What is the extremal of the functional: 

​$$J[y] = \int_{-1}^0 \left( 12xy - (y')^2 \right) dx$$

subject to y(0) = 0 and y(−1) = 1 ?​​

Correct option is C

​$$\text{The functional is given as:} \\[10pt]J[y] = \int_{-1}^0 \left( 12xy - (y')^2 \right) dx. \\[10pt]\text{Let:} \\[10pt]F = 12xy - (y')^2. \\[10pt]\text{Using the Euler-Lagrange equation:} \\[10pt]\frac{\partial F}{\partial y} - \frac{d}{dx} \left( \frac{\partial F}{\partial y'} \right) = 0. \\[10pt]\textbf{Compute derivatives} \\[10pt]\frac{\partial F}{\partial y} = 12x, \quad \frac{\partial F}{\partial y'} = -2y'. \\[10pt]\frac{d}{dx} \left( \frac{\partial F}{\partial y'} \right) = \frac{d}{dx}(-2y') = -2y''. \\[10pt]\text{Substituting into the Euler-Lagrange equation:} \\[10pt]12x - (-2y'') = 0 \implies 12x + 2y'' = 0. \\[10pt]\textbf{Solve the differential equation} \\[10pt]y'' = -6x. \\[10pt]\text{Integrating once:} \\[10pt]y' = \int -6x \, dx = -3x^2 + A, \\[10pt]\text{where } A \text{ is a constant.} \\[10pt]\text{Integrating again:} \\[10pt]y = \int (-3x^2 + A) \, dx = -x^3 + Ax + B, \\[10pt]\text{where } B \text{ is another constant.} \\[10pt]\textbf{Apply boundary conditions} \\[10pt]1. \text{From } y(0) = 0: \\[10pt]0 = -0^3 + A(0) + B \implies B = 0. \\[10pt]2. \text{From } y(-1) = 1: \\[10pt]1 = -(-1)^3 + A(-1) + 0 \implies 1 = 1 - A \implies A = 0. \\[10pt]\text{Thus, the solution is:} \\[10pt]y = -x^3. \ (\textbf{Option C}) \\[10pt]\textbf{Final Answer:} \\[10pt]\textbf{Option (C): } y = -x^3.$$

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