Natural-Forced and Transient-SteadyState pairs of solutions We have the following circuit, 
where, $u(0)=V_{0}$.
The ode that describes this circuit that has $V_{s}$ as input and the voltage $u(t)$ of the capacitor as output is the following:
$\dot{u} + \tau u = \tau V_{s}$, where $\tau=\frac{1}{RC}$.

*

*If I solve the ode, this way:
for the homogeneous part, we have
$u_{h}(t)=ce^{\lambda t}$.
After replacing it to the initial equation, we find that $\lambda=-\tau$, and we get
$u_{h}(t)=ce^{-\tau t}.$
To find the particular solution, we consider
$u_{p}(t)=AV_{s}$
and after the replacement we get $A=1$, which makes $u_{p}(t)=V_{s}$
Their sum is:
$u(t)=ce^{-\tau t}+V_{s}$
, and by applying the initial condition, we get that $c=V_{0}-V_{s}$
Therefore, the total solution is:
$u(t)=(V_{0}-V_{s})e^{-\tau t}+V_{s}$
the homogeneous solution coincides with the transient solution, while the particular
one coincides with the steady-state solution

*If I solve it, that way:
The homogeneous part it's the same as before, so:
$u_{h}(t)=ce^{-\tau t}$.
We find the particular solution by using directly the following equation:
$u_{p}(t)=e^{-\int_{0}^{t}\tau  dt }\int_{0}^{t} (\tau V_{s} e^{\int_{0}^{t}\tau  dt })dt=e^{-\tau t}V_{s}(1-e^{\tau t})=V_{s}(1-e^{\tau t})$
Their sum is:
$u(t)=ce^{-\tau t}+V_{s}(1-e^{\tau t})$
, and by applying the initial condition, we get that $c=V_{0}$
Therefore, the total solution is:
$u(t)=V_{0}e^{-\tau t}+V_{s}(1-e^{\tau t})$
the homogeneous solution coincides with the natural solution, while the particular
one coincides with the forced solution
Of course the total solutions are identical but I would be interested in knowing why the one way yields directly the $u_{natural}$-$u_{forced}$ pair, while the other way yields directly the $u_{transient}$-$u_{steady-state}$ pair.
 A: Before answering the main question, let's first check out where the formula appearing in the second method comes from, with just a little bit more generality. The idea is multiplying the left-hand side of the ODE
$$ \dot u + \tau u = \tau V _ s \tag 0 \label 0 $$
with a function $ v $ so that it results in a derivative of a function. This idea comes from the knowledge of the product rule
$$ \frac { \mathrm d } { \mathrm d t } ( u v ) = \dot u v + u \dot v \text . $$
So, if we choose $ v $ such that $ \dot v = \tau v $, we will have
$$ ( \dot u + \tau u ) v = \dot u v + u ( \tau v ) = \dot u v + u \dot v = \frac { \mathrm d } { \mathrm d t } ( u v ) \text , $$
which results in the ODE
$$ \frac { \mathrm d } { \mathrm d t } ( u v ) = \tau v V _ s \text . $$
We can then use the fundamental theorem of calculus to conclude that we must have
$$ u ( t ) v ( t ) = A + \int _ 0 ^ t \tau v ( r ) V _ s ( r ) \ \mathrm d r \tag 1 \label 1 $$
for some constant $ A $, and in case we were lucky enough to find a $ v $ that takes only nonzero values,
$$ u ( t ) = \frac 1 { v ( t ) } \left( A + \int _ 0 ^ t \tau v ( r ) V _ s ( r ) \ \mathrm d r \right) \text . $$
To find a suitable $ v $, we use our knowledge of the properties of the exponential function (which is what helped us finding solutions of the homogeneous part) to see that we must have
$$ v ( t ) = B e ^ { \tau t } \tag 2 \label 2 $$
for some constant $ B $ (the expression $ e ^ { \int _ 0 ^ t \tau ( r ) \ \mathrm d r } $ appearing in your formula is for the more general case where $ \tau $ is not a constant but a function of $ t $; we don't need to complicate things in the case of our problem, and it's enough to note that in case $ \tau $ is constant, we have $ \int _ 0 ^ t \tau ( r ) \ \mathrm d r = \tau t $). Note that in case $ B = 0 $, $ v $ is constantly zero, and when multiplying \eqref{0} by $ v $, we lose all information about $ u $. So we're only interested in the case $ B \ne 0 $, and luckily, in this case $ v $ only takes nonzero values, and we end up with
$$ u ( t ) = \frac 1 { B e ^ { \tau t } } \left( A + \int _ 0 ^ t \tau B e ^ { \tau r } V _ s ( r ) \ \mathrm d r \right) \text , $$
or simply
$$ u ( t ) = e ^ { - \tau t } \left( C + \tau \int _ 0 ^ t e ^ { \tau r } V _ s ( r ) \ \mathrm d r \right) \text , \tag 3 \label 3 $$
where $ C = \frac A B $ is a constant. It's then straightforward to verify that any function $ u $ of this form satisfies \eqref{0}, and thus we've found all the possible solutions (that are regular enough so that we could use the mentioned tools, in particular the fundamental theorem of calculus).

Back to the main question, we know that given any particular solution $ u _ p $ of \eqref{0}, one can find all solutions of \eqref{0} by summing up $ u _ p $ with a solution of the corresponding homogeneous ODE. So, it's all about choosing that particular solution $ u _ p $, and the form of $ u $ as a sum of two terms will be determined. Also, given an initial condition $ u ( t _ 0 ) = V _ 0 $, the solution will be unique. So, for a given initial value problem, the difference can only happen in the way it is presented as sum of two terms, and it is not about the solution itself, which you've already mentioned.
As we know that any particular solution of \eqref{0} is of the form \eqref{3}, it is in fact all about choosing the constant $ C $. One might ask "well, I have to calculate the integral appearing in \eqref{3} anyway; so why bother choosing a nonzero $ C $?". That's one option, and results in decomposing the solution to the natural and forced components, as I will discuss in a moment. But that's not necessarily the most "natural" choice, given a physical interpretation of the problem. From a mathematical point of view, note that for example we could write
$$ u ( t ) v ( t ) = \tilde A + \int _ { \tilde a } ^ t \tau v ( r ) V _ s ( r ) \ \mathrm d r $$
or even
$$ u ( t ) v ( t ) = \hat A - \int _ { - t } ^ { \hat a } \tau v ( - r ) V _ s ( - r ) \ \mathrm d r $$
instead of \eqref{1}, where $ \tilde A $, $ \tilde a $, $ \hat A $ and $ \hat a $ are some constants. These are equivalent to \eqref{1}, which can be seen by setting $ \tilde A = A + \int _ 0 ^ { \tilde a } \tau v ( r ) V _ s ( r ) \ \mathrm d r $ and $ \hat A = A + \int _ 0 ^ { - \hat a } \tau v ( r ) V _ s ( r ) \ \mathrm d r $. The fact that the latter is "unnatural" is because one cannot interpret the dummy variable $ r $ properly (it's something like time in reverse, which is not something "natural"), but what about the former? Similarly, we could consider the expression
$$ v ( t ) = \tilde B e ^ { \tau \left( t - \tilde b \right) } $$
instead of \eqref{2}, which is again equivalent to it, as can be seen by setting $ \tilde B = B e ^ { \tau \tilde b } $. So, why would one choose \eqref{1} and \eqref{2} over these? Equivalently, why would one prefer $ \tilde a = 0 $ and $ \tilde b = 0 $? The answer is again in the interpretation of the problem. The initial condition was given to us with $ t _ 0 = 0 $; i.e. our system is supposed to start to work at $ t = 0 $ (think of it like we constructed the circuit at that moment of time). The integral term in \eqref{3} tells us that the value of $ u $ at time $ t $ is determined by "summing up" some effects from the beginning of the system working.
So, back to the definitions:

*

*The natural response of the system is the solution of the homogeneous equation corresponding to \eqref{0}, where the initial condition is exactly the one given. That means it's what the system would look like if we didn't force it's behavior to change through giving it an input.

*The forced response of the system is the solution of \eqref{0} with zero initial condition. That means it's what the system with the same input would look like if it started from not working (not exactly, but something like that).

*The transient response of the system is the part of solution which fades over time. That means it will tend toward zero as time goes to infinity.

*The steady-state (or the permanent response) of the system is the part of the solution that does not fade over time. That means as time goes on, the actual solution becomes more and more similar to it.

From the above explanations, it should be obvious why you get the natural and forced solutions by your second method: putting $ C = 0 $ in \eqref{0} is exactly considering the system with zero initial condition, since the second term is already zero at $ t = 0 $ (no previous effects are summed up at the beginning).
But there's something inaccurate about the definitions of the transient and permament responses: they are not unique. For example, in case the input $ V _ s $ of out problem is constant over time, you can see that the particular solutions in both your methods can be considered as a permanent response, and the corresponding homogeneous solutions can be considered as a transient response. But in practice, when people talk about the permanent and transient responses, it's usually in the case that the characteristics of the system and the input are such that there is a constant or periodic solution, and they prefer that over the others. In your first method, you guessed that there is a constant solution, and luckily enough, there was. For another example, you can see that if $ V _ s ( t ) = V _ * \cos ( \omega t + \phi ) $ for some constants $ V _ 0 $, $ \omega $ and $ \phi $, then there exists a particular solution which is itself of the form $ u _ p ( t ) = V ^ * \cos ( \omega t + \psi ) $, for some constants $ V ^ * $ and $ \psi $. You can then use tools like Fourier series to generalize this to all (regular enough) periodic inputs. The examples we discussed are in fact of the type appearing in DC and AC circuits, which are why in the context of electrical engineering it's often talked about the permanent and transient responses.
