=lnâ£uâ£-2lnâ£2u+1â£=lnâ£xâ£+lnc;
å³æln[â£uâ£/(2u+1)²]=ln(câ£xâ£)ï¼æ
å¾ â£uâ£/(2u+1)²=câ£xâ£;
å°u=y/x代å
¥å¾ â£y/xâ£/[(2y/x)+1]²=câ£xâ£
åç®å¾ y=c(2y+x)²; 代å
¥åå§æ¡ä»¶y(0)=1å¾c=1/4ï¼
æ
ç¹è§£ä¸ºï¼4y=(2y+x)²; æåæ 4y²+4xy-4y+x²=0.
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