∵y=Cx (C是常數)是齊次方程x(1+x^2)dy=(1+x^2)ydx的通解∴設原方程的解為y=C(x)x (C(x)是關于x的函數)∵代入原方程,化簡得 C'(x)(1+x^2)=-1==>C'(x)=-1/(1+x^2)==>C(x)=-∫dx/(1+x^2)=C-arctanx (C是常數)∴y=C(x)x...代入原方程化簡等于-1那段不是很明白代入哪個方程額代入原方程x(1+x^2)dy=(y+x^2y-x^2)dx這個怎么帶出來的可以寫下嗎這地方沒明白你怎么運算能力這么差喲？
y=C(x)x ==>dy=[C'(x)x+C(x)]dx
==>x(1+x^2)[C'(x)x+C(x)]dx=(y+x^2y-x^2)dx

==>x^2(1+x^2)C'(x)+x(1+x^2)C(x)=x(1+x^2)C(x)-x^2
==>x^2(1+x^2)C'(x)=-x^
==>(1+x^2)C'(x)=-1謝謝！晚上腦袋不轉了呵呵