Q:

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 $c_2$.

