Consider the ODE $$y'' + py' + qy = e^{ax} \tag{4}.$$ We have that $$y_p = Ae^{ax}\tag{5},$$ might be a particular solution of $(4)$. Here $A$ is the undetermined coefficient that we want to determine in such a way that $(5)$ will actually satisfy $(4)$.
Upon substituting $(5)$ into $(4)$, we get $$A(a^2 + pa + q)e^{ax} = e^{ax},$$ so $$A=\frac{1}{a^2+pa+q} \tag{6}$$ This value of $A$ will make $(5)$ a solution of $(4)$ except when the denominator on the right of $(6)$ is zero. The source of this difficulty is easy to understand, for the exception arises when a is a root of the auxiliary equation $$m^2 + pm + q = 0 \tag{7}$$
Prove:- If $a$ is not a root of the auxiliary equation $(7)$, then $(4)$ has a particular solution of the form $Ae^{ax}$; if a is a simple root of $(7)$, then $(4)$ has no solution of the form $Ae^{ax}$ but does have one of the form $Axe^{ax}$; and if $a$ is a double root, then $(4)$ has no solution of the form $Axe^{ax}$, but does have one of the form $Ax^2e^{ax}$.
The homogeneous solution is already in the form $e^{ax}$. Having a function of the same form in the non-homogeneous equation, unfortunately, requires a modification. Try the function $xe^{ax}$.
In general if your homogeneous solution takes the form $f(x)$ and the non-homogeneous equation has $f(x)$ also then try plugging in $xf(x)$.