Calculus of variations
This is again an equation with separable variables
c
q
jyjdy = dx
c 1 jyj^3 /^2 = x+c 2
y = c 33
q
(x+c 2 )^2 ,y( 0 )= 1 ,y( 1 )=^3
p
4
1 = c 33
q
c 22 ,^3
p
4 =c 33
q
( 1 +c 2 )^2
There are two solutions
c 3 = c 2 = 1 ,y=^3
q
(x+ 1 )^2
c 3 =^3
p
9 ,c 2 = ^1
3
,y=^3
p
93
s
x ^1
3
2
=^3
q
( 3 x 1 )^2.