Differential Equations Workbook For Dummies®

6.1. Distinctly Different: Working with Real and Distinct Roots

In this section, you practice the case where the characteristic equation has real and distinct roots first — that is, the roots aren't imaginary, and they're not the same.

Here's a linear second order differential equation that's homogeneous and has constant coefficients:

y″ + 3y′ + 2 = 0

Given this equation's form, you can safely bet that the solutions are something like

y = e^rx

Plugging that solution into the differential equation gives you

r²e^rx +3re^rx + 2e^rx = 0

and dividing by e^rx gives you

r² +3r + 2 = 0

Surprise! There's your characteristic equation, which you can solve with the quadratic equation to get

(r + 1)(r + 2) = 0

So the characteristic equation's roots are −1 and −2, giving you these two solutions:

y = e^−x and y = e^−2x

The process is similar for higher order differential equations, but the algebra is a little tougher because the characteristic equation is of a higher order. See what I mean in the following example problem and then try to solve a few of the practice problems on your own.

NOTE

EXAMPLE

Q. Find the solution to this differential equation:

y″′ − 6y″ + 11y′ − 6y = 0

with these initial conditions:

y(0) = 9

y′(0) = 20

y″(0) = 50

A. y = 2e^x + 3e^2x + 4e^3x

1. This differential equation has constant coefficients, so you can start by assuming a solution of the form

   y = e^rx

2. Plugging your attempted solution into the differential equation gives you

   r³e^rx − 6r²e^rx + 11re^rx − 6e^rx = 0

3. Canceling out e^rx leaves ...