We can solve for [tex]x(t)[/tex] first by rewriting the system of first-order ODEs as a single second-order ODE in [tex]x(t)[/tex]:
Taking the derivative of the first ODE gives
[tex]\dfrac{\mathrm dx}{\mathrm dt}=3x+2y+1\implies\dfrac{\mathrm d^2x}{\mathrm dt^2}=3\dfrac{\mathrm dx}{\mathrm dt}+2\dfrac{\mathrm dy}{\mathrm dt}[/tex]
while solving for [tex]2y[/tex] gives
[tex]\dfrac{\mathrm dx}{\mathrm dt}=3x+2y+1\implies2y=\dfrac{\mathrm dx}{\mathrm dt}-3x-1[/tex]
Then
[tex]\dfrac{\mathrm d^2x}{\mathrm dt^2}=3\dfrac{\mathrm dx}{\mathrm dt}+2(-2x-y+1)[/tex]
[tex]\dfrac{\mathrm d^2x}{\mathrm dt^2}=3\dfrac{\mathrm dx}{\mathrm dt}-4x-2y+2[/tex]
[tex]\dfrac{\mathrm d^2x}{\mathrm dt^2}=3\dfrac{\mathrm dx}{\mathrm dt}-4x-\left(\dfrac{\mathrm dx}{\mathrm dt}-3x-1\right)+2[/tex]
[tex]\implies\dfrac{\mathrm d^2x}{\mathrm dt^2}-2\dfrac{\mathrm dx}{\mathrm dt}+x=3[/tex]
which is linear with constant coefficients, so it's trivial to solve; the corresponding homogeneous ODE
[tex]x''-2x'+x=0[/tex]
has characteristic equation
[tex]r^2-2r+1=(r-1)^2=0[/tex]
with root [tex]r=1[/tex] (multiplicity 2), so the characteristic solution is
[tex]x_c=C_1e^t+C_2te^t[/tex]
For the non-homogeneous ODE, assume a particular solution of the form
[tex]x_p=a\implies{x_p}'={x_p}''=0[/tex]
Substituting these into the ODE gives
[tex]0-2\cdot0+a=3\implies a=3[/tex]
Then the general solution for [tex]x(t)[/tex] is
[tex]\boxed{x(t)=C_1e^t+C_2te^t+3}[/tex]
From here, we find
[tex]\dfrac{\mathrm dx}{\mathrm dt}=C_1e^t+C_2(t+1)e^t[/tex]
so that
[tex]2y=(C_1e^t+C_2(t+1)e^t)-3(C_1e^t+C_2te^t+3)-1[/tex]
[tex]\implies\boxed{y(t)=\left(\dfrac{C_2}2-C_1\right)e^t-C_2te^t-5}[/tex]
