Method of Undetermined Coefficients using X's on left hand side I have a couple questions on this question.
The question is asking me to find the general solution to
$$x'' + 6x' + 9x = \cos(2t) + \sin(2t).$$
Solving for the general solution, I got $$Y_c = C_1e^{-3x}+ C_2xe^{-3x}.$$
I was wondering if the fact that the left hand side uses $x$'s instead of $y$'s matters? For almost every question, it uses $y$'s instead of $x$'s.
Also, for $Y_p$, I got $Y_p = A\cos(2t) + B\sin(2t)$.
I ultimately got the answer to be $$C_1e^{-3x}+ C_2xe^{-3x} - (7\cos(2t)/169) +(17\cos(2t)/169),$$ which I do not feel to be correct.
 A: In this problem, $x$ is the dependant variable, and $t$ is the independant variable. It looks like for your homogenous solution, you solved the homogenous ODE
$$y_c''(x)+6y_c'(x)+9y_c(x)=0$$
With the result
$$y_c(x)=C_1e^{-3x}+C_2xe^{-3x}$$
All the work you did in solving this is totally valid, but you renamed both variables for the ODE. Using the original variables from the problem, we get,
$$x_c''(t)+6x_c'(t)+9x_c(t)=0$$
$$\Rightarrow x_c(t)=C_1e^{-3t}+C_2te^{-3t}$$
Your $y_p(t)$ is calculated correctly, but again should actually be named $x_p(t)$.
You can then get the final answer by adding $x_c(t)$ and $x_p(t)$.
A: The complementary solution is $x_c(t)=C_1e^{-3t}+C_2te^{-3t}$. Note: $x$ is the dependent variable and $t$ is the independent variable in this case.
Moreover, the complementary solution is the solution to the following second-order homogeneous differential equation: $$x_c''+6x_c'+9x_c=0.$$ Finally, in order to get the particular solution, one must guess it to be: $$y_p(t)=A\cos(2t)+B\sin(2t)$$ where $A$ and $B$ are to be found using the Method of Undetermined Coefficients.
After substitution and some algebraic manipulation, the equation becomes:
$$(5A+12B)\cos(2t)+(-12A+5B)\sin(2t)=\cos(2t)+\sin(2t)$$
This implies one gets the following system of two equations with two unknowns via Method of Undetermined Coefficients:

*

*$5A+12B=1$,

*$-12A+5B=1$.

Using Cramer’s rule to solve the system, the solution to the system is $A=-7/169$ and $B=17/169$.
This implies the particular solution is $$x_p(t)=-\frac7{169}\cos(2t)+\frac{17}{169}\sin(2t).$$
Hence, the general solution to the non-homogeneous second-order differential equation is:
$$X(t)=x_c(t)+x_p(t)=C_1e^{-3t}+C_2te^{-3t}-\frac7{169}\cos(2t)+\frac{17}{169}\sin(2t)$$ where $C_1$ and $C_2$ are constants.
