$ \int_{0}^{1} ( y''^{2} + x^{2} - y^{2} ) dx $

$f=y''^{2}-y^{2}+x^{2}$ --------------(1) Since it is a function of second order derivative, $\therefore$ Differentiating (1) w.r.t. y. $\cfrac{\delta f}{\delta y}=-2y$ --------------(2) $\therefore$ Differentiating (1) w.r.t. y''. $\cfrac{\delta f}{\delta y''}=2y''$ --------------(3) $\therefore$ Differentiating (1) w.r.t. y'. $\cfrac{\delta f}{\delta y'}=0$ --------------(4) Therefore, the equation becomes, $\cfrac{\delta f}{delta y}-\cfrac{d}{dx} \left( \cfrac{\delta f}{\delta y'} \right) + \cfrac{d^{2}}{dx^{2}} \left( \cfrac{\delta f}{\delta y''} \right) = 0$ --------------(5) Put (2), (3), (4) in (5), $-2y-\cfrac{d}{dx}(0)+\cfrac{d^{2}}{dx^{2}}(2y'')=0$ $-2y+\cfrac{d^{2}}{dx^{2}}\left(2 \cfrac{d^{2}y}{dx^{2}} \right)=0$ $-2y+2\cfrac{d^{4}y}{dx^{4}}=0$ $2\cfrac{d^{4}y}{dx^{4}}=2y$ $\cfrac{d^{4}y}{dx^{4}}=y$ $\cfrac{d^{4}y}{dx^{4}}-y=0$ --------------(6) This is a linear equation. $\therefore$ A. E. equation is given by (i.e. in terms of D), Equation (6) becomes, $\therefore D^{4}-1=0 \Longrightarrow [(D^{2})^{2}-(1)^{2}]=(D^{2}-1)(D^{2}+1)=0$ So, $D^{2}=1$ or $D^{2}=-1$ $\therefore D=\pm 1$ OR $D=\pm i$ $\therefore$ The solution is: $y=C_{1} e^{x}+ C_{2} e^{-x} + C_{3} \cos {x} + C_{4} \sin {x}$