How do you solve this differential equation? Though I've read questions on this site and really appreciate the quality of the answers, this is my first question, so I hope it follows the site's guidelines.
When working with potential energy curves from physics, I came across the equation 
$a(x(t)) = 6 x(t) + 5$ 
where $a(x(t))$ denotes acceleration as a function of position, and position is a function of time.
I was wondering how one would solve this equation, if possible, to find the particle in question's trajectory as a function of time, i.e. let $x = x(t)$ and solve the equation given that $a(t) = x''(t)$.
 A: We can rewrite the ODE as:
$$x'' -6x = 5$$
The homogeneous solution is found by noting:
$$m^2 - 6 = 0 \rightarrow m_{1,2} = \pm ~ \sqrt{6}$$
This gives:
$$x_h(t) = c_1 e^{\sqrt{6} t} + c_2 e^{-\sqrt{6} t}$$
For the particular, we choose $x_p = a$, and substitute into the ODE and arrive at $a = -\dfrac{5}{6}$.
Our final solution is:
$$x(t) = x_h(t) + x_p(t) = c_1 e^{\sqrt{6} t} + c_2 e^{-\sqrt{6} t} -\dfrac{5}{6}$$ 
A: What you have is 
$$
\frac{d^2 x}{d t^2} = 6 x  + 5 \tag 1$$
Since the equation is linear second order, the solution is of the form 
$$
x = A + B_1 e^{a_1 t}  + B_2 e^{a_2 t}$$
Differentiating
$$
x'= a_1 B_1 e^{a_1 t} + a_2 B_2 e^{a_2 t} \\
x''= a_1^2 B_1 e^{a_1 t} + a_2^2 B_2 e^{a_2 t}
$$
Substituting in (1) we get
$$
a_1^2 B_1 e^{a_1 t} + a_2^2 B_2 e^{a_2 t} = 6 A + 6 B_1 e^{a_1 t}  +6  B_2 e^{a_2 t}+5 \tag 2$$
Comparing coefficients in (2) we get
$$
A = -5/6\\
a_1^2 = 6 \\
a_2^2 = 6
$$
If $a_1 = a_2$ then the two terms merge into one, so we pick $a_1 \neq a_2$ giving
$$
a_1 = + \sqrt{6}, ~~a_2 = - \sqrt{6}$$
Hence the general solution is
$$
x(t)= -5/6 + B_1 e^{\sqrt{6} t} + B_2 e^{-\sqrt{6} t}$$
If you know the initial position and initial velocity (or any two conditions) then you can find $B_1$ and $B_2$.
A: If I well understand the new wording of the question, a manner to solve the ODE is shown in attachment :

