分类：教育/科学-理工学科-数学
问题：求解微分方程如图 
如图
回答：∫{x→a}√[1+f'(u)]du=[f(x)-xf'(x)]/c1.x=a==>f(a)-af'(a)=02.原式两边求导得:-√[1+f'(x)]=-xf''(x)/c==>x/x=d[f'(x)]/√[1+f'(x)]两边积分得:clnx+D=2√[1+f'(x)]3.两边平方得:f'(x)=[clnx+D]^2/4-1两边积分得:f(x)=[c^2/2-1]x+x[clnx+D]^2/4-cx[clnx+D]/2+E.4.f(a)-af'(a)=0==>[c^2/2-1]a+a[clna+D]^2/4-ca[clna+D]/2+E==a{[clna+D]^2/4-1}==>ac^2/2-ca[clna+D]/2+E=0   ((1))5.将4.的解代入原方程:c∫{x→a}√[1+f'(u)]du=[f(x)-xf'(x)]验证c∫{x→a}√[1+f'(u)]du==c∫{x→a}[clnu+D]/2du==ca[clna+D]/2-cx[clnx+D]/2-c^2/2∫{x→a}du==ca[clna+D]/2-cx[clnx+D]/2-c^2[a-x]/2   ((2))[f(x)-xf'(x)]==[c^2/2-1]x+x[clnx+D]^2/4-cx[clnx+D]/2-ac^2/2++ca[clna+D]/2-x[(clnx+D)^2/4-1]==c^2x/2-cx[clnx+D]/2-ac^2/2+ca[clna+D]/2  ((3))==>((2))=((3))==>f(x)=[c^2/2-1]x+x[clnx+D]^2/4-cx[clnx+D]/2--ac^2/2+ca[clna+D]/2为原方程的通解.注意:只研究c>0,clnx+D≥0的情况.若clnx+D<0,解类似.