Method of Undetermined Coefficients

We use the method of undetermined coefficients via complexification, which converts the trigonometric case into the exponential case, and then a direct computation for the exponential case using the structure of $p(r)$.

Step 1: Reduction to the complex exponential case. Write $\gamma = \alpha + i\beta \in \mathbb{C}$. Since

both $g_1(t) = P_n(t)e^{\alpha t}\cos\beta t$ and $g_2(t) = P_n(t)e^{\alpha t}\sin\beta t$ can be recovered as the real and imaginary parts of $G(t) := P_n(t) e^{\gamma t}$:

Since $a,b,c$ are real, if $Y(t) = U(t) + iV(t)$ (with $U,V$ real-valued) satisfies

then equating real and imaginary parts (using that differentiation and the real linear combination $a(\cdot)''+b(\cdot)'+c(\cdot)$ commute with taking real/imaginary parts) gives

So it suffices to solve the single complex equation $aY''+bY'+cY = P_n(t)e^{\gamma t}$ and then take real and imaginary parts; this handles both sinusoidal cases (and the pure polynomial/exponential case is the sub-case $\beta=0$, $\gamma=\alpha$ real, where we work directly with real $Y$).

Step 2: Solving $aY'' + bY' + cY = P_n(t) e^{\gamma t}$. Try $Y(t) = t^s W(t) e^{\gamma t}$ where $W(t) = w_0 + w_1 t + \cdots + w_n t^n$ is an undetermined polynomial of degree $n$ and $s$ is the multiplicity of $\gamma$ as a root of $p(r) = ar^2+br+c$. Substitute $Y = t^s W(t) e^{\gamma t}$ into the left side. Using the product rule repeatedly (or the identity below), one computes directly that for any smooth function $u(t)$,

where $p(\gamma) = a\gamma^2+b\gamma+c$ and $p'(\gamma) = 2a\gamma+b$. This identity follows by direct expansion: with $u = u(t)$,

so

confirming the claimed identity. Now set $u(t) = t^s W(t)$, a polynomial of degree $n+s$. We consider the three cases dictated by $s$. Case $s=0$ ($\gamma$ not a root, so $p(\gamma)\neq 0$). Then $u = W$, a degree-$n$ polynomial, and we need

Write $W(t) = \sum_{k=0}^n w_k t^k$. Because $p(\gamma) \neq 0$, the coefficient of $t^n$ on the left is $p(\gamma) w_n$ (the terms $W''$ and $W'$ contribute only degree $\le n-1$ and $\le n-2$ respectively), so matching the $t^n$ coefficient of $P_n(t)$ determines $w_n$ uniquely. Having fixed $w_n$, the coefficient of $t^{n-1}$ on the left is $p(\gamma)w_{n-1} + (\text{known quantity from } w_n)$, again solvable uniquely for $w_{n-1}$ since $p(\gamma)\neq 0$. Proceeding downward by finite induction on the degree $k = n, n-1, \dots, 0$, every $w_k$ is uniquely determined. Thus a polynomial $W$ of degree exactly $n$ (matching $P_n$) exists solving the equation, and $Y = We^{\gamma t}$ works. Case $s=1$ ($\gamma$ a simple root: $p(\gamma) = 0$, $p'(\gamma)\neq 0$). Then $u = tW(t)$, degree $n+1$, and $u'' = tW'' + 2W'$, $u' = tW' + W$. The equation $au''+p'(\gamma)u' + p(\gamma)u = P_n(t)$ becomes, using $p(\gamma)=0$,

i.e.

Collect by degree: the highest-degree term of $W'$ appearing multiplied by $t$ is $p'(\gamma) t W'$, whose leading contribution to degree $n$ comes from the degree-$(n-1)$ coefficient of $W'$ times $t$, and since $p'(\gamma)\neq 0$ this again allows solving degree-by-degree from top to bottom for the coefficients of $W$ (an ordinary linear triangular system with nonzero diagonal entries $p'(\gamma)$), exactly as in Case $s=0$ but shifted by one derivative. A polynomial $W$ of degree $n$ exists and is unique. Then $Y = tW(t)e^{\gamma t}$ solves the equation. Case $s=2$ ($\gamma$ a double root: $p(\gamma)=0$ and $p'(\gamma) = 2a\gamma + b = 0$, which for a monic-normalized real quadratic forces $\gamma = -b/2a$ and $a\gamma^2+b\gamma+c=0$). Then $u = t^2 W(t)$, and $u'' = t^2 W'' + 4tW' + 2W$. Since $p(\gamma)=p'(\gamma)=0$, the equation $au'' + p'(\gamma)u'+p(\gamma)u = P_n(t)$ collapses to

Here the coefficient of $t^n$ on the left, from the $2aW$ term (since $t^2W''$ and $4tW'$ only reach degree $n$ via lower coefficients of $W$ multiplied up — more precisely by matching top degree $n$: $2a w_n$ is the only degree-$n$ contribution from $2aW$, while $t^2 W''$ contributes at degree $n$ only through $w_n$ via $t^2 \cdot n(n-1)w_n t^{n-2}= n(n-1)w_n t^n$, and $4tW'$ contributes $4n w_n t^n$ from the $w_n$ term of $W'$ as well; summing, the coefficient of $t^n$ on the left is $a\big(n(n-1) + 4n + 2\big) w_n = a(n+1)(n+2) w_n$), and since $a\neq 0$ and $(n+1)(n+2) \ne 0$, this determines $w_n$ uniquely. As in the previous cases, proceeding by finite downward induction on degree, since the coefficient multiplying $w_k$ at each stage is $a(k+1)(k+2) \ne 0$, every coefficient of $W$ is uniquely determined. So a degree-$n$ polynomial $W$ exists, and $Y = t^2 W(t) e^{\gamma t}$ solves the equation.

Step 3: Assembling the real solution. In all cases we have produced $Y(t) = t^s W(t) e^{\gamma t}$ solving $aY''+bY'+cY = P_n(t)e^{\gamma t}$, with $W$ a (complex, in general) polynomial of degree $n$. Write $W(t) = Q_n(t) + iR_n(t)$ with $Q_n, R_n$ real polynomials of degree $\le n$ (degree exactly $n$ is not needed for the real decomposition to work — only for $W$ itself). Then

By Step 1, $U(t) = \operatorname{Re}Y(t) = t^s e^{\alpha t}(Q_n\cos\beta t - R_n\sin\beta t)$ solves $aU''+bU'+cU = g_1(t) = P_n(t)e^{\alpha t}\cos\beta t$, and $V(t)=\operatorname{Im}Y(t) = t^se^{\alpha t}(Q_n \sin\beta t + R_n\cos\beta t)$ solves $aV''+bV'+cV = g_2(t)$. Relabeling $-R_n$ as the new $R_n$ (or renaming appropriately in the sine case), each of $U$ and $V$ has exactly the claimed form

with $Q, R$ of degree at most $n$. This is precisely the asserted $y_p$ for $g_1$ or $g_2$ respectively, completing the proof for a single term $g$. $\blacksquare$

By linearity of the operator $L[y] = ay''+by'+cy$ (immediate from linearity of differentiation),