Calculus of variations
Again introduceu=y^0 .(It is always working whenyis missing.) Then
0 = u+xu^0.
^1
x
dx =^1
u
du
lnjxj = lnjuj+c
c
x
= u=y^0
y = c 1 lnx+c 2
Using the initial conditions the extremal solution isy=lnx.
Again introduceu=y^0 .(It is always working whenyis missing.) Then
0 = u+xu^0.
^1
x
dx =^1
u
du
lnjxj = lnjuj+c
c
x
= u=y^0
y = c 1 lnx+c 2
Using the initial conditions the extremal solution isy=lnx.