Calculus of variations
Again introduceu=y^0 .(It is always working whenyis missing.) Then0 = u+xu^0.
^1
xdx =^1
udu
lnjxj = lnjuj+c
c
x= u=y^0
y = c 1 lnx+c 2Using the initial conditions the extremal solution isy=lnx. Again introduceu=y^0 .(It is always working whenyis missing.) Then0 = u+xu^0.
^1
xdx =^1
udu
lnjxj = lnjuj+c
c
x= u=y^0
y = c 1 lnx+c 2Using the initial conditions the extremal solution isy=lnx.