Why don't phasors give the transient state?

Phasor analysis allows us to analyze a circuit's response to a sinusoidal steady-state response at a given single frequency. We represent a time-domain voltage \$V(t) = V_{0}cos(\omega t + \phi)\$ in phasor form by transforming it into a complex exponential via Euler's formula...

$$A\cos(\omega t + \phi) = Re\{Ae^{j( \omega t~+~\phi)}\} = Re\{Ae^{j\phi}e^{j\omega t}\}$$

...and then ignoring the frequency/time dependence (since we assume everything in the circuit is excited by a steady sinusoid of the same frequency). Thus,

$$V= Ae^{j\phi}$$

This \$V\$ is what we call a phasor. We can represent any current or voltage as a phasor. In order to recover a time-domain representation from a phasor, you can multiply it by \$e^{j\omega t}\$ and then take the real part. Note that sometimes for the sake of brevity/familiarity in calculating power, we also convert the amplitude of the phasor into RMS values (divide magnitude by \$\sqrt{2}\$ for a sinusoid). Phasors allow us to use analogous DC analysis techniques to recover transfer functions of linear circuits (by using impedances). Using superposition, we can use Fourier analysis to analyze a circuit's complete steady-state response as a sum of its steady-state response due to different frequency components.

It's useful to note the relation of phasors to the Laplace representation of a circuit. The Laplace representation of a circuit uses the variable \$s = \sigma + j\omega\$. Note that for \$\sigma = 0\$, the transfer function of the Laplace representation of a circuit reduces down to the phasor representation. This is a good indication that the real part of \$s\$ represents a transient response (and this can be easily observed by noting that \$e^{at}\$ for any real \$a\$ will result in a real value that either exponentially grows or decays). Note that the Laplace representation is a more general representation of a circuit that includes both transient and steady-state responses. Likewise, it's good to note that the Fourier transform is just a special case of the more general Laplace transform (the case where \$\sigma = 0\$ in \$s = j\omega\$).

Mathematically, nothing gets lost. Phasor analysis gives you the value of every voltage and current in the circuit for all time, in the form of cosine functions:

$$V_a = A_1 \cos (\omega t + \phi_1)$$ $$V_b = A_2 \cos (\omega t + \phi_2)$$ $$I_c = A_3 \cos (\omega t + \phi_3)$$ $$etc.$$

You specify \$\omega\$ and the amplitude and phase for at least one voltage or current, and phasor analysis gives you \$A\$ and \$\phi\$ for all the rest.

The problem isn't that the transient behavior gets lost, it's that you never put it in! By definition, phasor analysis works on unchanging eternal sinusoids at a single frequency applied to a linear time-invariant system -- the so-called sinusoidal steady state. "Steady state" is the opposite of "transient". You can extend this to cover exponential growth and decay, but again, this is eternal growth and decay. The math only works because your voltages and currents are complex exponentials, which aren't distorted by linear differential equations.

To describe a situation where you flip a switch at t = 0, you need to use a step function. Step functions cannot be represented by a single frequency, so phasor analysis breaks down. To handle this, you need to use Fourier analysis.

I know this is a late answer, but I want to give a different insight as to why phasors only give the steady state response.

Consider the well known RC circuit, with a driving source \$v_s=\text{V}\cos(\omega t + \phi)\$, then you have the differential equation:

$$ \text{RC}\dfrac{dv_c}{dt} + v_c=\cos(\omega t + \phi)$$

From a mathematics point of view, you can solve this differential equation by finding the homogeneous solution and a particular solution and when you add them up you get the general solution. So far so good.

Phasors give you a particular solution only (it doesn't give you the homogeneous solution, which is the transient solution) and the particular solution is what we call the steady state response.

In other words, the homogeneous solution (transient or natural response) is the solution to

$$ \dfrac{dv_c}{dt} + \dfrac{1}{\text{RC}}v_c=0$$

which you can find by the integrating factor method.

And the particular solution, using the fact that you can write the input source as \$\Re\{\text{V}e^{j\omega t}e^{j\phi}\}\$, where \$\Re\$ means the real part of. For the particular solution, we make a 'guess', based on the forcing function:

$$\dfrac{dv_c}{dt} + \dfrac{1}{\text{RC}}v_c= \dfrac{\text{V}}{\text{RC}}e^{j\phi}e^{j\omega t}$$

If you guess that your particular solution has the form \$v_{c,p}=\text{A}e^{j\omega t}\$, where \$A\$ will be a phasor too (it will have a magnitude and a phase in the end), just like \$\text{V}e^{j\phi}\$ is, by the definition of a phasor. $$j\omega \text{A}e^{j\omega t}+ \dfrac{1}{\text{RC}}\text{A}e^{j\omega t}= \dfrac{\text{V}}{\text{RC}}e^{j\phi}e^{j\omega t}$$ which you can simplify by dividing by \$e^{j\omega t}\$ and factoring the \$\text{A}\$ terms

$$\text{A}\bigg(j\omega+\dfrac{1}{\text{RC}}\bigg)= \dfrac{\text{V}}{\text{RC}}e^{j\phi}$$

$$ \text{A}=\dfrac{\text{V}e^{j\phi}}{j\omega\text{RC}+1}$$

And in the end, \$\text{A}\$, will be a phasor of the form:

$$\text{A}=|\text{A}|\angle{\theta} $$

So when you find \$v_{c,p}\$, you only have the particular solution (forced response, steady-state). You'd still need to find the solution to aforementioned homogeneous equation to have a full response.

In short, phasors give you a particular solution to the differential equation.