Discussion
Let \(y\) be a non-zero quadratic polynomial satisfying the differential equation
\[ (2 + x^{2})\frac{d^{2}y}{dx^{2}} + x\frac{dy}{dx} - ky = 0, \]
where \(k\) is a real constant. If \(y(1) = 1\), then the value of the integral \(\int_{0}^{1} 2y\,dx\) is
Answer: D
The value of the integral is \(\frac{4}{3}\).
Try \(y = ax^{2} + bx + c\) with \(a \neq 0\). Substituting \(y'' = 2a\) and \(y' = 2ax + b\):
\[ \begin{aligned} &(4a - ka)x^{2} + (b - kb)x \\ &\quad + (4a - kc) = 0 \end{aligned} \]
for all \(x\). The \(x^{2}\) term gives \(k = 4\); then \(-3b = 0\) gives \(b = 0\), and \(4a - 4c = 0\) gives \(c = a\). So \(y = a(x^{2} + 1)\), and \(y(1) = 2a = 1\) gives \(a = 1/2\):
\[ \begin{aligned} \int_{0}^{1} 2y\,dx &= \int_{0}^{1} (x^{2} + 1)\,dx \\ &= \frac{1}{3} + 1 = \frac{4}{3} \end{aligned} \]
\(\frac{1}{3}\) keeps only the \(x^{2}\) term, and \(\frac{2}{3}\) and 1 do not follow from \(y = (x^{2} + 1)/2\).
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