Problems

Solution: \begin{align*} && y &= x^2e^{-x^2} \\ \Rightarrow && y' &= 2xe^{-x^2} +x^2 \cdot (-2x)e^{-x^2} \\ &&&= e^{-x^2}(2x-2x^3) \\ &&&= 2e^{-x^2}x(1-x^2) \end{align*} Therefore the derivative is zero iff \(x = 0, \pm 1\) \begin{align*} && y &= \P(x) e^{-x^2} \\ \Rightarrow && y' &= e^{-x^2} (\P'(x)-2x\P(x)) \end{align*} Therefore we want \(\P'(x) - 2x\P(x) = Kx(x^2-a^2)(x^2-b^2)\) Since this has degree \(5\), we should look at polynomials degree \(4\) for \(\P\). We can also immediately see that \(0\) is a root of \(\P'(x)\), so \(\P(x) = a_4x^4+a_3x^3+a_2x^2+a_0\). WLOG \(a_4 = 1\) and \(K = -2\), so \begin{align*} && -2(x^5-(a^2+b^2)x^3+a^2b^2x) &= 4x^3+3a_3x^2+2a_2x- 2x(x^4+a_3x^3+a_2x^2+a_0) \\ &&&= -2x^5-2a_3 x^4+(4-2a_2)x^3+(2a_2-2a_0)x \\ \Rightarrow && a_3 &= 0 \\ && a^2+b^2 &= 2-a_2 \\ \Rightarrow && a_2 &= 2-a^2-b^2 \\ && a^2b^2 &= a_0-a_2 \\ \Rightarrow && a_0 &= a^2b^2 + 2-a^2-b^2 \\ \Rightarrow && \P(x) &= x^4+(2-a^2-b^2)x^2+(a^2-1)(b^2-1)x \end{align*}

Solution: \begin{align*} && y_0(x) &= e^{-\lambda x^2} \\ && \lim_{x \to \pm \infty} y_0(x) &= 0 \Leftrightarrow \lambda > 0 \\ && \lim_{x \to \pm \infty} y'_0(x) &= \lim_{x \to \pm \infty} 2x\lambda e^{-\lambda x^2} \\ &&&= 0\Leftrightarrow \lambda > 0 \\ && y''_0(x) &= 4x^2 \lambda^2 e^{-\lambda x^2} + 2\lambda e^{-\lambda x^2} \\ \\ && y''_0 - \omega^2 x^2 y_0+(2\cdot 0 + 1) \omega y_0 &= e^{-\lambda x^2} \l 4x^2 \lambda^2 + 2 \lambda - \omega^2 x^2 + \omega\r \\ &&&=0 \Leftrightarrow \lambda = \pm \frac{\omega}{2} \end{align*} Therefore \(y_0\) satisfies if \(\lambda = \frac{\omega}{2}\) Similarly for \(y_1\), \begin{align*} && y_1(x) &= xe^{-\lambda x^2} \\ && \lim_{x \to \pm \infty} y_1(x) &= 0 \Leftrightarrow \lambda > 0 \\ && \lim_{x \to \pm \infty} y'_1(x) &= \lim_{x \to \pm \infty} \l -2x^2 \lambda e^{-\lambda x^2} + e^{-\lambda x^2} \r \\ &&&= 0\Leftrightarrow \lambda > 0 \\ && y''_0(x) &= e^{-\lambda x^2} \l 4x^3 \lambda^2-4x\lambda - 2x\lambda \r \\ &&&= e^{-\lambda x^2} \l 4x^3 \lambda^2-6x\lambda \r \\ && y''_1 - \omega^2 x^2 y_1+(2\cdot 1 + 1) \omega y_1 &= e^{-\lambda x^2} \l 4x^3\lambda^2-6x\lambda-\omega^2x^3+3\omega x\r \\ &&&=0 \Leftrightarrow \lambda = \pm \frac{\omega}{2} \end{align*} Therefore \(y_1\) satisfies if \(\lambda = \frac{\omega}{2}\) \begin{align*} \frac{\mathrm{d}}{\mathrm{d}x}\left(y_{m}\frac{\mathrm{d}y_{n}}{\mathrm{d}x}-y_{n}\frac{\mathrm{d}y_{m}}{\mathrm{d}x}\right) &= y'_my'_n+y_my''_n - y'_ny'_m-y_ny''_m \\ &= y_my''_n - y_ny''_m \\ &= y_m(\omega^2 x^2 y_n - (2n+1)\omega y_n) - y_n(\omega^2 x^2 y_m - (2m+1)\omega y_m) \\ &= y_my_n (2m-2n)\omega \\ &= 2(m-n) \omega y_my_n \end{align*} Therefore: \begin{align*} \int_{-\infty}^{\infty} y_m(x)y_n(x) \d x &= \int_{-\infty}^{\infty} \frac{1}{2(m-n)} \frac{\mathrm{d}}{\mathrm{d}x}\left(y_{m}\frac{\mathrm{d}y_{n}}{\mathrm{d}x}-y_{n}\frac{\mathrm{d}y_{m}}{\mathrm{d}x}\right) \d x \\ &= \frac{1}{2(m-n)} \left [ y_{m}\frac{\mathrm{d}y_{n}}{\mathrm{d}x}-y_{n}\frac{\mathrm{d}y_{m}}{\mathrm{d}x}\right]_{-\infty}^{\infty} \\ &\to 0 \end{align*} This condition is known as Orthogonality. In fact this question is talking about a Sturm-Liouville orthogonality condition, in particular for the quantum harmonic oscillator, and the eigenfunctions are related to Hermite polynomials.