Note that the characteristic solutions to this ODE are [tex]e^{-2x/5}[/tex] and [tex]e^{2x}[/tex], so we can safely assume a particular solution of the form
[tex]y_p=(ax+b)e^{3x}[/tex]
with derivatives
[tex]{y_p}'=ae^{3x}+3(ax+b)e^{3x}=(3ax+a+3b)e^{3x}[/tex]
[tex]{y_p}''=3ae^{3x}+3(3ax+a+3b)e^{3x}=(9ax+6a+9b)e^{3x}[/tex]
Substitute these expressions into the ODE and solve for a and b. Notice that each term on either side contains a factor of [tex]e^{3x}[/tex], which we can cancel.
[tex]-\dfrac54(9ax+6a+9b)+2(3ax+a+3b)+(ax+b)=3x[/tex]
[tex]-\dfrac{17a}4x-\left(\dfrac{11a}2+\dfrac{17b}4\right)=3x[/tex]
[tex]\implies\begin{cases}-\frac{17a}4=3\\\frac{11a}2+\frac{17b}4=0\end{cases}[/tex]
[tex]\implies a=-\dfrac{12}{17}\text{ and }b=\dfrac{264}{289}[/tex]
So the particular solution is
[tex]y_p=\left(-\dfrac{12x}{17}+\dfrac{264}{289}\right)e^{3x}=\boxed{\dfrac{12}{289}(22-17x)e^{3x}}[/tex]