Exponential decaying Swept / chirp signal
Frequency is the derivative of phase, so you need to start with the expression for frequency, which in your case is
$f=200\exp(-\alpha*t)$

At time $t=0$ this will have the frequency of 200, but we need to solve for $\alpha$ so that the frequency at time $t=2$ is 20.  Solving for $\alpha$ gives $\alpha=1.1513$.
To get the correct expression for the phase you need to integrate the above expression for the phase i.e. solve

$f =\frac{1}{2\pi}\frac{d\phi}{dt}$

The following matlab code generates this signal.

fs=500; %sampling frequency
t=(0:2*fs)/fs; %Time vector
alpha=1.1513;
sig = exp(-j*2*pi*200/alpha*exp(-alpha*t));


Note that just because the instantaneous frequency behaves this way doesn't mean the spectrogram will follow it exactly. It will get better if you have a higher time-bandwidth product.