Undetermined Coefficients

Definition Let $L[y] = g$ be a linear equation of order $n$ on an open interval $I$ on which the leading coefficient never vanishes and the other coefficients and the forcing $g$, divided by it, are continuous. The complementary solution is the general solution of the homogeneous equation $L[y] = 0$,

where $y_1, \ldots, y_n$ is a fundamental set of solutions on $I$ and $C_1, \ldots, C_n$ are arbitrary constants. A particular solution $y_p$ is any one function with $L[y_p] = g$ on $I$. Every solution of $L[y] = g$ on $I$ equals $y_c + y_p$ for exactly one choice of the constants, because the difference of two solutions of the forced equation solves the homogeneous one. Initial conditions are imposed on the sum $y_c + y_p$, never on $y_c$ alone. In this lesson the operator has constant coefficients,

with real numbers $a_0, \ldots, a_n$ and $a_n \ne 0$, so every interval on which $g$ is continuous qualifies. Its characteristic polynomial is $P(r) = a_nr^n + \cdots + a_1r + a_0$, and $L[e^{rt}] = P(r)\,e^{rt}$.

Definition The method of undetermined coefficients finds a particular solution of $L[y] = g$ when $L$ has constant coefficients and $g$ is a finite linear combination of terms $t^ke^{\alpha t}\cos\beta t$ and $t^ke^{\alpha t}\sin\beta t$, with $k$ a nonnegative integer, $\alpha$ real and $\beta \ge 0$. Polynomials, exponentials, sines, cosines and their products are all of this kind, and so are their derivatives. The method has four steps.

1. Find $y_c$ from the roots of the characteristic polynomial.
2. Write a trial function with unknown coefficients for each forcing term, including the overlap correction described below.
3. Substitute the trial into $L[y] = g$ and equate the coefficients of each function $t^je^{\alpha t}\cos\beta t$ and $t^je^{\alpha t}\sin\beta t$ on the two sides.
4. Solve the resulting linear system for the coefficients, then add $y_c$ and apply any initial data to the sum.

With the trial chosen by the rules of the next two cards, this linear system always has exactly one solution. The coefficients are fixed by the equation, unlike the constants in $y_c$, which are fixed by initial data. The method does not cover forcing such as $\ln t$, $\tan t$ or $\sec t$, whose derivatives do not stay inside the span of any finite set of functions, and it does not cover equations with variable coefficients. Those equations still have solutions, and variation of parameters finds them.

Definition For a forcing term $t^ke^{\alpha t}\cos\beta t$ or $t^ke^{\alpha t}\sin\beta t$ with $\beta \ne 0$, the trial function is

where $A(t) = A_0 + A_1t + \cdots + A_kt^k$ and $B(t) = B_0 + B_1t + \cdots + B_kt^k$ have unknown coefficients and $m$ is the overlap multiplicity of the next card. For a term $t^ke^{\alpha t}$, the case $\beta = 0$, the trial is $t^me^{\alpha t}A(t)$. Both polynomials contain every power from $t^0$ to $t^k$, and the cosine and the sine both appear even when the forcing contains only one of them. Forcing terms with the same $\alpha$ and $\beta$ share one trial, whose degree is the largest $k$ among them. Terms with different $\alpha$ or $\beta$ get separate trials, and by superposition the particular solutions found for them are added. When $m = 0$ the common cases look like this.

Definition For a forcing term with exponent $\alpha$ and frequency $\beta$, let $\lambda = \alpha + i\beta$. The overlap multiplicity $m$ is the multiplicity of $\lambda$ as a root of the characteristic polynomial $P$, with $m = 0$ when $P(\lambda) \ne 0$. Because $P$ has real coefficients, the conjugate $\bar{\lambda}$ has the same multiplicity, so one value of $m$ serves for the cosine and the sine. The whole trial is multiplied by $t^m$, and no lower powers are added back. The reason is the shift identity

which holds for every differentiable function $f$, complex values allowed. Write $P(r) = (r - \lambda)^mQ(r)$ with $Q(\lambda) \ne 0$. The $m$ factors $D - \lambda$ differentiate the polynomial factor $m$ times, so they send $e^{\lambda t}$ times any polynomial of degree less than $m$ to zero. The factor $t^m$ lifts the trial polynomial above that range, and the image under $L$ is then $e^{\lambda t}$ times a polynomial of the same degree $k$ as in the forcing.

Definition A nonzero constant-coefficient operator $M(D)$ annihilates a function $g$ on an interval if $M(D)\,g = 0$ there. The operator $D^{k+1}$ annihilates every polynomial of degree at most $k$, the operator $(D - \alpha)^{k+1}$ annihilates $t^je^{\alpha t}$ for $0 \le j \le k$, and the operator

annihilates $t^je^{\alpha t}\cos\beta t$ and $t^je^{\alpha t}\sin\beta t$ for $0 \le j \le k$. If $M(D)\,g = 0$, applying $M(D)$ to both sides of $L[y] = g$ shows that every solution also solves the homogeneous equation $M(D)L[y] = 0$, whose characteristic polynomial is $M(r)P(r)$. The basis functions of this larger equation that are not already in the basis of $y_c$ form the trial function, with the overlap powers of $t$ built in. The larger equation is necessary but not sufficient, so the coefficients must still come from substituting into $L[y] = g$. No nonzero constant-coefficient operator annihilates $\ln t$ or $\tan t$, which is why the method stops at the class of forcing described above.