Can the Method of Undetermined Coefficients be used on $y''+2y'-y=t^{-1}e^t$? The question is not to solve, but to determine whether or not 
\begin{align}
y''+2y'-y=t^{-1}e^t\tag{1}
\end{align}
may be solved using the method of undetermined coefficients. I do not think so, as it is explicitly stated in my book (Nagle) that this method may only be used in cases where our forcing term $f\left(t\right)$ etc. is an exponential, sine or cosine, polynomial $p_n\left(t\right)$, or a product of these functions. Hence, in this case we would have
\begin{align}
f\left(t\right)=\frac{e^t}{t},\tag{2}
\end{align}
which is a quotient of a polynomial and exponential. But am I right?
On a side note, suppose instead
\begin{align}
f\left(t\right)=\alpha^{\beta t},\;\;\alpha,\beta\;\text{are constant.}\tag{3}
\end{align}
Then it seems this is also an exponential, and thus can be solved using the method of undetermined coefficients. Am I right in that assumption?
 A: Yes, this condition is right and correctly applied.
Note that $α^{βt} = e^{\ln(α)\,βt}$, so that there is nothing new gained by varying the base of the exponential function.
Also possible to solve with this method is the case where $f(t)=t·e^{βt}$ with the trial solution $(A+Bt)·e^{βt}$.
A: The method of indeterminate coefficients works well with the product of exponentials (including complex) and polynomials thanks to linear dependency.
When you plug a tentative function like $e^{Ct}t^k$ in the LHS, the derivatives
$$\left(e^{Ct}t^k\right)'=Ce^{Ct}t^k+e^{Ct}kt^{k-1},$$ are a linear combinations of an exponential times a power of equal or lower degree, and similarly for the higher derivatives. So when the RHS has the form $e^{Ct}R(t)$, by appropriately choosing the coefficients of a linear combination of the exponential times a power, from $0$ to the degree of $R$, you can reconstruct the RHS. The $r+1$ terms $e^{Ct}t^k$ form a basis on which you construct the solution.
This property does not necessarily hold for RHS of other forms. In particular, with a tentative function like $\dfrac{e^t}t$,
$$\left(\frac{e^t}t\right)'=\frac{e^t}t-\frac{e^t}{t^2},$$
$$\left(\frac{e^t}t\right)''=\frac{e^t}t-2\frac{e^t}{t^2}+2\frac{e^t}{t^3},$$
the degree of the denominator will go increasing with the order of the derivative, so that a basis would need an infinite number of members.
