Differential Equations Workbook For Dummies®

6.4. Answers to Homogeneous Linear Higher Order Differential Equation Problems

Here are the answers to the practice questions I provide throughout this chapter. I walk you through each answer so you can see the problems worked out step by step. Enjoy!

1 Find the solution to the following differential equation:

y″′ + 7y″ + 14y′ + 8y = 0

with these initial conditions:

y(0) = 3

y′(0) = −7

y″(0) = 14

Solution:

1. Because this differential equation has constant coefficients, start by assuming a solution of the following form:

   y = e^rx

2. Plug your attempted solution into the differential equation:

   r³e^rx + 7r²e^rx + 14re^rx + 8e^rx = 0

3. Then cancel out e^rx:

   r³ + 7r² + 14r + 8 = 0

4. Now you have a cubic equation, which you can factor into the following either by hand or by using the equation-solving tool at www.quickmath.com (see the chapter intro for specifics on accessing this tool):

   (r + 1)(r + 2)(r + 4) = 0

5. So the roots are

   r₁ = −1, r₂ = −2, and r₃ = −4

6. These roots are real and distinct, so the solutions are

   y₁ = e^−x, y₂ = e^−2x, and y₃ = e^−4x

7. Thus, the general solution is

   y = c₁e^−x + c₂e^−2x + c₃e^−4x

8. Now you can apply the initial conditions. But in addition to the form for y, you also need y′:

   y′ = −c₁e^−x − 2c₂e^−2x − 4c₃e^−4x

   and y″:

   y″ = c₁e^−x + 4c₂e^−2x + 16c₃e^−4x

9. From the initial conditions, your three simultaneous equations in c₁, c₂, and c₃ that you must solve to find the coefficients are

   y(0) = c₁ + c₂ + c₃