​$$\text{The Euler equations satisfied by the extremals of the functional} \\[10pt]I(y) = \int_{0}^{5} \left[ y^2 + x^3 y' \right] dx \\[10pt]\text{define a solution curve in the } (x, y) \text{-plane which is}$$​​

Correct option is B

​$$F = y^2 + x^3 y' \\[10pt]\text{Using the Euler equation:} \\[10pt]\frac{\partial F}{\partial y} - \frac{d}{dx} \left( \frac{\partial F}{\partial y'} \right) = 0 \\[10pt]\frac{\partial F}{\partial y} = 2y, \quad \frac{\partial F}{\partial y'} = x^3 \\[10pt]\text{Substitute into the Euler equation:} \\[10pt]2y - \frac{d}{dx}(x^3) = 0 \\[10pt]2y - 3x^2 = 0 \\[10pt]\Rightarrow y = x^2 \\[10pt]\text{Thus, } y \text{ is quadratic.}$$​​