Solving Second Order Homogeneous Differential Equations (Three case studies)


Consider a second order differential equation of the form:

$$ay''+by'+cy=0 $$

...we want to find two functions (y1,y2) such that {y1,y2} is a fundamental set.

We will use a substitution that involves:

$$y=e^{rx} $$

...which implies that:

$$y'=re^{rx} $$

$$y''=r^2e^{rx} $$

Substituting above expressions into the original form of our equation gives us:

$$ar^2e^{rx} +bre^{rx} + ce^{rx} = 0$$

...from which we can factor our the e^rx:

$$(ar^2 + br + c)e^{rx} = 0 \;,\;where \; e^{rx}\ne0 $$

...and divide

$$(ar^2 + br + c) = 0 $$

$$(ar^2 + br + c) \; is \; our \; "characteristic \; equation" $$

From here we want to find values of "r". We use the quadratic formula to do so:

$$r = \frac{-b \pm \sqrt{b^2=4ac}}{2a} $$

$$r_1 = \frac{-b - \sqrt{b^2=4ac}}{2a} $$

$$r_2 = \frac{-b + \sqrt{b^2=4ac}}{2a} $$

Using these values of "r" in our original substitution, we obtain a fundamental set where:

$$y_1 = e^{r_1x}$$

$$y_2 = e^{r_2x} $$

Also, note that if y1 and y2 are solutions to our differential equation, then y1 + y2 is also a solution. Furthermore, if a given function is a solution to a given differential equation, then any constant multiplied by that function is also a solution. Therefore, the general solution to our second order differential equation is given by:

$$ y(x) = C_1e^{r_1x} + C_2e^{r_2x} $$

From the above explanation we can have three possible outcomes:

  1. $\qquad r_1 \ne r_2 $
  2. $\qquad r_1 = r_2 \; (1 \; real \; repeated \; root)$
  3. $\qquad Complex \; root \; (b^2-4ac \lt 0)$

The following three pages of this section will investigate examples of the above three cases.

Continue on to case #1 (r1 != r2)