Revision as of 09:09, 5 October 2009 by Dknott01 (Talk | contribs)

Back to the MA366 course wiki

Separable Equations

If the differential equation in question can be written as

$ M(t)dt=N(y)dy $

then the equation is called separable. The name refers to your ability to separate the function y and the dependent variable t, with all of the t's and dt's on one side and all of the y's and dy's on the other. If, by using a little bit of algebra, you can get your equation into this form, then you can simply integrate both sides to find the solution. Integrate the left side with respect to t, and the right side with respect to y, and your differential equation will turn into something more familiar and useful. Unfortunately, very few equations are separable, and you'll soon find yourself pining for them as you're forced to use much more involved methods of solution.

In Section 1, we used this fact in solving the initial value problems. In the differential equations I used, it was intuitively obvious that "integrating both sides" was allowed, but this won't always be the case.

Suppose we are given the differential equation:

$ yy'+\frac{1}{t^3}=\frac{5}{t} $

This is one of those times where you should use the notation dy/dt instead of just y'. So let's use that notation.

$ y\frac{dy}{dt}+\frac{1}{t^3}=\frac{5}{t} $

The objective here is to get all of the y's and dy's on one side of the equation, and all t's and dt's on the other. As you know, dy/dt isn't actually a fraction, but you can sort of pretend it is and treat dy and dt as separate entities, and move them around with algebra. Let's do some algebra:

$ \begin{align} y\frac{dy}{dt}+\frac{1}{t^3}&=\frac{5}{t}\\ y\frac{dy}{dt}&=\frac{5}{t}-\frac{1}{t^3} \end{align} $

Alumni Liaison

Basic linear algebra uncovers and clarifies very important geometry and algebra.

Dr. Paul Garrett