求微分方程y"=arctan x 的通解
发布网友
发布时间：2024-04-27 08:50
共1个回答
热心网友
时间：2024-04-28 07:33
两边积分：y'=xactanx-∫x/(1+x^2)dx=xarctanx-1/2∫d(x^2+1)/(x^2+1)=xarctanx-1/2ln(x^2+1)+C1
两边积分：y=1/2∫arctanxd(x^2)-1/2xln(x^2+1)+1/2∫x*2x/(x^2+1)dx+C1x=1/2x^2arctanx-1/2∫x^2/(1+x^2)dx-1/2xln(x^2+1)+∫x^2/(x^2+1)dx+C1x=1/2x^2arctanx-1/2xln(x^2+1)+1/2∫(x^2+1-1)/(x^2+1)dx+C1x=1/2x^2arctanx-1/2xln(x^2+1)+1/2∫dx-1/2∫1/(x^2+1)dx+C1x=1/2x^2arctanx-1/2xln(x^2+1)+1/2x-1/2arctanx+C1x+C2=1/2x^2arctanx-1/2xln(x^2+1)-1/2arctanx+C1x+C2
