Solving Second Order Non-Homogeneous ODE I haven't solved an ODE in quite some time and I'm finding myself a bit rusty! If someone could help me solve the following ODE that would be great:
$$\frac{d^2x}{dt^2} + \frac{1}{c_1} \frac{dx}{dt} = c_2$$
I attempted using the method of undetermined coefficients and found the complementary solution as follows:
$$r^2 + \frac{1}{c_1}r = 0 \implies r = 0 \text{ or } r = -\frac{1}{c_1} \implies x_C(t) = A + Be^{-\frac{1}{c_1}t} \text{ , } A,B \in \Bbb{R}$$
From there I attempted to find the particular solution with a guess of $x(t)= C \in \Bbb{R}$ which would have resulted in $c_2 = 0$ which I don't believe to be right.
Any help would be great!
 A: $$
x_h(t)=A+B\mathrm{e}^{-t/c_1}
$$
is the general solution of the homogeneous equation
$$
x''+c_1x'=0.
$$
In order to find the general solution of the inhomogeneous one
$$
x''+c_1x'=c_2, \tag{1}
$$
you need first to find a particular solution of it. One such is
$$
x_p(t)=(c_2/c_1)t.
$$
Then the general solution of (1) is
$$
x(t)=x_h(t)+x_p(t)=A+B\mathrm{e}^{-t/c_1}+(c_2/c_1)t.
$$
Note. In general, if $x_h=c_1\varphi_1+\cdots+c_n\varphi_n$ is the general solution of
$$
x^{(n)}+a_{n-1}(t)x^{(n-1)}+\cdots+a_0(t)x^{(0)}=0,
$$
and $x_p$ is a particular solution of
$$
x^{(n)}+a_{n-1}(t)x^{(n-1)}+\cdots+a_0(t)x^{(0)}=h(t), \tag{2},
$$
then $x=x_h+x_p$ is the general solution of (2).
A: $$\frac{d^2x}{dt^2} + \frac{1}{c_1} \frac{dx}{dt} = c_2$$
For the particular solution try:
$$x_p'=A \implies A=c_2c_1$$
$$ \implies x_p=c_1c_2t+K$$
You can choose $K=0$. Note that you can integrate directly the DE:
$$\frac{d^2x}{dt^2} + \frac{1}{c_1} \frac{dx}{dt} = c_2$$
$$\frac{dx}{dt} + \frac{1}{c_1} x(t) = c_2t+K$$
Then it's just a first order differential equation that is easy to solve.
