作业帮求y''+arctanx=0通解
来源：学生作业帮 编辑：作业帮 分类：数学作业 时间：2024/05/04 12:53:24
求y''+arctanx=0通解
∵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).
==>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).