Find the infimum of the set $\{ \lambda: \lambda \in \mathbb{R}, x^2y''-3xy'-\lambda y =0, \lim_{x \to \infty } y(x)=0 \}$. We have a second order Cauchy-Euler ODE $ x^2y''-3xy'-\lambda y=0$. Assuming $x >0$ and letting $x =e^t$, we can solve this ODE.
The auxiliary equation becomes $r(r-1)-3r- \lambda=0$, which gives $r=\frac{4\pm \sqrt{16+4\lambda}}{2}$=$2\pm \sqrt{4+\lambda}$. Now, when $\lambda \ge-4$, we get real roots and when  $\lambda <-4$, we get complex roots. But I can't think of the use of the given limit in the question and how to find infimum using this information. Kindly help . Thanks.
 A: Here is how I'm interpreting the given set: It is the set of all those $\lambda \in \Bbb R$ such that any solution $y$ of the given ODE satisfies $\lim_{x \to \infty} y(x) = 0$.

Case 1. $\lambda > -4$.
Thus, the auxiliary equation has distinct real roots $r_1$ and $r_2$ which means that the general solution of the Cauchy Euler ODE is given as $$a x^{r_1} +b x^{r_2}.$$
This will have limit $0$ at $\infty$ iff $r_1$ and $r_2$ are both negative. But the root $2 + \sqrt{4 + \lambda}$ is positive.
Case 2. $\lambda = -4$.
Now, the roots are equal (and real). The general solution here is $$x^2(a + b \ln x).$$ Again, the general solution does not vanish at $\infty$.
Case 3. $\lambda < -4$.
The roots are of the form $2 \pm \iota \gamma$ for $\gamma = \sqrt{-(\lambda + 4)}$. Then, the general solution is written as $$x^2(a \sin(\gamma \ln x)) + b \cos(\gamma \ln x)).$$
Again, the general solution does not vanish at $\infty$.

Thus, the given set is empty and (depending on your convention) the infimum should be $+\infty$.
