Calculus of variations
First we solve the equation
2 yu(y) d
dy
u(y)+u^2 (y)= 0.
Ifu(y)=0 theny^0 (x)=0 theny(x)is a constant which is not a
solution. We can simplify byu
2 ydu
dy
+u = 0 ,^1
u
du= ^1
2 y
dy
lnu = ^1
2
lny+c
u = cp^1
jyj
,y^0 =cp^1
jyj