behavior of non-linear differential equation The non-linear differential equation is defined as following
$$\begin{cases}\frac{dx}{dt}=-\frac{1}{t^n}x+\frac{1}{t^m} (t>1)\\
x(1)=x_0 \end{cases}$$
where $n,m$ are positive integers. I want to find all pairs $(n,m)$ such that there exists $\lim_{t\rightarrow \infty}x(t)$. 
My idea: First I solved the linear differential equation 
$$ \frac{dx}{dt}=-\frac{1}{t^n}x $$
If $n=1$, the solution is $x(t)=\frac{A}{t} (A\in \mathbb{R})$ . Regarding $A$ as the function of $t$ I substituted the solution and got $A=-\frac{1}{mt^m}$ and $\lim_{t\rightarrow \infty}x(t)=0$. I am in trouble when $n\geq 2$. Similarly I got 
$$ x(t)=A\exp\left(\frac{1}{(n-1)t^{n-1}}\right) $$
as the solution of the above linear equation. Regarding $A$ as the function again, I got 
$$ \frac{dA}{dt}=\frac{1}{t^m}\exp\left(-\frac{1}{(n-1)t^{n-1}}\right)$$
But I think this differential equation can't be solved generally. When $\lim_{t\rightarrow \infty}x(t)$ converges? Please give me some advice. 
 A: First: the integral
\begin{equation}
 \int_1^t \tau^{-m} \text{exp} \left(-\frac{1}{(n-1)\tau^{n-1}}\right)\,\text{d}\tau
\end{equation}
exists, and can be expressed in terms of the incomplete Gamma function, see also DLMF and MathWorld. Introducing now
\begin{equation}
 s = \frac{t^{1-n}}{n-1} \quad \Longleftrightarrow \quad t = \left((n-1)s\right)^{\frac{1}{1-n}}, \tag{1}
\end{equation}
the solution you find is expressed as
\begin{equation}
 x(s) = e^s \left(c_1 + (n-1)^{\frac{m-n}{n-1}} \Gamma\left(\frac{m-1}{n-1},s\right)\right), \tag{2}
\end{equation}
with $c_1$ a constant (that can be expressed in terms of $x_0$). Note that you could also have found $(2)$ by using the variable transformation $(1)$ on the original ODE, using $-t^n \frac{\text{d} x}{\text{d} t} = \frac{\text{d} x}{\text{d} s}$. 
In any case, for $n>1$, the limit $t \to \infty$ is replaced by the limit $s \downarrow 0$. Now, using any of the links mentioned above, you can infer that
\begin{equation}
 \lim_{s \downarrow 0} \,\Gamma\left(\frac{m-1}{n-1},s\right) = \Gamma\left(\frac{m-1}{n-1}\right),
\end{equation}
as long as $\frac{m-1}{n-1}>0$. The (complete) Gamma function $\Gamma(z)$ is defined for all positive $z$, so the limit of $(2)$ exists for all $m>1$. 
However, for $m=1$, we have $\Gamma(0,s) = -\text{Ei}(-s)$ (see e.g. 1), and the exponential integral $\text{Ei}(x)$ has a logarithmic singularity as $x \to 0$. To conclude: the original limit $\lim_{t\to \infty} x(t)$ exists for all positive integers $m$, $n$, excluding the set $\{m=1,n>1\}$.
