Answer
$$
y^{\prime \prime}-2 y^{\prime}-3y=3e^{2t}
$$
The general solution of the given equation is
$$ y=c_{1} e^{3 t} +c_{2} e^{ -t} - e^{2t} $$
where $ c_{1} $and $c_{1}$ are arbitrary constants.
Work Step by Step
$$
y^{\prime \prime}-2 y^{\prime}-3y=3e^{2t} \quad\quad\quad \quad (1)
$$
First we find the general solution of the homogeneous equation as follows:
We assume that $ y = e^{rt}$, and it then follows that $r$ must be a root of the characteristic equation
$$
r^{2}-2r-3=(r-3)(r+1)=0,$$
so its roots are
$$r_{1}=3 ,\quad r_{2}=-1, $$
so the general solution of the homogeneous equation is
$$ y=c_{1} e^{3 t} +c_{2} e^{ -t} $$
where $ c_{1} $and $c_{1}$ are arbitrary constants.
Next, we guess that the particular solution has the form $ Y(t)= A e^{2t} $,
where A is the undetermined coefficient, then $ Y^{\prime}(t)= 2A e^{2t} $ and
$ Y^{\prime \prime }(t)= 4A e^{2t} $. Substituting the expressions for $ Y(t) $, $ Y^{\prime}(t) $, and $Y^{\prime \prime }(t) $ into the differential equation, we obtain
$$
4A e^{2t} -2 . 2A e^{2t}-3A e^{2t}=3e^{2t}
$$
$$
-3A e^{2t}=3e^{2t}
$$
Hence $A=-1$ , so a particular solution of Eq. (1) is $ Y(t)= - e^{2t} $
The general solution of the nonhomogeneous equation (1) is
$$ y=c_{1} e^{3 t} +c_{2} e^{ -t} - e^{2t} $$
where $ c_{1} $and $c_{1}$ are arbitrary constants.
