Suppose that there exist some constant $E$ such that
$$\frac{d^2}{dx^2}\Psi + \frac{1}{x^2}\Psi - \frac{1}{x}\Psi = E\Psi$$
and suppose that $\Psi$ is of the form
$$\Psi = ke^{ax^b}$$
for some chosen constants $k, a, b$.
Then,
$$\frac{d^2}{dx^2}\Psi + \frac{1}{x^2}\Psi - \frac{1}{x}\Psi$$ $$=\frac{d^2}{dx^2}(ke^{ax^b}) + \frac{1}{x^2}ke^{ax^b} - \frac{1}{x}ke^{ax^b}$$ $$=kab\frac{d}{dx}(x^{b-1}e^{ax^b}) + \frac{1}{x^2}ke^{ax^b} - \frac{1}{x}ke^{ax^b}$$
$$=kab(abx^{b-1}x^{b-1}e^{ax^b} + (b-1)x^{b-2}e^{ax^b}) + \frac{1}{x^2}ke^{ax^b} - \frac{1}{x}ke^{ax^b}$$ $$(a^2b^2x^{2b-2}+ ab(b-1)x^{b-2} + \frac{1}{x^2} - \frac{1}{x})ke^{ax^b}$$ $$=E\Psi$$
Is there a clever choice of constants $a, b$ such that
$$a^2b^2x^{2b-2}+ ab(b-1)x^{b-2} + \frac{1}{x^2} - \frac{1}{x} = E$$
can just be a constant number $E$ (with all its $x$ dependence removed)?
Alternatively, can there be other forms of $\Psi$ that can be a solution to this equation?