The extremal of the functional

\(J(y)=\int_0^1\left[2\left(y^{\prime}\right)^2+x y\right] d x,\) y(0) = 0, y(1) = 1, y ∈ C²[0, 1] is

Concept: If \(J(y)=\int_0^1[F(x,y,y')] d x,\) then its extremum is

\(\dfrac{\partial F}{\partial y}-\dfrac{d}{dx}(\dfrac{\partial F}{\partial y'})=0\)

Explanation:

\(J(y)=\int_0^1\left[2\left(y^{\prime}\right)^2+x y\right] d x,\)  y(0)=0, y(1) = 1,  \(y\in C^2[0,1]\)

According to the question 

\(F(x,y,y')=2(y')^2+xy\)

\(\dfrac{\partial F}{\partial y}-\dfrac{d}{dx}(\dfrac{\partial F}{\partial y'})\) = 0

\(\dfrac{\partial (2(y')^2+xy)}{\partial y}-\dfrac{d}{dx}(\dfrac{\partial (2(y')^2+xy)}{\partial y'})\) = 0

⇒ \( x-\dfrac{d}{dx}(4y')\) = 0

⇒ x - 4y'' = 0

\(y''=\dfrac{x}{4}\)

after integrating  with respect to x, we get

\(y'=\dfrac{x^2}{8}+c\)

again integrating with respect to x, we get

\(y(x)=\dfrac{x^3}{24}+cx+d\)

where c and d are constants of integration

now, \(y(0)=d=0\)

and \(y(1)=\dfrac{1}{24}+c=1⇒c=\dfrac{23}{24}\)

Hence our extremum would be \(y(x)=\dfrac{x^3}{24}+\dfrac{23x}{24}\)

Hence option (4) is correct