solving this 2nd order ODE Could someone help me with this problem?
I have a second order ODE as such:
$$\frac{d^2x}{dt^2}+\beta \frac{dx}{dt} = f(t) $$
I am not sure how to solve this linear ODE, was hoping someone could help.
does this look right?
$$e^{\beta t} \frac{dx}{dt}+\beta e^{\beta t}x=e^{\beta t} g(t)  $$
$$ \frac{d}{dt}( xe^{\beta t} )= e^{\beta t} g(t) $$
$$xe^{\beta t} = \int e^{\beta t} g(t)  $$
$$x(t) = \frac{\int e^{\beta t} g(t)}{e^{\beta t}}  $$
 A: Integrate once and we have
$$\frac{dx}{dt} + \beta x = \int_{t_0}^t f(t') \ dt'$$
Call the RHS $g(t)$. Now, do you know how to handle an equation of the sort 
$$\frac{dx}{dt} + \beta x = g(t)$$
?
A: To proceed we will first solve the associated homogeneous equation:
$$
\frac{d^2x_h}{dt^2}+\beta \frac{dx_h}{dt} = 0
$$
of course the solutions will be:
$$
x_h(t) = c_1 \cdot e^{\lambda_1}t + c_2 \cdot e^{\lambda_2}t 
$$
Where $ \lambda_{1,2} $ are the solutions of the second degree equation
$$
\lambda^2 + \beta \cdot \lambda = 0 
$$
Then, once you have $ x_h $, you then use variation of parameters (or any annihilator approach) to  find a particular (only one) $ x_p $ that solves:
$$
\frac{d^2x_p}{dt^2}+\beta \frac{dx_p}{dt} = f(t)
$$
and the general solution for: 
$$
\frac{d^2x}{dt^2}+\beta \frac{dx}{dt} = f(t)
$$
is 
$$
x(t) = x_h(t) + x_p(t)
$$
A: So for another answer, the problem to solve is:
$$
\frac{d^2x}{dt^2}+\beta \frac{dx}{dt} = f(t)
$$
let us use the substituion, $r(t) = \dfrac{dx}{dt}$ so that  and $\dfrac{dr}{dt} = \dfrac{d}{dt} \dfrac{dx}{dt} = \dfrac{d^2x}{dt^2} $. We have then:
$$
\dfrac{dr}{dt} + \beta r = f(t) 
$$
and this is a first order ode that can be solved using the integration factor method
