Let $y(x) = x^n [c_1 \cos(\ln x) + c_2 \sin(\ln x)]$ for $x 0$, where $c_1, c_2,$ and $n$ are real

Question

Let $y(x) = x^n [c_1 \cos(\ln x) + c_2 \sin(\ln x)]$ for $x > 0$, where $c_1, c_2,$ and $n$ are real constants. The function $y(x)$ satisfies a second-order linear differential equation of the form $x^2 y'' + p(n) xy' + q(n) y = 0$. What is the minimum value of $p(n) + q(n)$ as $n$ varies over all real numbers?

Professor PC Thomas Classes · Answer

Let $u = c_1 \cos(\ln x) + c_2 \sin(\ln x)$. Differentiating $u$ with respect to $x$, we get $u' = \frac{1}{x}[-c_1 \sin(\ln x) + c_2 \cos(\ln x)]$, so $xu' = -c_1 \sin(\ln x) + c_2 \cos(\ln x)$. Differentiating again, $x u'' + u' = \frac{1}{x}[-c_1 \cos(\ln x) - c_2 \sin(\ln x)] = -\frac{u}{x}$. Thus, $x^2 u'' + xu' + u = 0$. We are given $y = x^n u$, which means $u = y x^{-n}$. Taking derivatives, we get $u' = y' x^{-n} - n y x^{-n-1}$ and $u'' = y'' x^{-n} - 2n y' x^{-n-1} + n(n+1) y x^{-n-2}$. Substituting these expressions into the differential equation for $u$, we get $x^2(y'' x^{-n} - 2n y' x^{-n-1} + n(n+1) y x^{-n-2}) + x(y' x^{-n} - n y x^{-n-1}) + y x^{-n} = 0$. Dividing the entire equation by $x^{-n}$, we obtain $x^2 y'' - 2nxy' + n(n+1)y + xy' - ny + y = 0$, which simplifies to $x^2 y'' + (1-2n)xy' + (n^2+1)y = 0$. By comparing this with the given form $x^2 y'' + p(n) xy' + q(n) y = 0$, we identify $p(n) = 1-2n$ and $q(n) = n^2+1$. Therefore, the sum is $p(n)+q(n) = n^2-2n+2 = (n-1)^2+1$. Since $(n-1)^2 \ge 0$ for all real $n$, the minimum value of $p(n)+q(n)$ is $1$, which occurs when $n=1$.