Concept/Strategy:
Consider a 2nd order non-homogeneous differential equation of the form:
$$ay''+by'+cy=g(x) $$
We will first solve the above equation as if it were homogeneous and of the form:
$$ay''+by'+cy=0 \qquad(1)$$
Solving equation (1) is done using the methods described in the Homogeneous 2nd order pages.
After solving for the homogeneous case, we have a general solution (for the homogeneous case) of:
$$y_h = C_1 y_1 + C_2 Y_2 \qquad (2)$$
In the process of obtaining the above general solution, our "characteristic equation" also had two roots where roots r1 and r2 are some numbers m1 and m2 respectively:
$$r_1 = m_1 \;,\; r_2 = m_2 $$
Our original non-homogeneous differential equation can now be written as:
$$a(D-m_2)(D-m_1)y=g(x)$$
...where
$$y = \frac{1}{a}(D-m_1)^{-1} (D-m_2)^{-1} g(x) \qquad(3)$$
note: The proof of the above expression is not included here
We will also make use of the following theorem (also not proven here) in order to solve for y:
$$(D-m)^{-1} g(x) = e^{mx} \int{}{}e^{-mx}g(x)dx \qquad(4)$$
Theorem #4 will be used in a reiterative fashion until all of the D's are removed from equation (3) and we are left with a particular solution for y (designated as yp)
At this point we will have the general solution to our original non-homogeneous differential equation (designated as yg) that is the sum of our solution to the homogeneous case (eqn #2) and the particular solution (yp)
$$y_g = C_1 y_1 + C_2 Y_2 + y_p$$
In the next page we will take a look at an example
Continue on to Annihilator example (2nd Order)