∵y''+arctanx=0 　　==>y''=-arctanx 　　==>y'=-∫arctanxdx=(1/2)ln(1+x^2)-xarctanx+C1*(应用分部积分法,C1*是常数) 　　∴y=∫[(1/2)ln(1+x^2)-xarctanx+C1]dx 　　=(x/2)ln(1+x^2)-(x^2/2)arctanx+(1/2)arctanx+(C1*-1/2)x+C2(应用分部积分法,C2是常数) 　　=(x/2)ln(1+x^2)-(x^2/2)arctanx+(1/2)arctanx+C1x+C2(令C1=C1*-1/2) 　　故原方程的通解是y==(x/2)ln(1+x^2)-(x^2/2)arctanx+(1/2)arctanx+C1x+C2(C1,C2是常数).