Chapter 2: Problem 6
Solve the given nonhomogeneous ODE by variation of parameters or undetermined coefficients. Give a general solution. (Show the details of your work.) $$x^{2} y^{\prime \prime}-x y^{\prime}+y=x \ln |x|$$
Short Answer
Expert verified
The general solution is \(y = c_1 x + c_2 x^{-1} + \frac{x^2 \ln|x|}{3} - \frac{x^2}{3}.\)
Step by step solution
01
Identify the Type of Differential Equation
The given equation \(x^2 y'' - x y' + y = x \ln |x|\) is a second-order linear nonhomogeneous differential equation. It is nonhomogeneous because the right-hand side is not zero.
02
Find the Homogeneous Solution
First, we solve the corresponding homogeneous equation: \(x^2 y'' - x y' + y = 0\). This is a Cauchy-Euler equation. To solve it, we use the trial solution \(y_h = x^m\), yielding the characteristic equation \(m(m-1) + m - 1 = 0\), simplifying to \(m^2 - 1 = 0\). The solutions are \(m = 1\) and \(m = -1\). Therefore, the homogeneous solution is \(y_h = c_1 x + c_2 x^{-1}\).
03
Select Method for Nonhomogeneous Solution
To find the particular solution, choose between the method of variation of parameters or undetermined coefficients. Since the right-hand side is \(x \ln |x|\), which is not a simple polynomial, trigonometric, or exponential function, we'll use variation of parameters.
04
Compute Wronskian
To use variation of parameters, compute the Wronskian of the solutions \(x\) and \(x^{-1}\). The Wronskian \(W(x)\) of two functions \(u_1 = x\) and \(u_2 = x^{-1}\) is given by:\[W(x) = \begin{vmatrix} u_1 & u_2 \ u_1' & u_2' \end{vmatrix} = \begin{vmatrix} x & x^{-1} \ 1 & -x^{-2} \end{vmatrix} = -1 - (-1) = -2.\]
05
Determine Particular Solution using Variation of Parameters
The particular solution \(y_p\) is given by:\[y_p = -y_1 \int \frac{y_2(x)g(x)}{W(x)} \, dx + y_2 \int \frac{y_1(x)g(x)}{W(x)} \, dx.\]Here, \(g(x) = x \ln |x|\), \(y_1 = x\), \(y_2 = x^{-1}\), and \(W(x) = -2\):\[y_p = -x \int \frac{x^{-1} \cdot x \ln |x|}{-2} \, dx + x^{-1} \int \frac{x \cdot x \ln |x|}{-2} \, dx.\]Simplify and solve the integrals:\[y_p = \frac{x}{2} \int \ln |x| \, dx - \frac{x^{-1}}{2} \int x^2 \ln |x| \, dx.\]Using integration by parts on both integrals, we find:\[ \int \ln |x| \, dx = x \ln|x| - x \quad\text{and}\quad \int x^2 \ln|x| \, dx = \frac{x^3}{3} \ln|x| - \frac{x^3}{9}. \]Substitute back into \(y_p\) and simplify.
06
Simplify and Combine Solutions
Substitute the results from Step 5 into \(y_p\):\[y_p = \frac{x}{2}(x \ln|x| - x) - \frac{x^{-1}}{2} \left( \frac{x^3}{3} \ln|x| - \frac{x^3}{9} \right).\]Simplify the expression carefully:\[y_p = \frac{x^2 \ln|x|}{2} - \frac{x^2}{2} - \frac{x^2 \ln|x|}{6} + \frac{x^2}{18}.\]Combine like terms:\[y_p = \frac{x^2 \ln|x|}{3} - \frac{x^2}{3}.\]
07
State the General Solution
The general solution to the differential equation is the sum of the homogeneous solution and the particular solution:\[y = c_1 x + c_2 x^{-1} + \frac{x^2 \ln|x|}{3} - \frac{x^2}{3}.\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Cauchy-Euler Equation
The Cauchy-Euler equation is a specific type of differential equation that has a particular structure, allowing it to be transformed into a simpler form for solving. It often appears in the form \(x^n y^{(n)} + a_{n-1}x^{n-1} y^{(n-1)} + \ldots + a_0 y = f(x)\), where \(f(x)\) can be zero, making it homogeneous, or non-zero, making it nonhomogeneous.
One key characteristic of Cauchy-Euler equations is that their coefficients are powers of the independent variable, \(x\). This special form allows us to use a substitution approach, where we assume a solution of the form \(y = x^m\).
Once substituted, the differential equation simplifies down to a characteristic equation in the variable \(m\). Solving it gives us essential values for forming the homogeneous solution. For the exercise above, solving the Cauchy-Euler equation yielded a characteristic equation \(m^2 - 1 = 0\). It resulted in roots \(m = 1\) and \(m = -1\).
The general solution is then expressed in terms of these roots, resulting in the homogeneous solution \(y_h = c_1 x + c_2 x^{-1}\), where \(c_1\) and \(c_2\) are arbitrary constants.
One key characteristic of Cauchy-Euler equations is that their coefficients are powers of the independent variable, \(x\). This special form allows us to use a substitution approach, where we assume a solution of the form \(y = x^m\).
Once substituted, the differential equation simplifies down to a characteristic equation in the variable \(m\). Solving it gives us essential values for forming the homogeneous solution. For the exercise above, solving the Cauchy-Euler equation yielded a characteristic equation \(m^2 - 1 = 0\). It resulted in roots \(m = 1\) and \(m = -1\).
The general solution is then expressed in terms of these roots, resulting in the homogeneous solution \(y_h = c_1 x + c_2 x^{-1}\), where \(c_1\) and \(c_2\) are arbitrary constants.
Variation of Parameters
Variation of Parameters is a powerful method for finding particular solutions to nonhomogeneous differential equations. Unlike the method of undetermined coefficients, variation of parameters is versatile and can be utilized for a wider class of functions.
This method assumes that a solution for the homogeneous equation is already known. With a nonhomogeneous equation, such as \(x^2 y'' - x y' + y = x \ln |x|\), a particular solution \(y_p\) is added to this homogeneous solution.
The process involves computing solutions \(y_1\) and \(y_2\) of the homogeneous equation, forming a particular solution by setting \(y_p = -y_1 \int \frac{y_2(x)g(x)}{W(x)} \, dx + y_2 \int \frac{y_1(x)g(x)}{W(x)} \, dx\).
Here, \(g(x)\) is the nonhomogeneous part of the differential equation, and \(W(x)\) is the Wronskian. In our example, variation of parameters helped us find a solution when \(g(x) = x \ln |x|\) by strategically using integrals.
This method assumes that a solution for the homogeneous equation is already known. With a nonhomogeneous equation, such as \(x^2 y'' - x y' + y = x \ln |x|\), a particular solution \(y_p\) is added to this homogeneous solution.
The process involves computing solutions \(y_1\) and \(y_2\) of the homogeneous equation, forming a particular solution by setting \(y_p = -y_1 \int \frac{y_2(x)g(x)}{W(x)} \, dx + y_2 \int \frac{y_1(x)g(x)}{W(x)} \, dx\).
Here, \(g(x)\) is the nonhomogeneous part of the differential equation, and \(W(x)\) is the Wronskian. In our example, variation of parameters helped us find a solution when \(g(x) = x \ln |x|\) by strategically using integrals.
Wronskian
The Wronskian is a determinant used in the context of differential equations to determine the linear independence of a set of functions. For two functions \(u_1\) and \(u_2\), the Wronskian \(W(x)\) is calculated as:
For example, in the step-by-step solution provided, the Wronskian was computed for \(u_1 = x\) and \(u_2 = x^{-1}\) to help apply the variation of parameters method:
A non-zero Wronskian implies linear independence, which is crucial for determining the form of the particular solution in variation of parameters. In simple terms, the Wronskian being non-zero assured us that the functions \(x\) and \(x^{-1}\) were linearly independent and thus valid for constructing the solution.
- \[W(x) = \begin{vmatrix} u_1 & u_2 \ u_1' & u_2' \end{vmatrix} = u_1 u_2' - u_2 u_1'\]
For example, in the step-by-step solution provided, the Wronskian was computed for \(u_1 = x\) and \(u_2 = x^{-1}\) to help apply the variation of parameters method:
- \[W(x) = \begin{vmatrix} x & x^{-1} \ 1 & -x^{-2} \end{vmatrix} = -2\]
A non-zero Wronskian implies linear independence, which is crucial for determining the form of the particular solution in variation of parameters. In simple terms, the Wronskian being non-zero assured us that the functions \(x\) and \(x^{-1}\) were linearly independent and thus valid for constructing the solution.