Problem 17 - Differential Equations and Linear Algebra

Question

Use the variation-of-parameters method to find the general solution to the given differential equation. $$y^{\prime \prime}+4 y^{\prime}+4 y=\frac{4 e^{-2 x}}{1+x^{2}}+2 x^{2}-1$$

Step-by-Step Solution

Answer

The general solution to the given differential equation using the variation-of-parameters method is: \[y(x) = \left(C_1 + C\right) e^{-2x} + \left(C_2 + D\right) xe^{-2x} -\frac{1}{2} e^{-2x} \left[e^{4x}(1+x)-(1+2x^2)\right].\]

Step 1: Find the Complementary Function

To find the complementary function, solve the homogeneous version of the given differential equation: \[ y^{\prime\prime} + 4y^{\prime} + 4y = 0. \] To solve this, first find the characteristic equation: \[ r^2 + 4r + 4 = 0. \] This is a quadratic equation. Solve for r: \[ r = \frac{-4 \pm \sqrt{(-4)^2 - 4 \cdot 4}}{2} = -2. \] So we have a repeated root r = -2. Therefore, the complementary function is: \[ y_c(x) = C_1 e^{-2x} + C_2 xe^{-2x}. \]

Step 2: Find the Particular Solution

To find the particular solution using the variation-of-parameters method, write the given differential equation as: \[ y^{\prime\prime} + 4y^{\prime} + 4y = f(x), \] where \[ f(x) =\frac{4 e^{-2 x}}{1+x^{2}}+2 x^{2}-1. \] Choose new functions u_1(x) and u_2(x) such that: \[ y_p(x) = u_1(x) e^{-2x} + u_2(x) xe^{-2x}. \] To find u_1(x) and u_2(x), we need to find the Wronskian of the complementary functions and solve the following two equations: \[ \begin{cases} u'_1(x)e^{-2x}+u'_2(x)xe^{-2x} = 0 \\ -u_1(x)(2e^{-2x})-u_2(x)(-2xe^{-2x}+e^{-2x}) = f(x) \end{cases} \] First, find the Wronskian: \[ W = \begin{vmatrix} e^{-2x} & xe^{-2x} \\ -2e^{-2x} & -2xe^{-2x}+e^{-2x} \end{vmatrix} = -2x(e^{-4x}), \] Now, by solving the two equations: \[ \begin{cases} u'_1(x) = 0 \\ u'_2(x) = \frac{f(x)}{-2x(e^{-4x})} \end{cases}. \] Integrate u'_1(x) and u'_2(x) to find u_1(x) and u_2(x): \[ \begin{cases} u_1(x) = C \\ u_2(x) = -\frac{1}{2} \int \frac{f(x)}{x} e^{4x} dx \end{cases}. \] Integrating the second equation yields: \[ u_2(x) = -\frac{1}{2} \left[e^{4x}(1+x)-(1+2x^2)\right] + D, \] where C and D are constants. Now, substituting u_1(x) and u_2(x) in the equation for y_p(x), we get: \[ y_p(x) = Ce^{-2x} -\frac{1}{2} e^{-2x} \left[e^{4x}(1+x)-(1+2x^2)\right] + Dxe^{-2x}. \]

Step 3: Combine the Complementary Function and the Particular Solution

Combine the complementary function y_c(x) and the particular solution y_p(x) to find the general solution: \[ y(x) = y_c(x) + y_p(x) = \left(C_1 + C\right) e^{-2x} + \left(C_2 + D\right) xe^{-2x} -\frac{1}{2} e^{-2x} \left[e^{4x}(1+x)-(1+2x^2)\right]. \] The general solution to the given differential equation is: \[ y(x) = \left(C_1 + C\right) e^{-2x} + \left(C_2 + D\right) xe^{-2x} -\frac{1}{2} e^{-2x} \left[e^{4x}(1+x)-(1+2x^2)\right]. \]

Key Concepts

Complementary Function

In solving non-homogeneous linear differential equations, the concept of the complementary function is crucial. The complementary function is the solution to the associated homogeneous differential equation. This is done by setting the right-hand side to zero and solving for the characteristic equation. For example, with the equation: \[ y'' + 4y' + 4y = 0, \] the characteristic equation becomes \( r^2 + 4r + 4 = 0 \). Solving gives a repeated root \( r = -2 \), indicating that the complementary function involves both \( e^{-2x} \) and \( xe^{-2x} \). Hence, the complementary function is: \[ y_c(x) = C_1 e^{-2x} + C_2 xe^{-2x}, \] where \( C_1 \) and \( C_2 \) are constants determined by initial conditions.

Particular Solution

The particular solution addresses the non-homogeneous part of the differential equation. In this context, we use the method of variation of parameters, a technique allowing us to find a specific solution that fits the non-homogeneous equation. Given the form of our solution: \( y_p(x) = u_1(x) e^{-2x} + u_2(x) xe^{-2x} \), functions \( u_1(x) \) and \( u_2(x) \) need to be determined. By solving the equations

• \( u'_1(x) e^{-2x} + u'_2(x) xe^{-2x} = 0 \)
• \(-u_1(x)(2e^{-2x}) - u_2(x)(-2xe^{-2x}+e^{-2x}) = f(x) \)

And integrating, we calculate the particular solution. It's a tailored solution adapting with the specific non-homogeneous part \( f(x) = \frac{4 e^{-2x}}{1+x^2} + 2x^2 - 1 \). This provides the piece of the puzzle we need to construct the general solution.

Wronskian

The Wronskian is a determinant used to check the linear independence of solutions to differential equations. For our case, where complementary solutions are \( e^{-2x} \) and \( xe^{-2x} \), the Wronskian is calculated as follows: \[ W = \begin{vmatrix} e^{-2x} & xe^{-2x} \ -2e^{-2x} & -2xe^{-2x}+e^{-2x}\end{vmatrix}, \] resulting in \(-2xe^{-4x}\). This non-zero determinant confirms that our solutions are indeed linearly independent. A linearly independent set is necessary to solve for the functions \( u_1(x) \) and \( u_2(x) \) in the variation of parameters, ensuring we've captured all behavior of the solution basis.

General Solution

The general solution of a differential equation is the combination of the complementary function and the particular solution. It encapsulates all possible solutions of the equation. After determining \( y_c(x) \) and \( y_p(x) \), the general solution becomes: \[ y(x) = (C_1 + C)e^{-2x} + (C_2 + D)xe^{-2x} - \frac{1}{2}e^{-2x}[e^{4x}(1+x)-(1+2x^2)]. \]Here, the terms \( C_1 + C \) and \( C_2 + D \) are constants reflecting initial conditions or boundary values that further specify the solution. The general solution expresses every possible instantiation the differential equation can allow, given different constant values. This completeness is what makes understanding and deriving the general solution so fundamental in differential equation problems.