Consider the equation for the charge on a capacitor in an LRC circuit $$\left({\frac{1}{2}}\right)\frac{d^2q}{dt^2}+5\frac{dq}{dt}+12q = E$$ which is linear with constant coefficients.

First we will work on solving the corresponding homogeneous equation. Divide through the equation by the coefficient on $\frac{d^2q}{dt^2}$, and find the auxiliary equation (using m as your variable) $= 0$ which has roots .

The solutions of the homogeneous equation are

Now we are ready to solve the nonhomogeneous equation $\frac{d^2q}{dt^2}+10\frac{dq}{dt}+24q = 2 E$. We will use $e^{-6t}$ to do reduction of order (it doesn't matter which one of the homogenous solutions we choose), which will give us a particular solution of the nonhomogeneous equation. With this choice the particular solution has the form
$y_p = ue^{-6 t}$.
Then (using the prime notation for the derivatives)
$y_p^\prime =$
$y_p^{\prime\prime} =$
So, plugging $y_p$ into the left side of the differential equation, and reducing, we get


Using this reduced form, moving the term $e^{-6t}$ to the right side of the equation, we can rewrite $y_p^{\prime\prime}+10 y_p^\prime + 24 y_p= 2 E$ in terms of $u^{\prime\prime}$ and $u^\prime$ as


At this point to solve for u we want to integrate both sides, but to do that we need to integrate E. We will look at two special cases for E so that you get the idea how to proceed.

Case 1: E is a constant (DC voltage).
Then integrating both sides
$u' - 2 u =$ (we don't use a constant when we do this integration, it turns out to be redundant)
This is a linear equation with integrating factor
Solving this equation, using b as our constant, we get that
$u =$
Plugging this back into $y_p$ we finally find that
$y_p =$
Making the general solution $y = a e^{-6t} + y_p = a e^{-6t} + b e^{-4t} + \frac{2 E}{24}$. Notice that we already knew that $e^{-6t}$ and $e^{-4t}$ where solutions to the homogeneous equation, we just did all that work to figure out that the constant $\frac{2 E}{24}$ is a solution of the nonhomogeneous equation, which is kind of obvious when you look at it. Now lets do an example where the answer will not be so obvious.

Case 2: E = sin(pt) (AC voltage).
You might want to break out your TI-89, Maple, Mathematica, etc for this one. Or you can practice up your integration by parts, a lot.
Starting with the equation from above, $u''-2u' = 2Ee^{6t}$, integrating both sides (again without using a constant) we get
$u' - 2 u =$
This is a linear equation with integrating factor , just like the example above. Solving this equation, using b as our constant, we get
$u =$
Plugging this back into $y_p$ we finally find that
$y_p =$
Making the general solution $y = a e^{-6t} + y_p = a e^{-6t} + be^{-4t}+\frac{2}{\left(36+p^{2}\right)\!\left(16+p^{2}\right)}\!\left(24\sin\!\left(pt\right)-6p\cos\!\left(pt\right)-4p\cos\!\left(pt\right)-p^{2}\sin\!\left(pt\right)\right)$
Since we already knew that $e^{-6t}$ and $e^{-4t}$ were solutions to the homogeneous equation, what we really worked out here is that the particular solution is $\frac{2}{\left(36+p^{2}\right)\!\left(16+p^{2}\right)}\!\left(-6p\cos\!\left(pt\right)+24\sin\!\left(pt\right)+\left(-4\right)p\cos\!\left(pt\right)-p^{2}\sin\!\left(pt\right)\right)$. Going back to look at our calculations, we could have done the integrations above without using constants, and pieced the solution together using superposition.