x（1+x^2）dy=（y+x^2y-x^2）dx通解

∵y=Cx (C是常数)是齐次方程x(1+x^2)dy=(1+x^2)ydx的通解∴设原方程的解为y=C(x)x (C(x)是关于x的函数)∵代入原方程,化简得 C'(x)(1+x^2)=-1==>C'(x)=-1/(1+x^2)==>C(x)=-∫dx/(1+x^2)=C-arctanx (C是常数)∴y=C(x)x...代入原方程化简等于-1那段不是很明白代入哪个方程额代入原方程x(1+x^2)dy=(y+x^2y-x^2)dx这个怎么带出来的可以写下吗这地方没明白你怎么运算能力这么差哟？
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谢谢！晚上脑袋不转了呵呵