Discussion
The functional
\(\int_{0}^{1} \left(y'^{2} + (y + 2y')y'' + kxyy' + y^{2}\right) dx\), \(y(0) = 0, y(1) = 1, y'(0) = 2, y'(1) = 3\)
is path independent if \(k\) equals
Answer: B
The functional is path independent when \(k = 2\).
A functional is path independent when its integrand is an exact derivative \(dG/dx\), since then the integral depends only on the end values. Group the terms:
\[ \begin{aligned} y'^{2} + yy'' &= (yy')' \\ 2y'y'' &= (y'^{2})' \\ y^{2} + 2xyy' &= (xy^{2})' \end{aligned} \]
So the integrand is \((yy' + y'^{2} + xy^{2})'\) exactly when \(kxyy' = 2xyy'\), that is \(k = 2\). For any other \(k\), the leftover term \((k - 2)xyy'\) is not a total derivative (its Euler-Lagrange expression \(-(k-2)y\) does not vanish), so the value would depend on the path. Hence 1, 3 and 4 are wrong.
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