解微分方程时积分常数的最后形式到底是什么?

2025-04-02 17:01:56
推荐回答(1个)
回答1:

xy'-y*ln(y)=0很显然y>0所以dy/(y*lny)-dx/x=0化成这个形式讨论一下:分母不能为零,所以x≠0,y*lny≠0;因为y>0,亦即x≠0,lny≠0但原方程里面没有这些限制;因此这里出现了不允许的情况应单独作为补充说明。 亦dlny/(lny)-dx/x=0亦dln|lny|-dln|x|=dln|lny/x|=0积分一次ln|lny/x|=A,这里A为任意实数变形为|lny/x|=e^{A},即lny/x=±e^{A}=B在令上述B=±e^{A}中实际上限定了B≠0,这恰是lny≠0的体现;因此若将lny=0补充进来则lny/x=C,这里的C应该为任意实数注意在这里仍有x≠0,若再变形为lny=Cx这就把x=0也补充进来了,因为原方程只有一个限定条件y>0;所以解题过程中多余的约束应当适时解除若再变形y=e^{Cx}=e^{C}*e^{x}=D*e^{x}注意这里C取任意实数,正对应着D>0因此最后这个积分常数应当受到限制,这也恰是y>0的体现!

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