How is replacing $D^2$ with $-a^2$ justified when finding the particular integral?

Question

I read in my engineering mathematics textbook that if we have a differential equation of the form $f(D)y=\cos(ax)$ or $f(D)y = \sin(ax)$ where $f(D)$ is a polynomial in $D$ i.e. $\frac{d}{dx}$ then we can find its particular integral using the following rule:

• Write P.I as $$\frac{1}{f(D)} \cos(ax)$$ or $$\frac{1}{f(D)} \sin(ax)$$

• Replace all occurrences of $D^2$ in $f(D)$ with $-a^2$. For example if $f(D)$ is $2D^3+D^2+D$ then write it as $2D(-a^2)-a^2+D$.

• Then use the general methods to find evaluate the particular integral as usual.

But I'm not sure how the step in which we replace $D^2$ by $-a^2$ justified mathematically.

Answer 1

Because ultimately, $D^2$ is getting applied to $\sin(ax)$ and $\cos(ax)$, meaning that applying $D^2$ is the same as multiplying with $-a^2$.

Answer 2

Actually, what you are using is the identity \begin{align} f(D)(e^{\lambda x}) = f(\lambda)e^{\lambda x}. \end{align} For instance, \begin{align} \left(\left( \frac{d}{dx}\right)^2-2\frac{d}{dx} -3\right)e^{\lambda x} = (\lambda^2-2\lambda-3)e^{\lambda x}. \end{align} Any how, if $f(\lambda) \neq 0$ then we have \begin{align} f(D)\left( \frac{e^{\lambda x}}{f(\lambda)}\right) = e^{\lambda x} \ \ \ \Longleftrightarrow \ \ \ f(D)^{-1}e^{\lambda x}=\frac{e^{\lambda x}}{f(\lambda)} \end{align} which means $f(\lambda)^{-1}e^{\lambda x}$ is a particular solution to the differential equation with forcing term $e^{\lambda x}$. Moreover, if $\lambda = ia$ then we have \begin{align} f(D)\left(\frac{e^{iax}}{f(ia)} \right) = e^{iax}. \end{align} Since $f(D) = 2D^3+D^2+D$ then we have \begin{align} f(D)\left(\frac{e^{iax}}{-a^2-ia}\right) = e^{iax} \ \implies \ \ f(D)\left(\operatorname{Re}\left(\frac{e^{iax}}{-a^2-ia} \right) \right)=\cos ax. \end{align} Hence we see that \begin{align} \operatorname{Re}\left(\frac{e^{iax}}{-a^2-ia}\right)=\operatorname{Re}\left(\frac{-e^{iax}(a^2-ia)}{a^4+a^2}\right) = -\frac{a^2\cos ax+a\sin ax}{a^4+a^2} \end{align} is a particular solution.

Edit: Another way to make rigorous the solution to \begin{align} P(D) y = f(x) \end{align} is given by \begin{align} y = \frac{f(x)}{P(D)} \end{align} is by using Laplace transform. Since I only want a particular solution, I will impose an artifical initial condition $y(0) = y'(0) = \ldots = y^{(n-2)}(0) =y^{(n-1)}(0)=0$ if $P$ is degree $n$. Any how, by Laplace transform, we see that \begin{align} \mathcal{L}(P(D)y)= P(s)Y(s) = F(s) \end{align} where $Y(s)$ is the Laplace transform of $y(x)$ and $F(s)$ the Laplace tranform of $f(x)$. In particular, you get that \begin{align} Y(s) = \frac{F(s)}{P(s)} \ \ \implies \ \ \ y(t) = \mathcal{L}^{-1}\left( \frac{F(s)}{P(s)}\right) \end{align}