There is a way to get the complete solution directly. It helps that all our roots are real. There is a way to "factor" the left side like a polynomial. This method attempts to get the equation into the form
$z'+az=f(x)$
which can be solved through the use of an integrating factor.
You have already that the characteristic polynomial is $(r-5)(r+1)(r-1)$, which I will rewrite as $(r-5)(r^2-1)$. Now let $z=y''-y$, which you'll notice has the characteristic polynomial $r^2-1$. Now the original equation can be rewritten.
$y'''-y'-5y''+5y=(y''-y)'-5(y''-y)=z'-5z=3e^{-x}$
You'll notice that the characteristic polynomial for this equation in $z$ corresponds to the other factor $r-5$. This is of course solved my multiplying through by the integrating factor $e^{-5x}$
$e^{-5x}z'-5e^{-5x}z=(e^{-5x}z)'=3e^{-6x}$ $e^{-5x}z=-\frac12e^{-6x}+k_1$ $z=-\frac12e^{-x}+k_1e^{-5x}$
As long as we don't get our equation into the form
$u'+u=f(x)$
the $e^{-x}$ term will not cancel out prior to integration and introduce an $xe^{-x}$ term, which would make further integration more complicated. So now we have
$z=y''-y=-\frac12e^{-x}+k_1e^{-5x}$
Again, we choose a substitution based on a factor of the characteristic polynomial. To avoid the problem mentioned above, we make the substitution $u=y'+y$. This yields
$y''-y=(y'+y)'-(y'+y)=u'-u=-\frac12e^{-x}+k_1e^{-5x}$ $e^{-x}u'-e^{-x}u=(e^{-x}u)'=-\frac12e^{-2x}+k_1e^{-6x}$ $e^{-x}u=\frac14e^{-2x}+k_2e^{-6x}+k_3,k_2=-\frac{k_1}6$ $u=y'+y=\frac14e^{-x}+k_2e^{-5x}+k_3e^x$
And finally we can solve for $y$.
$e^xy'+e^xy=(e^xy)'=\frac14+k_2e^{-4x}+k_3e^{2x}$ $e^xy=\frac14x+k_4e^{-4x}+k_5e^{2x}+k_6$ $y=\frac14xe^{-x}+k_4e^{-5x}+k_5e^x+k_6e^{-x}$