TITLE: A funtion equation $2f(x)+f''(x)=xf'(x)$
QUESTION [5 upvotes]: Suppose $f(x)\in C^2(\mathbb R)$ solves the equation $$
2f(x)+f''(x)=xf'(x),\forall x\in\mathbb R
$$
Assume that $(f(0),f'(0))\neq (0,0)$. Prove that $(f(x),f'(x))$ is unbounded.

REPLY [4 votes]: Actually $f(x)$ itself is unbounded as long as $f(0)$ and $f'(0)$ are not both $0$.  One solution to the DE is $f_1(x) = x^2-1$.  Writing $f(x) = u(x) f_1(x)$, the DE becomes $(x^2-1) u''(x) + (- x^3 + 5 x) u'(x) = 0$, which has general solution
$$ u'(x) = C \ \exp \int \left(\dfrac{x^3-5x}{x^2-1}\right)\; dx $$
Now the integrand $ > x/2$, say, for large enough $x$.  Thus
if $C > 0$, $u'(x) > C (\exp(x^2/4) + D)$ for some constant $D$, and in particular $u'(x) > C \exp(x)$ for sufficiently large $x$.  Integrate and
we have $u(x) > C \exp(x) + (constant)$ for sufficiently large $x$, and then $f(x) = u(x) (x^2 - 1)$ is unbounded.  Similarly if $C < 0$.  For $C = 0$, the solution is
a constant times $f_1(x)$, which is unbounded unless it is identically $0$.