The problem I encounter is that: find the function $y(x)$ with $y(1)=1$ $y'(1)=-2$ $y(2)=1/4$ and $y'(2)=-1/4 $ that minimises $ \int_1^2{x^4[y''(x)]^2dx}.$
I know the equation I should use is $$\frac{\partial f}{\partial y}-\frac{d}{dx}\frac{\partial f}{\partial y'}+\frac{d^2}{dx^2}\frac{\partial f}{\partial y''}=0.$$
In this case, I think the only remaining part is$$\frac{d^2}{dx^2}\frac{\partial f}{\partial y''}$$
while $$\frac{d}{dx}(2x^4y'')=8x^3y''+2x^4y^{(3)}=0.$$
I think something is wrong with it, and I also don't know how to solve this differential equation.
Could anyone give me some advice on it? Thanks.
Developping Ninad Munshi's comment, if $(2x^4y'')'' = 0$, then it means that $(2x^4y'')'$ is a constant, so that the successive integrations are (with $C_1,C_2,C_3,C_4$ some constants) : $$ \begin{array}{rcl} (2x^4y'')'' &=& 0 \\ (2x^4y'')' &=& C_1 \\ 2x^4y'' &=& C_1x + C_2 \\ y'' &=& \displaystyle \frac{C_1}{2x^3} + \frac{C_2}{2x^4} \\ y' &=& \displaystyle -\frac{C_1}{4x^2} - \frac{C_2}{6x^3} + C_3 \\ y &=& \displaystyle \frac{C_1}{4x} + \frac{C_2}{12x^2} + C_3x + C_4 \\ y &=& \displaystyle Ax + B + \frac{C}{x} + \frac{D}{x^2} \end{array} $$ after relabelling the constants of integration, which you will determine thanks to the initial/boundary conditions.