Solution:

$\int_{x_0}^{x_1}2xy+y^{''2}$----(1)

diff. w.r.t y

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

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

$\frac {\delta f}{\delta y'}=0$---(3)

diff. w.r.t. y''

$\frac {\delta f}{\delta y''}=-2y''$----(4)

By using equation

$\frac {\delta f}{\delta y}-\frac{d}{dx}(\frac {\delta f}{\delta y'})+\frac{d^2}{dx^2}(\frac {\delta f}{\delta y''}) = 0$--------------(5)

put (2),(3),(4) in (5)

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

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

$2x-2(\frac {d^4 y}{d x^4})=0$

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

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

$(\frac {d^4 y}{d x^4})-x=0$

This is linear differential equation of the fourth order..

A.E. is $D^4=0$

i.e.$D=0, 0, 0, 0$

the C.F. is y=$c_1+c_2x+c_3x^2+c_4x^3$

P.I. , $y= \frac{1}{D^4} x$

$y= \frac{1}{D^3}\int x dx=\frac{1}{D^3} \frac{x^2}{2}$

$y= \frac{1}{D^2}\int \frac{x^2}{2} dx =\frac{1}{D^2} \frac{x^3}{3.2}$

$y= \frac{1}{D}\int \frac{x^3}{6} dx=\frac{1}{D} \frac{x^4}{6.4}$

$y= \int \frac{x^4}{24} dx=\frac{x^5}{24.5}$

$y=\frac{x^5}{120}$

Hence solution is

$y=c_1+c_2x+c_3x^2+c_4x^3+\frac{x^5}{120}$