x(1+x^2)dy=(y+x^2y-x^2)dx通解
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x(1+x^2)dy=(y+x^2y-x^2)dx通解
x(1+x^2)dy=(y+x^2y-x^2)dx通解
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=x(C-arctanx)
故原方程的通解是y=x(C-arctanx).
x(1+x^2)dy=(y+x^2y-x^2)dx通解
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