Question

求通解(1+x^2)y'+y=arctanx

Answer 1

∵(1+x^2)y'+y=arctanx

==>[(1+x^2)y'+y]e^(arctanx)/(1+x^2)=arctanx*e^(arctanx)/(1+x^2)

(等式两端同乘e^(arctanx)/(1+x^2))

==>e^(arctanx)dy+ye^(arctanx)dx/(1+x^2)=arctanx*e^(arctanx)dx/(1+x^2)

==>e^(arctanx)dy+yd(e^(arctanx))=arctanxd(e^(arctanx))

==>d(ye^(arctanx))=arctanxd(e^(arctanx))

==>ye^(arctanx)=arctanx*e^(arctanx)-e^(arctanx)+C (应用分部积分法,C是常数)

==>y=arctanx-1+Ce^(-arctanx)

∴原方程的通解是y=arctanx-1+Ce^(-arctanx).