1.æ±å¾®åæ¹ç¨xy'-3y+x⁴=0满足åå§æ¡ä»¶y(1)=2çç¹è§£ã
解ï¼å
æ±é½æ¬¡æ¹ç¨xy'-3y=0çé解ï¼
å离åéå¾dy/y=(3/x)dxï¼
积åä¹å¾lny=3lnx+lnC'=ln(C'x³)ï¼
æ
å¾y=C'x³ï¼æC'æ¢æ为xçå½æ°uï¼å¾y=ux³..........(1)
å°(1)对xå导æ°å¾dy/dx=3x²u+x³(du/dx).............(2)
å°(1)å(2)代å
¥åæ¹ç¨å¾x[3x²u+x³(du/dx)]-3ux³+x⁴=0
åç®å¾x⁴(du/dx)+x⁴=x⁴(du/dx+1)=0
å 为xâ 0ï¼æ
å¿
ædu/dx+1=0ï¼å³ædu=-dxï¼æ
u=-x+C
代å
¥(1)å¼å³å¾åæ¹ç¨çé解为y=(-x+C)x³=-x⁴+Cx³
å°åå§æ¡ä»¶y(1)=2代å
¥å¾2=-1+Cï¼æ
å¾C=3ï¼
äºæ¯å¾åæ¹ç¨çç¹è§£ä¸ºy=-x⁴+3x³.
(æªå®ï¼å¾
ç»ï¼è¯·å«ä¸æçé¢ç¨åº)
追é®æå¾
追ç2.æ±æ²é¢e^x-z+xy=3å¨ç¹(2ï¼1ï¼0)å¤çåå¹³é¢åæ³çº¿æ¹ç¨ã
解ï¼z=e^x+xy-3ï¼
∂z/∂x=e^x+yï¼∂z/∂y=xï¼æ
∂z/∂xâ£[x=2ï¼y=1]=e²+1ï¼∂z/∂yâ£[x=2ï¼y=1]=2ï¼
æ
åé¢æ¹ç¨ä¸º(e²+1)(x-2)+2(y-1)-z=0
æ³çº¿æ¹ç¨ä¸º(x-2)/(e²+1)=(y-1)/2=z/(-1)