<P>请问这样的微分方程组这么解算啊?</P>
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<P><FONT face="Times New Roman"> </FONT></P><FONT face="Times New Roman">dy1/dx</FONT>=<FONT face="Times New Roman">f (x, y1, y2)</FONT>;<BR><FONT face="Times New Roman">dy2/dx</FONT>=<FONT face="Times New Roman">g (x, y1, y2)</FONT>;
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<P> 区间为(<FONT face="Times New Roman">0, L</FONT>);</P>
<P> 边界条件为:<BR><FONT face="Times New Roman"> x</FONT>=<FONT face="Times New Roman">0</FONT>时,<FONT face="Times New Roman">y1</FONT>=<FONT face="Times New Roman">a </FONT>;<BR><FONT face="Times New Roman"> x</FONT>=<FONT face="Times New Roman">L</FONT>时,<FONT face="Times New Roman">y2</FONT>=<FONT face="Times New Roman">b </FONT>;</P>
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<P><FONT face="Times New Roman"></FONT></P>
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<P>我查了很多的资料,一般资料上的算例所给的边界条件为都是(<FONT face="Times New Roman">x</FONT>=<FONT face="Times New Roman">0</FONT>时,<FONT face="Times New Roman">y1</FONT>=<FONT face="Times New Roman">a </FONT>;<FONT face="Times New Roman">x</FONT>=<FONT face="Times New Roman">0</FONT>时,<FONT face="Times New Roman">y2</FONT>=<FONT face="Times New Roman">b </FONT>)。而对于边界条件为(<FONT face="Times New Roman">x</FONT>=<FONT face="Times New Roman">0</FONT>时,<FONT face="Times New Roman">y1</FONT>=<FONT face="Times New Roman">a </FONT>;<FONT face="Times New Roman">x</FONT>=<FONT face="Times New Roman">L</FONT>时,<FONT face="Times New Roman">y2</FONT>=<FONT face="Times New Roman">b </FONT>)时的基本没有。希望那位知道的朋友能够指点一下,先谢谢了啊!</P>