Differential Equations Workbook For Dummies®

7.4. Answers to Nonhomogeneous Linear Higher Order Differential Equation Problems

Following are the answers to the practice questions presented throughout this chapter. Each one is worked out step by step so that if you messed one up along the way, you can more easily see where you took a wrong turn.

1 What's the solution to this differential equation?

y″′ − 6y″ + 11y′ − 6y = 18e^4x

Solution: y = c₁e^x + c₂e^2x + c₃e^3x + 3e^4x

1. First off, get the homogeneous version of the equation:

   y″′ − 6y″ + 11y′ − 6y = 0

2. Looks like the homogeneous version has constant coefficients, so go ahead and assume a homogeneous solution of the form

   y = e^rx

3. Plug your attempted solution into the differential equation:

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

4. Cancel out e^rx:

   r³ − 6r² + 11r − 6 = 0

5. Now you have a cubic equation that can be factored into

   (r − 1)(r − 2)(r − 3) = 0

   Note: For some help factoring cubic (or higher!) equations, you can always turn to www.numberempire.com/equationsolver.php.

6. So the roots are

   r₁ = 1, r₂ = 2, and r₃ = 3

7. Hmmm. Those roots are real and distinct, which makes the solutions

   y₁ = e^x

   y₂ = e^2x

   y₃ = e^3x

8. Therefore, you can calculate that the homogeneous solution is

   y_h = c₁e^x + c₂e^2x + c₃e^3x

9. Well done. Now you need to assume a solution of the following form in order to find your particular solution:

   y_p = Ae^4x

10. Substitute y_p into the differential equation to get

    64Ae^4x − 96Ae^4x + 44Ae^4x − 6Ae^4x = 18e^4x

11. Canceling out e^4x gives you

    6A = 18

    which is actually

    A = 3

12. The particular solution is therefore

    y_p = 3e^4x

13. Add the ...