Recently the topic of second order differential equations and their particular solutions is thought in our college , and they done some messy calculation and arrived at some complex formulas to Solve for particular solutions (they refer it as particular integral) , here is a example of one of those
$\frac{1}{f\left(D^{2}\right)} \sin a x=\frac{\sin a x}{f\left(-a^{2}\right)}$
But I am familiar with the method which uses trial solutions to find particular solution
So it is very hard to comprehend that how the term ($\frac{1}{f\left(D^{2}\right)} \sin a x$) can refer to particular solution of a differential equation with sinusoidal forcing term
Someone plese help me to understand it
Best Answer
Set $D:=\frac{\mathrm d}{\mathrm dx}$
$D(\sin ax)=a\cos ax, D^2(\sin ax)=(-a^2)\sin ax$
It serves for notational convenience. Given an ODE with constant coefficients, $f(D)y=p(x)$ then a particular solution is denoted by $y_p=\frac 1{f(D)}p(x).$ Think of it as $\overbrace{\frac{\mathrm d f}{\mathrm dx}(x)=g(x)}^{Df(x)=g(x)}\implies \overbrace{f(x)=\int g(t)\,dt}^{f(x)=\frac 1{D} g(x)}. $
Note that $D^{2r}(\sin ax)=(-a^2)^r\sin ax, r\ge 1$.
Suppose that the given ODE is:
$(D^{2r}+a_{2r-2}D^{2r-2}+...+a_2D^2)y=\sin ax\implies y=\frac 1{f(D^2)}\sin ax,\tag 1$ where $f(D^2):=D^{2r}+a_{2r-2}D^{2r-2}+...+a_2D^2$.
Note that: $(D^{2r}+a_{2r-2}D^{2r-2}+...+a_2D^2)\sin ax=f(D^2)\sin ax=f(-a^2)\sin ax\tag 2$
Use $(2)$ in $(1)$ to get: $y=\frac 1{f(-a^2)}\sin ax,$ provided that $f(-a^2)\ne 0$.