asymptotic analysis to solve differenial equation

Question

$$ \frac{df(x)}{dx}+\bigg(1+\frac{1}{x}\bigg)f(x)=0 $$

how to solve above differential equation using asymptotic analysis ?

Does that give an exact solution ?

Answer 1

For large values of $|x|$, the term $\frac{1}{x}$ vanishes and the solution to the given differential equation becomes close to the solution to the following differential equation

$$\frac{d g(x)}{d x}+g(x)=0 \tag{1}$$

i.e. for $|x|\rightarrow \infty $ we have $f(x)\rightarrow g(x)$. $(1)$ is a linear differential equation with constant coefficients and the solution is $g(x)=e^{-x}$.

Now we can try to express $f(x)$ as $f(x)=h(x)g(x)=h(x)e^{-x}$ and see if by substituting this form of $f(x)$ to the original differential equation gives a simpler equation:

$$\frac{d}{dx}(h(x)e^{-x})+(1+\frac{1}{x})h(x)e^{-x}=0$$

After simplification we get

$$\frac{dh(x)}{dx}+\frac{1}{x}h(x)=0$$

which is not that much of a simplification since this equation is also separable as the initial one. Either way, separating the variables gives

$$ \frac{dh}{h}=-\frac{dx}{x}$$

and the solution is $h(x)=-\frac{C}{x}$, where $C$ is the constant of integration.

Therefore the solution to the initial differential equation is

$$f(x)=h(x)g(x)=\frac{Ce^{-x}}{x}$$

But this result also follows directly from integrating

$$\frac{df}{f}=-(1+\frac{1}{x})dx$$