# Particular solution for a 2nd order linear ODE - finding the sign of the constant?

Background

I am considering the following 2nd order ODE from my vibrations analysis textbook:

$$M\ddot{x}+c\dot{x} +kx =F_0\sin(\omega t)$$

where $$M$$ is the mass, $$k$$ is the spring stiffness, and $$c$$ is the damping coefficient.

The solution to this ODE consists of the homogeneous solution plus the particular solution. The homogeneous solution corresponds to transient vibrations that eventually die out, so we're most interested in the particular solution. Assume that $$x_p(t)=X\sin(\omega t-\phi)$$, where $$X$$ and $$\phi$$ are constants. Furthermore, represent the particular solution in complex form to simplify some of the algebra/derivatives. \begin{align*} x_p(t) &= Xe^{i(\omega t - \phi)}\\ \dot{x}_p(t)&=X\omega i e^{i(\omega t - \phi)}\\ \ddot{x}_p(t)&=-X\omega^2e^{i(\omega t - \phi)} \end{align*} Substituting these expressions into the original ODE allows us to solve for $$X$$ and $$\phi$$.

\begin{align*} \phi&=\arctan\left(\frac{c\omega}{k-M\omega^2}\right)\\ X&=\pm\frac{|F_0|}{\sqrt{(k-M\omega^2)^2+(c\omega)^2}} \end{align*}

Question

Finally to my questions.

1. As far as I can tell, there's no reason to choose $$X$$ to be positive over $$X$$ being negative. In fact, if you work through the algebra, you'll find you can write $$X=F_0\sin(\phi)/(c\omega)$$, which makes it seem like the sign of $$X$$ depends on $$\phi$$, since $$F_0, c, \omega$$ are typically positive. Yet my textbook consistently chooses $$X$$ to be positive. Why can we take $$X$$ to be positive?

2. My textbook often uses the trick of representing $$x_p(t)$$ in complex form. In this case, $$x_p(t) = X\sin(\omega t - \phi)$$, so at the end of the analysis we take the imaginary part as the solution we actually care about. Why does this work? I know that if $$y_1$$ and $$y_2$$ are the solutions to an ODE, then $$y_1+y_2$$ is also a solution. But this is more like knowing $$y_1+y_2$$ is a solution and assuming that $$y_1$$ and $$y_2$$ separately are solutions.

1. We can choose $$X$$ to be positive because $$(-X)e^{i(\omega t - \phi)} = X e^{i(\omega t - (\phi + \pi))}$$. Since adding $$\pi$$ to $$\phi$$ doesn't change $$\tan\phi$$, we can always choose $$\phi$$ such that $$X$$ is positive.
2. This works because the differential operator (call it $$D$$ for brevity) on the LHS is real and linear. Since it's real, we have $$D[x]^* = D[x^*]$$. Since it's linear, we have $$i(D[x] - D[x^*]) = D[i(x - x^*)]$$. Put these together and you get $$\operatorname{Im}[D[x]] = D[\operatorname{Im}[x]]$$ (and similarly $$\operatorname{Re}[D[x]] = D[\operatorname{Re}[x]]$$). This trick does not work if either of these conditions isn't satisfied.
• It seems like the negative on $X$ simply changes the "start point" of our particular solution. I agree with your mathematics, and I see that we can choose $X$ to be positive, but for a given physical situation how do we know if $X$ should be positive? I suppose our choice has an affect on the other constants when we use the initial conditions to find the complete function $x(t)=x_h(t)+x_p(t)$? Commented May 3, 2022 at 19:58
• @nwsteg That's a bit outside of the realm of mathematics. It may be that choosing $X$ to be negative makes more sense in a particular physical situation, but that would be specific to that case. The important mathematical point is that you can choose $X$ to be positive or negative. Commented May 3, 2022 at 21:54