- 求解微分方程如图
- 如图
- ∫{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.