You say "consider the non-linear system (1)" but don't show what that is!

Is it \(\displaystyle \frac{d\begin{bmatrix}x_1 \\ x_2\end{bmatrix}}{dt}= \begin{bmatrix}-x_1 \\ x_1^2+ x_2 \end{bmatrix}\)?

If so then write it as two separate differential equations:

\(\displaystyle \frac{dx_1}{dt}= -x_1\) and

\(\displaystyle \frac{dx_2}{dt}= x_1^2+ x^2\).

The first can be written as \(\displaystyle \frac{dx_1}{x_1}= -dt\) and integrated:

\(\displaystyle ln(x_1)= -t+ C\) for some constant C. Taking the exponential of both sides, \(\displaystyle x_1= C'e^{-t}$ where $C_1= e^C \) is another constant.

Then the second equation can be written \(\displaystyle \frac{dx_2}{t}= x_1^2+ x_2= C'^2e^{-2t}+ x_2\). Fortunately that is now linear and we can write it as \(\displaystyle \frac{dx_2}{dt}- x_2= C'^2e^{-2t} \). The "associated homogeneous equation" is \(\displaystyle \frac{dx_2}{dt}- x_2= 0\) which has "characteristic equation" $r- 1= 0$ and "characteristc value $r= 1$. The general solution to the associated homogeneous equation is \(\displaystyle x_2= De^{t}\). To find a solution to the entire equation, try something of the form \(\displaystyle x_2= Ae^{-2t}\). Then \(\displaystyle \frac{dy}{dt}= -2Ae^{-2t}\) so the equation becomes \(\displaystyle -2Ae^{-2t}+ Ae^{-2t}= -Ae^{-2t}= C'^2e^{-2t}\) so \(\displaystyle A= -C'^2/2\).

That gives \(\displaystyle x_2= De^t+ (C'^2/2)e^{-2t}\).

So we have \(\displaystyle x_1= C'e^{-t}\) and \(\displaystyle x_2= De^t+ (C'^2/2)e^{-2t}\). The initial condition \(\displaystyle x_1(0)= C'= c_1\) and \(\displaystyle x_2= D+ C'^2/2= D+ c_1^2/2= c_2\) so \(\displaystyle D= c_2- c_1^2/2 \).