System of linear differential equations

In matrix form, this equation is $\left[\begin{array}{c} x\\ y\end{array}\right]' =\underbrace{\left[\begin{array}{cc} 1&1\\ 1&1\end{array}\right]}_{A}\underbrace{\left[\begin{array}{c} x\\ y\end{array}\right]}_{v}.$ Now we have $e^{tA}=\sum_{n=0}^{\infty}\frac{1}{n!} \left[\begin{array}{cc} t&t\\ t&t \end{array}\right]^n=I+\sum_{n=1}^{\infty} \frac{1}{n!}\left[\begin{array}{cc} 2^{n-1}t^n&2^{n-1}t^n\\2^{n-1}t^n&2^{n-1}t^n\end{array}\right],$ which becomes $I+\frac{1}{2}\left[\begin{array}{cc} e^{2t}-1&e^{2t} -1\\ e^{2t}-1&e^{2t}-1\end{array}\right]=\frac{1}{2}\left[\begin{array}{cc} e^{2t}+1&e^{2t} -1\\ e^{2t}-1&e^{2t}+1\end{array}\right].$ Our solution is $v(t)=\frac{1}{2}\left[\begin{array}{cc} e^{2t}+1&e^{2t} -1\\ e^{2t}-1&e^{2t}+1\end{array}\right]\left[\begin{array}{c} 1\\1 \end{array}\right],$ and multiplying out gives that $x(t)=y(t)=e^{2t}$ . Then, $x(2)y(2)=e^8$ , so the answer is 8.

Let, x=f(t) ;y=g(t) Given that,f'(t)=g'(t)=f(t)+g(t) which implies. (f'(t)/f(t)+g(t))=1.....(1) (g'(t)/f(t)+g(t))=1 ......(2) and dy/dx=1.(3) Adding (1) and(2) we get (f'(t)+g'(t)/f(t)+g(t)) =2 Integrating both sides w.r.t "t", and using given value we get ln(f(t)+g(t))=2t+ln2 Now,since y=x(using (3)), lnx=lny=2t, ln(xy)=2ln(x)=4t this implies at t=2 ln(xy)=4(2)=8