4.5; the surprisingly short yet very important section.

Section 4.5 is entitled “Inhomogeneous Equations; the Method of Undetermined Coefficients.”

An inhomogeneous equation is of the form y” + py’ + qy = f, where p, q, and f are functions of the independent variable (this variable is usually t). Recall that in the last section (or the section before that, or something) that f is the inhomogeneous term, and we called it the forcing term.

Here’s a theorem for you:

“Suppose that y_p is a particular solution to the inhomogeneous equation [above], and that y₁ and y₂ form a fundamental set of solutions to the associated homogeneous equation

Then the general solution to the inhomogeneous equation [above] is given by

where C₁ and C₂ are arbitrary constants” (164).

We can rewrite this general solution as y = y_p + y_h, where y_h is the general solution to the associated homogeneous equation, i.e. C₁y₁ + C₂y₂.

The method we’ll be looking at this section is called “the method of undetermined coefficients.” We’ll be looking at the inhomogeneous equation y” + py’ + qy = f, where p/q are constants and f = f(t) is a function in terms of the independent variable (in this case, t). It’s important to note that this method works only if the coefficients are constants. The key idea is that “if the forcing term f has a form that is replicated under differentiation, then look for a solution with the same general form as the forcing term” (165). The easiest example is an exponential of the form f(t) = e^at. Then the first derivative f’(t) is also an exponential. There would also be trigonometric forcing terms as well. Here are some examples:

 

Click on it to make it larger. Hooray internet! :D
EDIT: JK, you can't make it larger. Sorry 'bout that.

Something interesting to note when the forcing term is trigonometric is that there is a complex method of solving for particular solutions. If you don’t understand the complex method, this probably won’t be the fastest way for you to solve for stuff. Here’s a website, nonetheless, and following this will be an example: http://tutorial.math.lamar.edu/Classes/DE/ComplexRoots.aspx

 

If the forcing term is a polynomial, we just use this method of undetermined coefficients and solve for our coefficients. Hooray!

Sometimes you’ll get some exceptional cases, like if you get 0 if you insert your solution of a particular form (say, an exponential solution) of the form ae^-t. If this happens, “we look for a solution of the form y(t) = ate^-t. We multiply the usual general form by t. This is the way to find a solution whenever the usual form does not work. If this method does not work, multiply by t once more and try again” (169).

A theorem for you:

“Suppose that y_f(t) is a solution to the linear equation

and y_g(t) is a solution to

Then y(t) = αy_f(t) + βy_g(t) is a solution to…” (170).

Finally, I have a picture summarizing some of the stuff we saw during this section. We can always have more complicated forcing terms, thus paving a narrower path into the realm of “all work, no sleep.” Ah, that blasted place. How I loathe that blasted place.

In any event, I’ll probably see you in 4.6. 

As seen from the 172nd page.