How to solve $\ddot{x} = x + 8e^{3t}$ without Laplace transform? How do you solve this diff-eq without using laplace transforms?
$\ddot{x} = x + 8e^{3t}$
That $8e^{3t}$ is throwing me off...Also, I need to get two constants in the answer so I can solve for initial values.
 A: The undetermined coefficients method works neatly.
The complementary solution is $$x''-x=0 \iff x= c_1e^{-t}+c_2e^t$$
Then $$x_p=a_1 e^{3t}, x_p \;''=9a_1e^{3t}$$
From that you get $a=1$ and $$x=x_p+x_c=e^{3t}+c_1e^{-t}+c_2e^{t}$$
A: Let's go through the steps of solving such an equation like this one.
$$
\ddot x - x = 8e^{3t}
$$
This problem actually has two parts. Assuming that this equation describes some physical problem, the homogenous equation $\ddot x - x = 0$ describes inherent qualities of the system itself (e.g. the way a pendulum of known mass and length will move) while the right-hand side describes some form of input (e.g. force exerted on the pendulum).
The first step is to solve the homogenous equation. This will give the general solution, which contains a number of arbitrary constants. The existence of these constants means that this solution describes a family of functions, which is where it gets its name general solution. The constants take particular values only when there are initial conditions to the problem. For ODEs with constant coefficients, the solution is of the form $x(t)=e^{at} \Rightarrow \ddot x =a^2 e^{at}$. Substitute in the equation:
$$
a^2 e^{at} - e^{at} = 0 \\
(a^2 -1) e^{at} = 0 \\
a^2 - 1 = 0 \\
(a-1)(a+1) = 0 \\
a_1 = 1, \quad a_2=-1
$$
So the general solution is
$$
x_g(t) = c_1 e^t + c_2 e^{-t}
$$
This function completely describes the system, but not the system's reaction to the input $f(t)=8e^{3t}$. Unlike the homogenous solution, this solution $x_p$ will be a single function (i.e. no arbitrary constants). The generality of the complete solution $x=x_g+x_p$ will stem from the generality of the homogenous solution. This makes things easier because any one function that solves the initial equation is enough, unlike the previous step where we always have to assume the most general form of solution. In this example, the particular solution will be of the form $x=Ae^{3t} \Rightarrow \ddot x = 9Ae^{3t}$. So,
$$
9Ae^{3t} - A e^{3t} = 8e^{3t} \\
8Ae^{3t} = 8e^{3t} \\
A = 1
$$
Note that this particular solution $x_p = e^{3t}$ is mostly dependent on the form of the non-homogenous term $f(t)$ and not the form of the differential equation (you might notice that the diff. equation only influenced the value of $A$ and not the form of $x_p$).
The complete solution is
$$
x(t) = c_1 e^t + c_2 e^{-t} + e^{3t}
$$
You can apply initial conditions at this point if needed.
A: First, solve the homogeneous problem $\ddot{x}_h = x_h$. This has solutions $x_h = A e^{-t}+B e^t$. 
Then, guess a particular solution of the form $x_p = Ce^{3t}$, and determine the value of $C$ that works. 
Finally, the general solution is $x = x_h+x_p$. 
A: Hint: This equation is of the form $(D^2 - 1)x = 8e^{3t}$. A simple way to find a particular solution is to guess $x_p = Ae^{3t}$ and solve for $A$ in $(D^2 - 1)x_p = 8e^{3t}$. Then the complete solution is $x_p + x_h$ where $x_h$ satisfies $(D^2 - 1)x_h = 0$. You should have two degrees of freedom in $x_h$.
