Finding the magnitude of the Transfer Function
Well, we have the following circuit:







simulate this circuit &ndash; Schematic created using CircuitLab

When analyzing a transistor we need to use the following relations:


$$\text{I}_\text{E}=\text{I}_\text{B}+\text{I}_\text{C}\tag1$$
Transistor gain \$\beta\$:
$$\beta=\frac{\text{I}_\text{C}}{\text{I}_\text{B}}\tag2$$


Using KCL, we can write:

$$
\begin{cases}
\text{I}_\text{y}=\text{I}_\text{c}+\text{I}_4\\
\\
\text{I}_\text{t}=\text{I}_3+\text{I}_4\\
\\
\text{I}_\text{x}=\text{I}_2+\text{I}_3\\
\\
\text{I}_8=\text{I}_5+\text{I}_7\\
\\
\text{I}_7=\text{I}_6+\text{I}_\text{y}\\
\\
\text{I}_1=\text{I}_5+\text{I}_6\\
\\
\text{I}_\text{x}=\text{I}_1+\text{I}_2\\
\\
\text{I}_8=\text{I}_\text{t}+\text{I}_\text{c}\\
\\
\beta=\frac{\text{I}_\text{c}}{\text{I}_\text{t}}
\end{cases}\tag3
$$

Using KVL, we can write:

$$
\begin{cases}
\text{I}_\text{x}=\frac{\text{V}_\text{x}-\text{V}_5}{\text{R}_1}\\
\\
\text{I}_2=\frac{\text{V}_5}{\text{R}_2}\\
\\
\text{I}_3=\frac{\text{V}_5-\text{V}_4}{\text{R}_3}\\
\\
\text{I}_4=\frac{\text{V}_\text{y}-\text{V}_4}{\text{R}_4}\\
\\
\text{I}_5=\frac{\text{V}_1}{\text{R}_5}\\
\\
\text{I}_7=\frac{\text{V}_1-\text{V}_2}{\text{R}_6}\\
\\
\text{I}_7=\frac{\text{V}_2}{\text{R}_7}\\
\\
\text{V}_4-\text{V}_1=\alpha
\end{cases}\tag4
$$

Now, you can solve your problem using:

FullSimplify[
 Solve[{Iy == Ic + I4, It == I3 + I4, Ix == I2 + I3, I8 == I5 + I7, 
   I7 == I6 + Iy, I1 == I5 + I6, Ix == I1 + I2, \[Beta] == Ic/It, 
   I8 == It + Ic, Ix == (Vx - V5)/R1, I2 == (V5)/R2, 
   I3 == (V5 - V4)/R3, I4 == (Vy - V4)/R4, I5 == (V1)/R5, 
   I7 == (V1 - V2)/R6, I7 == V2/R7, V4 - V1 == \[Alpha]}, {Ix, I2, I3,
    It, I1, I4, Ic, I8, I5, I7, I6, Iy, V5, V4, V3, V1, V2}]]




EDIT

Let's assume \$\alpha=\frac{7}{10}\$, \$\beta=100\$ and \$\text{V}_\text{x}=10\$. In the (complex) s-domain we need to solve the following system of equations (I used Mathematica 12.0 to solve it):

FullSimplify[
 Solve[{Iy == Ic + I4, It == I3 + I4, Ix == I2 + I3, I8 == I5 + I7, 
   I7 == I6 + Iy, I1 == I5 + I6, Ix == I1 + I2, I8 == It + Ic, 
   100 == Ic/It, Ix == ((10/s) - V5)/600, 
   I2 == V5/(((((s*20*10^(-6)))*((1/(s*2*10^(-9)))))/(((1/(s*2*10^(-9)\
))) + ((s*20*10^(-6)))))), I3 == (V5 - V4)/(1/(s*1*10^(-9))), 
   I4 == ((18/s) - V4)/100, I5 == V1/2000, 
   I7 == (V1 - V2)/((1/(s*100*10^(-9)))), I7 == V2/100, 
   V4 - V1 == (7/10)/s}, {V1, V2, V4, V5, Ic, It, Ix, Iy, I1, I2, I3, 
   I4, I5, I6, I7, I8}]]


Now, I used inverse Laplace transform to plot the voltage \$\text{V}_2\left(t\right)\$:



Where:


$$\lim_{t\to0}\text{V}_2\left(t\right)=-\frac{7}{10}\tag5$$
$$\lim_{t\to\infty}\text{V}_2\left(t\right)=0\tag6$$


The maximum occurs when \$t\approx1.20098\space\mu\text{s}\$ and is then equal to:

$$\hat{\text{V}}_2\left(1.20098\right)\approx16.6022\space\text{V}\tag7$$