Notice how you have so much capability to analyze and derive solutions now. Consider where you were a few years ago. Because I know you have these abilities, I wanted to you to think in these terms.
So yes, basically that is it. And then you can do the same for the RC circuit. Actually, you don't even need to take it as far as you did, if you know the basics of a first order differential equation, which you know now. You know the solution to a first order equation with constant coefficients is an exponential type change, as you just derived. You should learn to recognize the form instantly whenever it occurs in your work, because it shows up in so many places. The same can be said for the second order system, although that one is more complicated.
For example, consider the magnetizing current in an induction motor, and how it evolves from the d-axis current which energized the flux in the rotor.
d Im/dt =-(R/L) * Im + (R/L) *Id, where R is the rotor resistance and L is the rotor inductance. In other words the rotor time constant is L/R. Notice the negative sign in front of the R/L which indicates stability because the pole is in the left side of the complex plane with Re(P)<0. Notice there is an input signal that drives the system, which in this case is Id.
So, for a series RC circuit driven by constant voltage source Vin, you would derive d Vc/dt = -(1/RC) * Vc + (1/RC) * Vin. The negative sign instantly tells you it is stable and the solution won't blow up. It's clear that Vin is the input that drives the system.
And, for a series RL circuit driven by a constant voltage source Vin, you would derive d I/dt = -(R/L) * I + (1/L) * Vin
Linear first order equations with constant coefficients, which are stable systems, always have this form. The pole locations are always obvious too because the Laplace transform can be done by inspection if you remember that derivatives are equivalent to multiplication by "s". The poles occur at s=-1/tau, where tau is the time constant, and the low pass filter transfer function form is always T=A * ( w/(s+w)), where w=1/tau, and A depends on the type of input that drives the system.
These are basic things that should be committed to memory. You should recognize them immediately, much the same way a sailor should know how to spot the north star.
Now let me give you a hint about a powerful trick to solving higher order systems. Consider a state space formulation. It actually looks like a first order system, except the object of the solution is a vector and not a variable. For a state space system you have dx/dt=A x + B u, where x and u are vectors and A and B are matrices. The x vector is the state or system variables and the u vector is the input variables. It is possible to solve such systems using the same basic solution you just used, but you have to be careful about the order of operations because the objects are vectors/matrices and not scalars. This is called the state transition matrix approach, and you can look it up in your books (e.g. p. 666 of Ogata). However, you recently saw an example of this when I derived the solution for a Buck converter using this method. I was then able to write a simple Matlab program to simulate buck converters in about 15 minutes, using the solution. In other words, think of the simple first order system you are asking about as a prototype for more advanced and complicated systems.
EDIT: check the sign of the exponential in your solution. You should get 1-exp(t/tau) type solutions.