Calculus of variations
This is again an equation with separable variablescq
jyjdy = dx
c 1 jyj^3 /^2 = x+c 2
y = c 33q
(x+c 2 )^2 ,y( 0 )= 1 ,y( 1 )=^3p
41 = c 33q
c 22 ,^3p
4 =c 33q
( 1 +c 2 )^2There are two solutionsc 3 = c 2 = 1 ,y=^3q
(x+ 1 )^2c 3 =^3p
9 ,c 2 = ^1
3,y=^3p
93s
x ^1
32
=^3
q
( 3 x 1 )^2.