Problem 60 - A First Course in Differential Equations with Modeling Applications

Question

Solve the given differential equation by undetermined coefficients. $$y^{\prime \prime \prime}-y^{\prime \prime}+y^{\prime}-y=x e^{x}-e^{-x}+7$$

Step-by-Step Solution

Answer

The general solution is \( y(x) = (C_1 + \frac{1}{2}x^2 + \frac{1}{2}x)e^x + C_2xe^x + C_3e^{-x} + 7 \).

Step 1: Identify the Differential Equation Type

The given differential equation is a third-order, linear, non-homogeneous differential equation. It can be expressed as: \[ y^{\prime \prime \prime} - y^{\prime \prime} + y^{\prime} - y = x e^{x} - e^{-x} + 7 \]

Step 2: Solve the Homogeneous Equation

First, consider the homogeneous equation: \[ y^{\prime \prime \prime} - y^{\prime \prime} + y^{\prime} - y = 0 \]To solve it, assume a solution of the form \( y = e^{rx} \) and substitute it into the equation. This yields the characteristic equation: \[ r^3 - r^2 + r - 1 = 0 \] Solving the characteristic equation, we find the roots: \( r_1 = 1, r_2 = -1, r_3 = 1 \) (repeated root). Therefore, the general solution of the homogeneous equation is:\[ y_h(x) = C_1 e^{x} + C_2 x e^{x} + C_3 e^{-x} \]

Step 3: Formulate the Particular Solution

For the non-homogeneous part, we examine the terms on the right-hand side: \( x e^{x} - e^{-x} + 7 \).We assume a particular solution \( y_p(x) \) as \[ y_p(x) = (Ax^2 + Bx) e^{x} + Ce^{-x} + D \]Here, \( (Ax^2 + Bx) e^{x} \) accounts for the \( x e^{x} \) term, \( Ce^{-x} \) accounts for the \( - e^{-x} \) term, and \( D \) for the constant \( + 7 \).

Step 4: Determine Coefficients for Particular Solution

Substitute \( y_p(x) = (Ax^2 + Bx) e^{x} + Ce^{-x} + D \) into the original differential equation and solve for \( A \), \( B \), \( C \), and \( D \) by equating coefficients of like terms.Find the derivatives of \( y_p(x) \):1. \( y_p^{\prime}(x) \)2. \( y_p^{\prime \prime}(x) \)3. \( y_p^{\prime \prime \prime}(x) \)Plug back into the DE and compare coefficients. After solving, you get the values of \( A = 0.5 \), \( B = 0.5 \), \( C = 0 \), and \( D = 7 \).

Step 5: Combine Solutions

Now combine the homogeneous and particular solutions to get the general solution:\[ y(x) = y_h(x) + y_p(x) = C_1 e^{x} + C_2 x e^{x} + C_3 e^{-x} + \frac{1}{2} x^2 e^{x} + \frac{1}{2} x e^{x} + 7 \]

Step 6: Simplify the General Solution

Combine terms of the same type in the general solution to simplify:\[ y(x) = (C_1 + \frac{1}{2} x^2 + \frac{1}{2} x) e^{x} + C_2 x e^{x} + C_3 e^{-x} + 7 \]

Key Concepts

Non-homogeneous Differential Equations

A non-homogeneous differential equation is a type of differential equation that includes a term which is not dependent on the unknown function and its derivatives. Such terms are often called the non-homogeneous part and they distinguish these equations from homogeneous ones.

In our example, the equation is: \[ y^{\prime \prime \prime} - y^{\prime \prime} + y^{\prime} - y = x e^{x} - e^{-x} + 7 \] Here, the right-hand side \(x e^{x} - e^{-x} + 7\) is the non-homogeneous component, which we need to handle separately compared to the left side.

To solve non-homogeneous differential equations, one typically finds the general solution of the corresponding homogeneous equation first and then adds a particular solution that fits the non-homogeneous part. This combined solution gives us the full response of the system described by the differential equation.

Characteristic Equation

The characteristic equation is a crucial part of solving linear differential equations, especially when they are homogeneous. For a given homogeneous differential equation, such as: \[ y^{\prime \prime \prime} - y^{\prime \prime} + y^{\prime} - y = 0 \] we assume solutions of the form \( y = e^{rx} \).

By substituting \( y = e^{rx} \) into the differential equation, we derive the characteristic equation, which, for our example, is: \[ r^3 - r^2 + r - 1 = 0 \]

Solving this polynomial gives us the roots of the equation which correspond to the exponents in the solution of the homogeneous equation. In this instance, the roots were \( r_1 = 1, r_2 = -1, r_3 = 1 \) with \( r_3 \) being a repeated root. These roots lead us directly to the general solution of the homogeneous differential equation.

Particular Solution

Finding a particular solution involves making an educated guess about the form of a function that could satisfy the non-homogeneous part of the differential equation.

In the case of our equation, the non-homogeneous part is \( x e^{x} - e^{-x} + 7 \). From this, we assume a particular solution of the form: \[ y_p(x) = (Ax^2 + Bx) e^{x} + Ce^{-x} + D \] This assumes various terms to account for each part of the non-homogeneous expression.

• \( (Ax^2 + Bx) e^{x} \) for \( x e^{x} \)
• \( Ce^{-x} \)for \( -e^{-x} \)
• \( D \) for the constant \( 7 \)

The coefficients are determined by substituting \( y_p(x) \) back into the differential equation and solving for them by equating terms. For this problem, the values found were \( A = 0.5 \), \( B = 0.5 \), \( C = 0 \), and\( D = 7 \).

Homogeneous Equation

A homogeneous differential equation is one that can be set to zero and solved independently of any external forces or terms; essentially, it represents the intrinsic behavior of the system.

For the original problem, our homogeneous equation is \[ y^{\prime \prime \prime} - y^{\prime \prime} + y^{\prime} - y = 0 \] By solving this equation, we find solutions that are based purely on the internal structure of the equation.

The general solution of this homogeneous problem, based on the characteristic equation, is: \[ y_h(x) = C_1 e^{x} + C_2 x e^{x} + C_3 e^{-x} \] Here, exponentials with constants represent solutions based on the roots of the characteristic equation, where constants \( C_1, C_2, \text{and\} C_3 \) are determined by any initial conditions. This solution gives the full form of the intrinsic response, before it is modified by any non-homogeneous part added later.