#### Answer

$$
y^{\prime \prime}-6 y^{\prime}+9 y=0
$$
The general solution of that equation is given by
$$ y=c_{1}t e^{3t} +c_{2} e^{3t} $$
where $ c_{1} $ and $c_{2}$ are arbitrary constants.

#### Work Step by Step

$$
y^{\prime \prime}-6 y^{\prime}+9 y=0 \quad \quad (1)
$$
We assume that $ y = e^{rt}$, and it then follows that $r$ must be a root of the characteristic equation
$$
r^{2}- 6r+9=(r-3)^{2}=0,$$
so $r_{1}=r_{2}=3$. Therefore one solution of Eq. (1) is $y_{1}=e^{3t}$. To find the general solution of Eq. (1), we need a second solution that is not a multiple of $y_{1}$. This second solution can be found, by using D’Alembert's method, by replacing $c$ by a function $v(t)$ and then trying to determine $ v(t)$ so that the product $v(t) y_{1}(t) $ is also a solution of Eq. (1).
Now we substitute $ y = v(t)y_{1}(t)$ in Eq. (1) and use the resulting equation to find $v(t)$.
$$ y=v(t) y_{1}(t)=v(t) e^{ 3t} \quad \quad (2)$$
we have
$$y^{\prime}=v^{\prime}(t) e^{3t}+3 v(t) e^{3 t} \quad \quad (3)$$
and
$$
y^{\prime \prime}=v^{\prime \prime}(t) e^{ 3t}+6 v^{\prime}(t) e^{3t}+9 v(t) e^{ 3t} \quad \quad (4) $$
By substituting the expressions in Eqs. (2), (3), and (4) in Eq. (1) and collecting terms, we obtain
$$ v^{\prime \prime}(t)=0 $$
Therefore
$$v(t)=c_{1}t+c_{2} $$
where $ c_{1} $ and $c_{2}$ are arbitrary constants. Finally, substituting for $v(t)$ in Eq. (2), we obtain
$$ y=c_{1}t e^{3 t} +c_{2} e^{3 t} \quad \quad (5)$$
The second term on the right side of Eq. (5) corresponds to the original solution $y_{1}(t) = e^{3t}$, but the first term arises from a second solution, namely, $y_{2}(t) = t e^{3t}$. We can verify that these two solutions form a fundamental set by calculating their Wronskian: $W(y_{1}, y_{2})(t)\ne0$
Therefore $y_{1}(t) = e^{3t}$, $y_{2}(t) = t e^{3t}$ form a fundamental set of solutions of Eq. (1), and the general solution of that equation is given by
$$ y=c_{1}t e^{3t} +c_{2} e^{3t} $$
where $ c_{1} $ and $c_{2}$ are arbitrary constants.