Chapter 8: Problem 13

Identify each of the differential equations as type (for example, separable, linear first order, linear second order, etc.), and then solve it. $$y^{\prime \prime}+4 y^{\prime}+5 y=26 e^{3 x}$$

Short Answer

Expert verified

It's a linear non-homogeneous second-order differential equation. The solution is $ y = e^{-2x}(C_1 \cos x + C_2 \sin x) + e^{3x} $.

Step by step solution

- Identify the Type of Differential Equation

The given differential equation is $ y^{\text{''}} + 4y' + 5y = 26e^{3x} $. It is a linear non-homogeneous second-order differential equation.

- Find the Complementary Solution

First, solve the corresponding homogeneous equation: $ y'' + 4y' + 5y = 0 $. The characteristic equation is $ r^2 + 4r + 5 = 0 $. Solving this quadratic equation using the quadratic formula $ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $, we get: $ r = \frac{-4 \pm \sqrt{16 - 20}}{2} = \frac{-4 \pm \sqrt{-4}}{2} = -2 \pm i $. Therefore, the complementary solution is $ y_c = e^{-2x}(C_1 \cos x + C_2 \sin x) $.

- Find the Particular Solution

To find a particular solution, $ y_p $, assume a solution of the form $ y_p = Ae^{3x} $. Substitute $ y_p $, its first and second derivatives into the non-homogeneous equation: $ y_p' = 3Ae^{3x} $ and $ y_p'' = 9Ae^{3x} $. Plugging these into the equation gives: $ 9Ae^{3x} + 4(3Ae^{3x}) + 5(Ae^{3x}) = 26e^{3x} $. Simplifying, we obtain: $ 9A + 12A + 5A = 26 $ which simplifies to $ 26A = 26 $, hence $ A = 1 $. Thus, the particular solution is $ y_p = e^{3x} $.

- Combine Solutions

The general solution to the differential equation is the sum of the complementary solution and the particular solution: $ y = y_c + y_p = e^{-2x}(C_1 \cos x + C_2 \sin x) + e^{3x} $.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

complementary solution

The complementary solution is formed from the solutions of the homogeneous equation and captures the behavior of the system.

particular solution

Note that different forms of non-homogeneous terms require different trial solutions (e.g., polynomials, exponentials, trigonometric functions). The key is trying out an appropriate form that matches the non-homogeneous term.

characteristic equation

The roots (real or complex) determine the form of the complementary solution. In our example, since the roots are complex, the complementary solution involves both exponential and trigonometric functions in:
\[ y_c = e^{-2x}(C_1 \cos x + C_2 \sin x) \].

homogeneous differential equation

The complementary solution derived from the homogeneous differential equation captures the free response of the system without external forcing. For our example, it is:
\[ y_c = e^{-2x}(C_1 \cos x + C_2 \sin x) \]
When combined with the particular solution, we achieve the full solution to the non-homogeneous differential equation.

The total solution is the sum of the complementary and particular solutions:
\[ y = e^{-2x}(C_1 \cos x + C_2 \sin x) + e^{3x} \].