Solution:

$F=y'^2 -y^2+2xy$-----(1)

It contain both the explicity,

diff. (1) w..r.t.y.

$\frac {\delta f }{\delta y} =0-2y+2x$

$\frac {\delta f }{\delta y} =-2y+2x$------(2)

diff.(1) w.r.t y'

$\frac {\delta f }{\delta y} =2y'-0+0$

$\frac {\delta f }{\delta y'} =2y$ -----(3)

It contains both the explicitly ,

by using eulers equation,

$\frac {\delta f }{\delta y} -\frac {d }{d x} (\frac {\delta f }{\delta y'}) =0$

$-2y+2x-\frac {d }{d x} (2y') =0$

$-2y+2x-\frac {d }{d x} (2\frac{dy}{dx}) =0$

$\frac {d^2 y }{d x^2}+y=x$------(4)

The algebric equation is

$D^2+1=0$

$D^2=-1$

$D=i \ and \ -i $

The C.F.is $y=c_1cos x +c_2sin x $

$P.I.= \frac {1}{D^2+1}x$

$\frac {1}{1+D^2} x $

=$x(1+D^2)^{-1}$

$=x(1-D^2 +D^4+-----)=x$

the complete solution is ,

$y=c_1cos x +c_2 sin x + x$------(5)

when x=0 ,y=0 in eqn (5)

$0= c_1 cos 0 + c_2 sin 0 +0$

$0= c_1(1) + 0 +0 $

$0=c_1$

when x=$\frac{\pi }{2} ,y=0 , in (1) $

$0= c_1cos\frac{\pi}{2} +c_2 sin\frac{\pi}{2} +\frac{\pi}{2} $

$0=0+c_2(1)+\frac{\pi}{2} $

$c_2=-\frac{\pi}{2} $

Put $c_1 \ and \ c_2$ values in eqn (5)

$y= (0) cos x -\frac{\pi}{2} sin x + x $

$y=-\frac{\pi}{2} sin x + x$