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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).

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    For the particular solutions of linear ODEs, you don't have to be picky about how you get them. If it solves the equation, then it's fine.2012-12-11

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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).

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    no, you would assume generally $y_p=(Ax^2+Bx+C)x^2$ (however, how it should be clear from your particular problem $y_p=Ax^4$ will suffice). (see also my edit of the answer)2012-12-11
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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 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.

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    Did you not mean $x^{n+k}$ as leading term. It contradicts with above observation.2012-12-11