5.æ±æç©çº¿y=x²ä¸ç´çº¿x=1åxè½´æå´å¾å½¢ç»yè½´æ转ææç«ä½çä½ç§¯ã
6.æ±å¾®åæ¹ç¨ (1+y²)dx-xy(1+x²)dy=0满足åå§æ¡ä»¶y(1)=2çç¹è§£ã
解ï¼xy(1+x²)dy=(1+y²)dx
å离åéå¾ï¼[y/(1+y²)]dy=[1/x(1+x²)]dx
å积åå¾ â«[y/(1+y²)]dy=â«[1/x(1+x²)]dx
(1/2)â«d(1+y²)/(1+y²)=â«[(1/x)-x/(1+x²)]dx
(1/2)ln(1+y²)=lnx-(1/2)â«d(1+x²)/(1+x²)=lnx-(1/2)ln(1+x²)+lnc=ln[cx/â(1+x²)]
æ å¾éæ§é解ï¼â(1+y²)=cx/â(1+x²)
å³1+y²=c²x²/(1+x²)ï¼y²=[c²x²/(1+x²)]-1;
ä»£å ¥åå§æ¡ä»¶y(1)=2ï¼å¾4=(c²/2)-1ï¼c²=10ï¼
æ 满足åå§æ¡ä»¶çç¹è§£ä¸ºï¼y²=[10x²/(1+x²)]-1=(9x²-1)/(1+x²).