How did $-4$ come in $y(t)?$ Write down the general solution of the linear system:
$$\frac{dx}{dt} = x + y \tag1$$
$$\frac{dy}{dt} = 4x − 2y.\tag 2$$
My attempt : Here i found $x(t)=c_1e^{-3t} + c_2 e^{2t}$
To find  $y(t)$ we have
$\frac{dy}{dt} = 4x − 2y.$
$\frac{d^2y}{dt^2} = 4\frac{dx}{dt} − 2\frac{dy}{dt}.\tag 3$
From $(1)$ and $(3)$, we have
$\frac{d^2y}{dt^2} = 4x +4y − 2\frac{dy}{dt}.\tag 5$
from $(1)$ and $(5)$, we have
$$\frac{d^2y}{dt^2} +\frac{dy}{dt} - 6y=0$$
now convert them into auxiliary equation , then we have
$y(t)= c_1e^{2t} + c_2e^{-3t}$
But orginal answer given is $y(t) = −4c_1e^{−3t} + c_2e^{2t}$
My confusion is that How did that $-4$ come in $y(t)?$
 A: As the $-4$ is multiplying the unknown constant $c_1$ it doesn't  matter.  Your $c_2$ is just the book's $-4c_1$.  Something in the book's solution made it appear before the $c_1$ was applied.
A: After finding $x(t)$, $y(t)$ easily follows from (1):
$$y(t) = -x+\frac{dx}{dt}=-c_1e^{-3t} - c_2 e^{2t}-3c_1e^{-3t} + 2c_2 e^{2t}=-4c_1e^{-3t}+ c_2 e^{2t} .$$
Note that the solutions $x(t)$ and $y(t)$ depend on TWO arbitrary constants $c_1$ and $c_2$.
A: We already have some short repeatedly edited answers from high rated user, answering why both answers are true, and why your answer is better. But let me give the answer for the puzzle with the $-4$ since it is not so easy to guess. So here is a longer answer, because there is no such longer comment. There are many methods to solve the given differential equation. One of the methods is to diagonalize the given matrix, let us call it $A$. So we search for $D,T$ with $D$ diagonal, $T$ with inverse, and
$$A=TDT^{-1}\ .$$
Well, here is the decomposition:
$$
\underbrace{
\begin{bmatrix}
1 & 1\\
4  & -2
\end{bmatrix}}_A
=
\underbrace{
\begin{bmatrix}
1 & 1\\
1  & -4
\end{bmatrix}}_T
\underbrace{
\begin{bmatrix}
2 & \\
  & -3
\end{bmatrix}}_D
\underbrace{
\begin{bmatrix}
4/5 & 1/5\\
1/5  & -1/5
\end{bmatrix}}_{T^{-1}}
\ .
$$
So instead of solving for $x,y$ in
$$
\begin{bmatrix}
x' \\ y'
\end{bmatrix}
=
\begin{bmatrix}
1 & 1\\
1  & -4
\end{bmatrix}
\begin{bmatrix}
2 & \\
  & -3
\end{bmatrix}
\underbrace{
\begin{bmatrix}
4/5 & 1/5\\
1/5  & -1/5
\end{bmatrix}
\begin{bmatrix}
x \\ y
\end{bmatrix}}_{\begin{bmatrix}X\\ Y\end{bmatrix}}
$$
we are solving for $X,Y$ in
$$
\begin{bmatrix}
X \\ Y
\end{bmatrix}
=
\begin{bmatrix}
2 & \\
  & -3
\end{bmatrix}
\begin{bmatrix}
X \\ Y
\end{bmatrix}
$$
The solution is clear, $X=ae^{2t}$, $Y=be^{-3t}$ for some constants... and we only need to pass to $x,y$ via...
$$
\begin{bmatrix}
x \\ y
\end{bmatrix}
=
T
\begin{bmatrix}
X \\ Y
\end{bmatrix}
=
\begin{bmatrix}
1 & 1\\
1  & -4
\end{bmatrix}
\begin{bmatrix}
ae^{2t} \\ be^{-3t}
\end{bmatrix}
=
\begin{bmatrix}
ae^{2t} + be^{-3t}
\\
be^{2t} \color{red}{-4} be^{-3t}
\end{bmatrix}
\ .
$$
The $c_1$ and $c_2$ from the OP are $b$, and respectively $a$.
So this "explains" (or rather guesses) why we get the $-4$ factor for the $y$... (And also why we have no question about $x$.)

A shorter answer would have been:
If you solve for $x$ instead of $y$, thus using the $c_1,c_2$ as the coefficients of the linear combination of $x$ w.r.t. the two exponential functions, which is the formula for $y$?
