Chapter 8: Problem 18
Obtain the general solution. $$ \left(D^{2}-1\right) y=8 x e^{x} $$
Short Answer
Expert verified
The general solution is \( y = c_1 e^x + c_2 e^{-x} + (4x - 4)xe^x \).
Step by step solution
01
Identify the Differential Equation Type
The given differential equation is \((D^2 - 1)y = 8xe^x\). This is a non-homogeneous linear differential equation with constant coefficients.
02
Solve the Homogeneous Equation
Consider the homogeneous part \( (D^2 - 1)y = 0 \). The auxiliary equation is \(m^2 - 1 = 0\). Solving for \(m\), we have \(m = \pm 1\). The general solution of the homogeneous equation is \( y_h = c_1 e^x + c_2 e^{-x} \), where \(c_1\) and \(c_2\) are arbitrary constants.
03
Find a Particular Solution
Next, find a particular solution \(y_p\) for the non-homogeneous equation. The right-hand side \(8xe^x\) suggests using the method of undetermined coefficients. Assume \(y_p = (Ax + B)x e^x\). Differentiate \(y_p\) twice and substitute in the original differential equation to find \(A\) and \(B\).
04
Differentiate the Particular Solution Assumption
Assume \(y_p = (Ax + B)x e^x \). Find the first and second derivatives.1. First derivative: \(y'_p = (A x + B)' x e^x + (A x + B)(x e^x)' = (A)(x e^x) + (Ax + B)(e^x + xe^x)\)2. Second derivative: Simplify and differentiate \( y'_p \) to get \( y''_p \).
05
Substitute and Solve for Coefficients
Substitute \(y_p\), \(y'_p\), and \(y''_p\) obtained from Step 4 into the differential equation \((D^2 - 1) y_p = 8x e^x\) and collect terms. Solve for \(A\) and \(B\) by equating coefficients for \(xe^x\) and \(e^x\). After solving, \(A = 4\) and \(B = -4\). Then, \(y_p = (4x - 4)xe^x\).
06
Write the General Solution
Combine the solutions from Step 2 and Step 5. The general solution is:\[ y = y_h + y_p = c_1 e^x + c_2 e^{-x} + (4x - 4)xe^x \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Undetermined coefficients method
The undetermined coefficients method is a straightforward approach to tackling non-homogeneous linear differential equations with constant coefficients. When faced with such an equation, one first identifies the form of the non-homogeneous part, also called the forcing function. In our example, the function is
- Type: Polynomial times an exponential, i.e., \(8x e^x\).
- \(8xe^x\), assume a form like \(y_p = (Ax + B)xe^x\),
General solution
The general solution of a differential equation embodies the complete set of solutions that satisfy the equation. It is typically made up of two components: a homogeneous solution and a particular solution. This is because any linear combination of these components will solve the original equation. In our case, the general solution
- combines \(y_h\) and \(y_p\),
- \(y_h\) is the solution to the associated homogeneous equation
- \(y_p\) is a specific solution to the non-homogeneous equation.
- \(y = c_1 e^x + c_2 e^{-x} + (4x - 4)xe^x\),
Homogeneous equation
A homogeneous equation is key to understanding the behavior of differential equations, representing the portion without the external function or forcing term. In our case, it is
- \((D^2 - 1)y = 0\),
- \(m^2 - 1 = 0\).
- \(m = 1\) and \(m = -1\),
- \(y_h = c_1 e^x + c_2 e^{-x}\).
Particular solution
The particular solution tackles the part of the differential equation that accounts for external influences or inhomogeneities. Unlike the homogeneous equation, the presence of a non-zero forcing term requires special attention. Here,
- \((D^2 - 1)y = 8xe^x\),
- \(y_p = (Ax + B)xe^x\).
- \(A = 4\)
- and \(B = -4\),
- \(y_p = (4x - 4)xe^x\).