Answer
$y=e^t$
Work Step by Step
$$y''-5y'+6y=2e^t$$
Consider first the homogeneous LHS. Let $y=e^{\lambda{t}}$ so that $(lny)'=\lambda$.
$${\lambda}^2-5\lambda+6=0$$
$$(\lambda-3)(\lambda-2)=0$$
$$\lambda_{1,2}=3,2$$
$\therefore$ $e^{3t}$ and $e^{2t}$ form a basis for solutions. Let $u(t)$ and $v(t)$ be the external functions associated with the solutions in the inhomogeneous case.
$$\therefore{y}=u(t)e^{3t}+v(t)e^{2t},\quad{y'}=u'(t)e^{3t}+v'(t)e^{2t}+3u(t)e^{3t}+2v(t)e^{2t}$$
To avoid over-complicating the inhomogeneous equation with variables, let us set up expression $y'$ in such a way that it would be most apparent that the homogeneous solutions $e^{3t}$ and $e^{2t}$ are twice differentiable.
$$\therefore{u'(t)}e^{3t}+v'(t)e^{2t}=0(\Rightarrow\frac{v'(t)}{u'(t)}=-e^t)$$
$$\therefore{y'}=3u(t)e^{3t}+2v(t)e^{2t}\Rightarrow{y''}=3u'(t)e^{3t}+2v'(t)e^{2t}+9u(t)e^{3t}+4v(t)e^{2t}$$
Substituting this into the original equation,
$$e^{3t}(3u'(t)+9u(t))+e^{2t}(2v'(t)+4v(t))-5(3u(t)e^{3t}+2v(t)e^{2t})+6(u(t)e^{3t}+v(t)e^{2t})=2e^t$$
$$3u'(t)e^{3t}+2v'(t)e^{2t}=2e^t$$
$$\therefore(\begin{matrix} e^{3t} & e^{2t}\\3e^{3t} & 2e^{2t}\end{matrix})(\begin{matrix}u'(t)\\v'(t)\end{matrix})=(\begin{matrix}0\\2e^t\end{matrix})$$
$$(\begin{matrix}u'(t)\\v'(t)\end{matrix})=-\frac{1}{e^{5t}}(\begin{matrix} 2e^{2t} & -e^{2t}\\-3e^{3t} & e^{3t}\end{matrix})(\begin{matrix}0\\2e^t\end{matrix})=(\begin{matrix}2e^{-2t}\\-2e^{-t}\end{matrix})$$
$$\therefore{u(t)}=\int{2e^{-2t}}dt=-e^{-2t}+C_{1},\quad{v(t)}=-\int{2e^{-t}}dt=2e^{-t}+C_{2}$$
$$\therefore{y}=(-e^{-2t}+C_{1})e^{3t}+(2e^{-t}+C_{2})e^{2t}=e^t+C_{1}e^{3t}+C_{2}e^{2t}$$
Comparing with the RHS of the original equation, $C_{1}=C_{2}=0$.
$$\therefore{y}=e^t$$
(Note #2: By initially choosing to disregard the terms $u'(t)$ and $v'(t)$, we leave only the terms $3e^{3t}$ and $2e^{2t}$. We then recognise that both these terms are in the form $\lambda{e^{\lambda{t}}}$, which would become ${\lambda}^2{e^{\lambda{t}}}$ when differentiated again. As such, when the terms $u'(t)$ and $v'(t)$ show up again alongside these terms in the expansion of y'', we factor out both terms of $e^{\lambda{t}}$ and end up with two expressions of the homogeneous equation in the LHS of the original. Both such expressions amount to zero and we are thus left with the terms in $u'(t)$ and $v'(t)$, with which we can state alongside the particular condition for $u'(t)$ and $v'(t)$ which was set earlier, to form an easily solvable system of equations. Hence, the initially "not yet clear" motive of disregarding the terms $u'(t)$ and $v'(t)$ becomes known.)
Substitution method (Checking):
Let $y=ke^t$ for some $k\in\mathbb{R}$.
$$\therefore{k}e^t-5ke^t+6ke^t=2ke^t=2e^t\Rightarrow{k}=1$$
$$\therefore{y}=e^t$$ as required.
