I'm having trouble applying the method of undetermined coefficients, as explained in Apostol's Calculus, to second order linear differential equations with constant coefficients containing trigonometric functions. For second-order linear differential equations with constant coefficients of the form $y'' + ay' + by = p(x)e^{mx}$, where $p$ is a polynomial, we assume that $y = u(x)e^{mx}$ for some function $u$ to arrive at a polynomial in $u$, which is easily solved. However, this method isn't working as expected when applied to equations of the form $y'' + ay' + by = p(x)e^{mx}\sin\alpha x$ or $y'' + ay' + by = p(x) e^{mx} \cos\alpha x$. I begin by assuming that $y = e^{mx}u(x)\sin\alpha x$ for some function $u$. Differentiating twice, we obtain an equation of the form
$$e^{mx}\sin\alpha x (\phi_1 u'' + \phi_2 u' + \phi_3 u + \phi_4) + e^{mx}\cos \alpha x (\psi_1 u'' + \psi_2 u' + \psi_3 u + \psi_4) = p(x)e^{mx}\sin \alpha x$$
where $\phi_i$ and $\psi_i$ depend only on $a$, $b$, $m$ and $\alpha$ for $1 \leq i \leq 4$. Dividing by $e^{mx}\sin\alpha x$ gives rise to a system of differential equations equations of the form
$$\phi_1 u'' + \phi_2 u' + \phi_3 u + \phi_4 = p(x) \\
\cot \alpha x (\psi_1 u'' + \psi_2 u' + \psi_3 u + \psi_4) = 0$$
A similar situation arises when I assume that $y = e^{mx}(u(x)\sin\alpha x + v(x) \cos \alpha x)$ for some functions $u$ and $v$, or when $\sin \alpha x$ is replaced by $\cos \alpha x$. I have tried dividing through by $\cot \alpha x$ in the second equation and solving the resultant system, but my answers in example problems aren't agreeing with those given in the text. In addition, my method leads to quite complex formulas. This leads me to believe that I'm applying the method of undetermined coefficients incorrectly. Thus, I'm looking for some help applying this method to equations of the indicated type, preferably using only tools and techniques developed by Apostol through Section 8.16 of Calculus (Vol. I).
Edit: A particular example is $y'' – 3y' = 2e^{2x}\sin x$. Letting $y = ue^{2x}\sin x$ for some function $u$, I obtain $e^{2x}\sin x (u'' – 2u) + e^{2x}\cos x(u'' + 3u' + 3u) = 2e^{2x}\sin x$. This gives rise to the pair of equations $u'' – 2u = 2$ and $\cot x (u'' + 3u' + 3u) = 0$. Proceeding from this point, I obtain an incorrect solution (although I admit, my problem could be simple algebra).
Best Answer
Assuming you know the method called undetermined coefficients, I think we should consider different cases. These cases are as follows. Let you are facing $$a_ny^{(n)}+\cdots a_0y=Q(x),~~~a_n\neq 0, ~~Q(x)\not\equiv 0~(x\in I)$$ You may know that for finding the particular solution $y_p(x)$, using this method, we have to compare the terms of $Q(x)$ and the terms in $y_c(x)$ such that all terms are independent on interval $I$. Now we can consider some cases which one of these are as follows:
For example, in your latest sample, the axillary equation is $m^2-3m=0$ and so $$y_c(x)=C_1+C_2e^{3x}$$ There is no term in $y_c$ similar to term in $\sin x e^{2x}$ so we set $y_p=A\sin x e^{2x}+B\cos x e^{2x}$ for proper values of $A$ and $B$.