I'm trying to solve an optimal stopping problem which led me to an obstacle problem involving the following family of ODE's


$$(x^2+d)y'(x)-2xy(x) = 1.$$


For simplicity I first considered the case $d = 0$ before moving to the much more relevant case $d > 0$. This revealed a somewhat unexpected phenomenon to me: The solutions are of the form
$C(x^2+d) + f(x)$, where $C$ is some constant. For $d = 0$ we have $f(x) = -\frac{1}{3x}$ which is negative for $x > 0$. On the other hand, for $d > 0$ we have
$$f(x) = \frac{x}{2 d}+ \frac{(x^2+d)\tan^{-1}\left(\frac x {\sqrt d}\right)}{2 d^{3/2}},$$
which is positive and $f(x)$ diverges as $d\to 0$; instead of converging to, say $x\mapsto -\frac 1 {3x}$.


Is there some theoretic result that shows that this must be the case? If not is there some result that would show convergence, but we can verify that its assumptions are violated?



(Non-) Convergence of solutions in a family of linear ODE's
Using the expansion for $\arctan(t)$ as $t\to +\infty$ in the form
$$\arctan(t) = \frac{\pi}{2}-\frac{1}{t}+\frac{1}{3t^3}+\cdots$$
we get, for fixed $x>0$ and positive $d\to 0$:
$$ f\_d(x) \sim \frac{\pi(x^2+d)}{4d^{3/2}} =: C\_d(x^2+d) $$
and further (that's where the magic cancellation happens) $$g\_d(x):=f\_d(x)-C\_d(x^2+d)\sim \frac{x}{2d}-\frac{(x^2+d)}{2d^{3/2}}\left( \frac{\sqrt{d}}{x}-\frac{d^{3/2}}{3x^3} \right) \\ = \frac{d^2-2dx^2}{6dx^3}\to -\frac{1}{3x}$$


There's nothing deep happening here, except the necessity to keep track of the constant $C$ as the parameter varies (and also of the intervals definition for maximal solutions, which hasn't been checked above).


