Differential Equations Workbook For Dummies®

5.4. Answers to Nonhomogeneous Linear Second 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 general solution to this nonhomogeneous second order differential equation?

y″ + 3y′ + 2y = 6e^x

Solution: y = c₁e^−x + c₂e^−2x + e^x

1. First, find the homogeneous version of the original equation:

   y″ + 3y′ + 2y = 0

2. Assume that the solution to the homogeneous differential equation is of the form y = e^rx. When you substitute that solution into the equation, you get the characteristic equation

   r² + 3r + 2 = 0

3. Go ahead and factor that out as follows:

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

4. If you determine that the roots, r₁ and r₂, of the characteristic equation are −1 and −2, you know that

   y₁ = e^−x and y₂ = e^−2x

5. Thus, the solution to the homogeneous differential equation is given by

   y = c₁e^−x + c₂e^−2x

6. Now you need a particular solution to the differential equation:

   y″ + 3y′ + 2y = 6e^x

   Note that g(x) has the form e^x here, so assume that the particular solution has the form

   y_p(x) = Ae^x

7. Substitute y_p(x) into the equation:

   Ae^x + 3Ae^x + 2Ae^x = 6e^x

8. Cancel out the e^x term:

   A + 3A + 2A = 6

   or 6A = 6, so A = 1.

9. Your particular solution is

   y_p(x) = e^x

10. Because the general solution of the nonhomogeneous equation that you started with is the sum of the corresponding homogeneous equation's general solution and a particular solution ...