Particular Solution of Differential Equations While solving second order non-homogenous differential equations of the form
$y''+y'=5$, I realized that unlike while solving the ones of the form $y''+y'+y=Ax^n$ where we assume, $y_p=Ax^n+Bx^{n-1}+\cdots$, we assume in this case that $y_p=Ax$ instead of $A$. Can some one give me some insight. Or if this is wrong, how to find particular solution of this form (even though this is very basic).
 A: The reason is that any constant term will disappear when you differentiate it, so even if you put $y_p=Ax+B$, the term $B$ wouldn't appear at all when you wrote $y''_p + y'_p = 5$ in terms of $x$. Since you're looking for a particular solution, rather than a complete family of solutions, it's fine to drop it.

More generally, if you had
$$a_n\dfrac{d^ny}{dx^n} + a_{n-1} \dfrac{d^{n-1}y}{dx^{n-1}} + \cdots + a_{n-k}\dfrac{d^{k}y}{dx^{k}} = c$$
(or even more general equations than this), where $0 \le k<n$ you'd need only substitute
$$y_p = b_n x^n + \cdots + b_k x^k$$
since terms in $x^{k-1}$ or lower powers of $x$ all disappear when differentiated $\ge k$ times.
A: You are looking for this particular solution because your right hand side has the form
$$
f(x)=5e^{0x},
$$
and zero is a root of your characteristic polynomial $p(r)=r^2+r$. If it was not a root, then the form is $y_p=A$.
Added: The general rule as follows: if your right hand side has the form 
$$
f(x)=P_n(x)e^{\alpha x},
$$
where $P_n(x)$ is a polynomial of degree $n$ and $\alpha$ is a constant, then you should look for a particular solution in the form
$$
y_p(x)=Q_n(x)x^{k}e^{\alpha x},
$$
where $Q_n(x)$ is a polynomial of degree $n$ with undetermined coefficients, $k$ is the multiplicity y of $\alpha$ as a root of characteristic polynomial (it can be also zero it is not a root). 
