求微分方程(y''')^2+(y'')^2=1满足所给初始条件y|x=0=0,y'|x=0=1,y''|x=0=0的特解
æ±å¾®åæ¹ç¨(y''')^2+(y'')^2=1满足æç»åå§æ¡ä»¶y|x=0=0,y'|x=0=1,y''|x=0=0çç¹è§£
解ï¼y'''=dy''/dxï¼y''=dy'/dxï¼å æ¤åå¼å¯å为(dy''/dx)²+(dy'/dx)²=1...........(1)
令dy'/dx=pï¼åy'''=dy''/dx=[d(dy'/dx)]/dx=dp/dxï¼ä»£å ¥(1)å¼å¾ï¼
(dp/dx)²+p²=1ï¼æ dp/dx=â(1-p²)ï¼
å离åéå¾dp/â(1-p²)=dxï¼ç§¯åä¹å¾arcsinp=x+C₁ï¼å³p=sin(x+C₁)ï¼
ä»£å ¥åå§æ¡ä»¶ï¼x=0æ¶y''=p=0ï¼å¾C₁=0ï¼æ p=sinx..........(2)
ç±(2)å¾y''=dy'/dx=sinxï¼å³ædy'=sinxdxï¼ç§¯åä¹å¾y'=â«sinxdx=-cosx+C₂ï¼
ä»£å ¥åå§æ¡ä»¶ï¼x=0æ¶y'=1ï¼å¾C₂=2ï¼æ æy'=2-cosx........(3)
ç±(3)å¾y=â«(2-cosx)dx=2x-sinx+C₃ï¼ä»£å ¥åå§æ¡ä»¶ï¼x=0æ¶y=0å¾C₃=0;
æ åæ¹ç¨çç¹è§£ä¸ºy=2x-sinx.
解ï¼y'''=dy''/dxï¼y''=dy'/dxï¼å æ¤åå¼å¯å为(dy''/dx)²+(dy'/dx)²=1...........(1)
令dy'/dx=pï¼åy'''=dy''/dx=[d(dy'/dx)]/dx=dp/dxï¼ä»£å ¥(1)å¼å¾ï¼
(dp/dx)²+p²=1ï¼æ dp/dx=â(1-p²)ï¼
å离åéå¾dp/â(1-p²)=dxï¼ç§¯åä¹å¾arcsinp=x+C₁ï¼å³p=sin(x+C₁)ï¼
ä»£å ¥åå§æ¡ä»¶ï¼x=0æ¶y''=p=0ï¼å¾C₁=0ï¼æ p=sinx..........(2)
ç±(2)å¾y''=dy'/dx=sinxï¼å³ædy'=sinxdxï¼ç§¯åä¹å¾y'=â«sinxdx=-cosx+C₂ï¼
ä»£å ¥åå§æ¡ä»¶ï¼x=0æ¶y'=1ï¼å¾C₂=2ï¼æ æy'=2-cosx........(3)
ç±(3)å¾y=â«(2-cosx)dx=2x-sinx+C₃ï¼ä»£å ¥åå§æ¡ä»¶ï¼x=0æ¶y=0å¾C₃=0;
æ åæ¹ç¨çç¹è§£ä¸ºy=2x-sinx.
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