Well there is also another method to find [imath]e^{\bold{A}t}[/imath] but it's not perfect with all matrices.
[imath]\displaystyle e^{\bold{A}t} = \sum_{n=0}^{\infty}\bold{A}^n \frac{t^n}{n!}[/imath]
It works with any [imath]\bold{A}[/imath] that is [imath]n \times n[/imath] matrix. But it works perfectly in [imath]n \times n[/imath] diagonal matrices. It can work in the OP matrix but because it is not a diagonal matrix, we will have a hard time until we recognize the pattern produced by the infinite series. Therefore, it's better to use the inverse Laplace Transform to avoid the headache of thinking.
Guess, I should review the Laplace transformation calculus.
Baker-Campbell-Hausdorff formula to correct the difference [imath] e^{X}e^{Y} - e^{X+Y}. [/imath] It is the formal reason for what you called a "headache of thinking".
I had a doubt 1% that my solution was not correct, but thanks for confirming the 100% success
My reviewed solution (and final confirmation of the correct result) is, for the sake of completeness:
1.) Elimination of [imath] x=x_1 [/imath] and [imath] x'=x'_1 [/imath] :
[math]\begin{array}{lll}
\begin{pmatrix}e^{-t}\\0\end{pmatrix}&=
\begin{pmatrix}x'\\y'\end{pmatrix}+\begin{pmatrix}6&-2\\2&10\end{pmatrix}\cdot \begin{pmatrix}x\\y\end{pmatrix}\, , \,x(0)=1\, , \,y(0)=-3\\[16pt]
0&=y'+2x+10y \\
e^{-t}&=x'+6x-2y\\
x&=-5y-0.5y'\\
x'&=-5y'-0.5y''\\
e^{-t}&=-5y'-0.5y''-30y-3y'-2y=-32y-8y'-0.5y''
\end{array}[/math]
2.) Ansatz to solve for [imath] y=x_2 [/imath] :
[math]\begin{array}{lll}
0&=(6-t)(10-t)+4=t^2-16t+64=(t+8)^2\\[16pt]
y&=Ae^{-8t}+Bte^{-8t}+Ce^{-t}\ ,\ A+C=-3\\
y'&=-8Ae^{-8t}+Be^{-8t}-8Bte^{-8t}-Ce^{-t}\\
y'&=(B-8A)e^{-8t}-8Bte^{-8t}-Ce^{-t}\\
y''&=-8(B-8A)e^{-8t}-8Be^{-8t}+64Bte^{-8t}+Ce^{-t}\\
y''&=(-16B+64A)e^{-8t}+64Bte^{-8t}+Ce^{-t}
\end{array}[/math]
3.) Calculation of the coefficients [imath] A,C [/imath] from the initial value [imath] y(0)=-3 [/imath] :
[math]\begin{array}{lll}
0&=32y+8y'+0.5y''+e^{-t}\\
&=32Ae^{-8t}+32Bte^{-8t}+(32C+1)e^{-t}\\
&\phantom{=}+(8B-64A)e^{-8t}-64Bte^{-8t}-8Ce^{-t}\\
&\phantom{=}+(-8B+32A)e^{-8t}+32Bte^{-8t}+0.5Ce^{-t}\\
&=(24.5C+1)e^{-t}\text{ i.e. } C=-\dfrac{2}{49} \text{ and }A=-\dfrac{145}{49}
\end{array}[/math]
4.) Calculation of the coefficient [imath] B [/imath] from the initial value [imath] x(0)=1 [/imath] :
[math]\begin{array}{lll}
x(t)&=-5y(t)-0.5y'(t)\\
&=-5Ae^{-8t}-5Bte^{-8t}-5Ce^{-t}+(-0.5B+4A)e^{-8t}+4Bte^{-8t}+0.5Ce^{-t}\\
&=(-A-0.5B)e^{-8t}-Bte^{-8t}-4.5Ce^{-t}\\
x(0)&=1=-A-0.5B-4.5C=\dfrac{145+9}{49}-\dfrac{B}{2} \text{ and }B=\dfrac{210}{49}
\end{array}[/math]
5.) Summarizing the solution:
[math]\begin{array}{lll}
49\begin{pmatrix}x(t)\\y(t)\end{pmatrix}
&=\begin{pmatrix} 40 \\ -145\end{pmatrix}e^{-8t}
+\begin{pmatrix} -210 \\ 210 \end{pmatrix}te^{-8t}
+\begin{pmatrix} 9 \\ -2 \end{pmatrix}e^{-t}\\[16pt]
49\begin{pmatrix}x'(t)\\y'(t)\end{pmatrix}
&=\begin{pmatrix} -530 \\ 1370 \end{pmatrix}e^{-8t}
+\begin{pmatrix} 1680 \\ -1680 \end{pmatrix}te^{-8t}
+\begin{pmatrix} -9 \\ 2 \end{pmatrix}e^{-t}
\end{array}[/math]
6.) Confirmation of the correctness by checking the IVP:
[math]\begin{array}{lll}
&49\cdot\begin{pmatrix}6&-2\\2&10\end{pmatrix}\begin{pmatrix}x(t)\\y(t)\end{pmatrix}=\begin{pmatrix}6&-2\\2&10\end{pmatrix}\begin{pmatrix} 40\\ -145 \end{pmatrix}e^{-8t}
+\begin{pmatrix}6&-2\\2&10\end{pmatrix}\begin{pmatrix} -210 \\ 210 \end{pmatrix}te^{-8t}
+\begin{pmatrix}6&-2\\2&10\end{pmatrix}\begin{pmatrix} 9\\-2 \end{pmatrix}e^{-t}\\[16pt]
&=\begin{pmatrix} 530\\-1370\end{pmatrix}e^{-8t}
+\begin{pmatrix} -1680\\ 1680 \end{pmatrix}te^{-8t}
+\begin{pmatrix} 58\\-2 \end{pmatrix}e^{-t}
\end{array}[/math]
[math]\begin{array}{lll}
49\begin{pmatrix}x'(t)\\y'(t)\end{pmatrix}+49\cdot\begin{pmatrix}6&-2\\2&10\end{pmatrix}\begin{pmatrix}x(t)\\y(t)\end{pmatrix}=\begin{pmatrix} 49 \\ 0 \end{pmatrix}e^{-t}
\end{array}[/math]
