Particular solution

What effect does the forcing function $f$ have on the solution?
How might we solve for this effect, called the particular solution?
One answer to the latter question will be given in this lecture: using the method of undetermined coefficients.

Before we turn to this method, please recognize that other methods, such as the method of variation or Laplace transforms apply to more general forms of forcing $f$ (although the integrals that accompany each may be unknown).
When choosing the method of undetermined coefficients, we limit the scope of applicability to systems subjected to forcing functions that are complex exponentials (which include sinusoids) or polynomials.
The principle of superposition, discussed in the Mechatronics course, allows us to construct solutions for linear systems subject to linear combinations of complex exponentials and polynomials.

Method of undetermined coefficients

The method is:

Based on the form of the forcing function, propose an appropriate solution that includes undetermined coefficients (being careful to propose a solution linearly independent of the homogeneous solution),
Substitute this proposed solution into the ODE, and
Determine the undetermined coefficients by solving the algebraic system of equations that results from equating terms on each side of the equation.

If there is, in fact, a solution to the algebraic system—that is, for the undetermined coefficients—our proposed solution is our particular solution, with coefficients now determined.
However, if there is no solution,¹, our proposed solution is not our particular solution.

Some suggested solution proposals

How can one propose a solution?
There are no clear answers other than “be clever or use known solutions.”
As remarkably unsatisfying as this is, we can still rejoice in being let off the hook, since we are certainly not clever.

As mentioned above, this method only really works if the forcing function is a complex exponential or a polynomial (but this can be extended, using superposition, to a large class of problems of interest).
(tab:undetermined_coefficients?) is provided as a guide, but it essentially boils down to: if $f$ is a complex exponential, propose that $y_p$ is a complex exponential; if $f$ is a polynomial, propose that $y_p$ is a polynomial.

Suggested particular solutions $y_p(t)$ (with undetermined coefficients) to propose for various forcing functions $f$.
Let $k$, $\lambda$, $\omega$, and $\phi$ be real constants and $n$ be a positive integer. Furthermore, let $K_i$ be the undetermined coefficients.

$f(t)$	proposed $y_p(t)$	test value
$k$	$K_1$	$0$
$k t^n$	$K_n t^n + K_{n-1} t^{n-1} + \dots + K_0$	$0$
$k e^{\lambda t}$	$K_1 e^{\lambda t}$	$\lambda$
$k e^{j \omega t}$	$K_1 e^{j \omega t}$	$j \omega$
$k \cos(\omega t + \phi)$	$K_1 \cos(\omega t) + K_2 \sin(\omega t)$	$j \omega$
$k \sin(\omega t + \phi)$	$K_1 \cos(\omega t) + K_2 \sin(\omega t)$	$j \omega$

The parenthetical caveat

The only caveat, here, is the parenthetical warning from the three-step method about choosing a linearly independent solution.
This is a result of a theorem we have not considered, here, but suffice it to say that, in order for our general solution to simply be the sum of the homogeneous and particular solutions, as we will propose in the next lecture, these two must be linearly independent.

We will not only skip the details of why this is the case, but also the details of how to deal with it, opting instead for a simple recipe.
The ``test values’’ in (tab:undetermined_coefficients?) are to test whether or not the particular solution is a component of the homogeneous solution.
If the test value is equal to any root of the characteristic equation of multiplicity $\mu$, then the proposed solution should be multiplied by $t^\mu$.

Example 7.2

A particular solution
Find the particular solution for the equation

$$ \frac{d^5 y}{d t^5} + 14 \frac{d^4 y}{d t^4} + 81 \frac{d^3 y}{d t^3} + 248 \frac{d^2 y}{d t^2} + 408 \frac{d y}{d t} + 288 y = f(t), $$

which is the same as that of [@ex:homogeneous1], with

f(t) = acos (ωt),

with a ∈ ℝ and ω = 5 rad/s.

We needn’t re-solve for the roots of the characteristic equation, since we did that in [@ex:homogeneous1].
They were

$$ \begin{aligned} \lambda_{11} &= -4, \\ \lambda_{21,22} &= -3, \\ \lambda_{31,41} &= -2 \pm j 2. \end{aligned} $$

We now consult [@tab:undetermined_coefficients].
The forcing function is in the fifth row with k = a, ω = ω, and ϕ = 0.
So our proposed particular solution is

y_p(t) = K₁cos (ωt) + K₂sin (ωt),

where K₁ and K₂ are our undetermined coefficients.
Note that our table indicates we should test that jω = j5 is not a root of the characteristic equation, which it is not (j5 ≠ λ_i).

We now substitute y ↦ y_p into the ODE:

$$ \frac{d^5 y_p}{d t^5} + 14 \frac{d^4 y_p}{d t^4} + 81 \frac{d^3 y_p}{d t^3} + 248 \frac{d^2 y_p}{d t^2} + 408 \frac{d y_p}{d t} + 288 y_p = a \cos(\omega t). $$

The five derivatives are:

$$ \begin{aligned} \frac{d y_p}{d t} &= - K_1 \omega \sin(\omega t) + K_2 \omega \cos(\omega t) \\ \frac{d^2 y_p}{d t^2} &= - K_1 \omega^2 \cos(\omega t) - K_2 \omega^2 \sin(\omega t) \\ \frac{d^3 y_p}{d t^3} &= K_1 \omega^3 \sin(\omega t) - K_2 \omega^3 \cos(\omega t) \\ \frac{d^4 y_p}{d t^4} &= K_1 \omega^4 \cos(\omega t) + K_2 \omega^4 \sin(\omega t) \\ \frac{d^5 y_p}{d t^5} &= - K_1 \omega^5 \sin(\omega t) + K_2 \omega^5 \cos(\omega t) \end{aligned} $$

Substituting into the ODE, we collect cosine and sine terms and equate them with the right-hand side:

$$ \begin{aligned} K_2 \omega^5 + 14 K_1 \omega^4 - 81 K_2 \omega^3 - 248 K_1 \omega^2 + 408 K_2 \omega + K_1 &= a \\ -K_1 \omega^5 + 14 K_2 \omega^4 + 81 K_1 \omega^3 - 248 K_2 \omega^2 - 408 K_1 \omega + K_2 &= 0 \end{aligned} $$

This is a linear system of two equations in two unknowns K₁ and K₂.
Solving it (with ω = 5 rad/s) gives:

K₁ ≈ (82.00⋅10⁻⁶)a, K₂ ≈ (−159.4⋅10⁻⁶)a

Since there is an algebraic solution, we know our proposed particular solution is, in fact, a particular solution!
And, of course, our coefficients K₁ and K₂ are now determined.

One should not simply throw up one’s hands at a certain point and declare “there’s no solution!” Rather, one should prove that there is none.↩︎

Online Resources for Section 7.4

No online resources.

\(f(t)\)	proposed \(y_p(t)\)	test value
\(k\)	\(K_1\)	\(0\)
\(k t^n\)	\(K_n t^n + K_{n-1} t^{n-1} + \dots + K_0\)	\(0\)
\(k e^{\lambda t}\)	\(K_1 e^{\lambda t}\)	\(\lambda\)
\(k e^{j \omega t}\)	\(K_1 e^{j \omega t}\)	\(j \omega\)
\(k \cos(\omega t + \phi)\)	\(K_1 \cos(\omega t) + K_2 \sin(\omega t)\)	\(j \omega\)
\(k \sin(\omega t + \phi)\)	\(K_1 \cos(\omega t) + K_2 \sin(\omega t)\)	\(j \omega\)