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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. – StuartHa Dec 11 '12 at 18:26
up vote 1 down vote accepted

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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I don't know, If I got it completely correct, but if I have $y"=cx^2$, would I assume $y_p=Ax^4+Bx^3+Dx^2+Ex+F$ – 007resu Dec 11 '12 at 19:43
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) – Artem Dec 11 '12 at 19:45

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.

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Did you not mean $x^{n+k}$ as leading term. It contradicts with above observation. – 007resu Dec 11 '12 at 20:23

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