Chapter 8: Problem 12
Use the variation-of-parameters method to find the general solution to the given differential equation. $$y^{\prime \prime}-y=2 \tanh x$$
Short Answer
Expert verified
The general solution to the given differential equation using the variation of parameters method is:
\[y(x) = c_1e^x + c_2e^{-x} + \frac{-k}{2e^{x}} + \frac{l}{2e^{-x}}\]
Step by step solution
01
Determine the complementary function
First, we need to solve the homogeneous equation associated with the given differential equation:
\[y'' - y = 0\]
The characteristic equation associated with this equation is:
\[r^2 - 1 = 0\]
Factoring, we get:
\[(r - 1)(r + 1) = 0\]
This yields the roots r = 1 and r = -1. Thus, the complementary function is given by:
\[y_c(x) = c_1e^x + c_2e^{-x}\]
02
Apply the variation of parameters method
Now, let's apply the variation of parameters method to find the particular solution. We will assume that the particular solution is of the form:
\[y_p(x) = u_1(x)e^x + u_2(x)e^{-x}\]
Taking the first derivative with respect to x, we get:
\[y_p'(x) = u_1'(x)e^x + u_2'(x)e^{-x} + u_1(x)e^x - u_2(x)e^{-x}\]
Taking the second derivative with respect to x, we get:
\[y_p''(x) = u_1''(x)e^x + u_2''(x)e^{-x} + 2u_1'(x)e^x - 2u_2'(x)e^{-x} + u_1(x)e^x + u_2(x)e^{-x}\]
Now, we will substitute y_p(x) and its derivatives into the given differential equation:
\[u_1''(x)e^x + u_2''(x)e^{-x} + 2u_1'(x)e^x - 2u_2'(x)e^{-x} + u_1(x)e^x + u_2(x)e^{-x} - u_1(x)e^x - u_2(x)e^{-x} = 2\tanh x\]
Simplifying the equation, we get:
\[u_1''(x)e^x + u_2''(x)e^{-x} + 2u_1'(x)e^x - 2u_2'(x)e^{-x} = 2\tanh x\]
Now we need to determine u_1'(x) and u_2'(x). To do this, we will compare the coefficients of \(e^x\) and \(e^{-x}\) on both sides of the equation. We obtain the two equations:
\[u_1''(x) + 2u_1'(x) = 0\] and \[u_2''(x) - 2u_2'(x) = 0\]
03
Solve for u_1'(x) and u_2'(x)
To solve for u_1'(x), we will make the following substitution: `v(x) = u_1'(x)`. Then, we have:
\[v'(x) + 2v(x) = 0\]
This is a linear homogeneous differential equation. We will use an integrating factor `e^(2x)` to solve it:
\[e^{2x}v'(x) + 2e^{2x}v(x) = 0\]
The derivative of `e^(2x)v(x)` is equal to the left-hand side:
\[\frac{d}{dx}(e^{2x}v(x)) = 0\]
Now, we integrate both sides concerning x:
\[e^{2x}v(x) = k\]
Now, solve for `v(x)`:
\[v(x) = u_1'(x) = \frac{k}{e^{2x}}\]
Similarly, for `u_2'(x)`, make the substitution `w(x) = u_2'(x)`:
\[w'(x) - 2w(x) = 0\]
The integrating factor is `e^(-2x)`:
\[e^{-2x}w'(x) - 2e^{-2x}w(x) = 0\]
Now we have:
\[\frac{d}{dx}(e^{-2x}w(x)) = 0\]
Integrating both sides concerning x:
\[e^{-2x}w(x) = l\]
Now, solve for `w(x)`:
\[w(x) = u_2'(x) = \frac{l}{e^{-2x}}\]
04
Find the particular solution
To find u_1(x) and u_2(x), we need to integrate u_1'(x) and u_2'(x) concerning x:
\[u_1(x) = \int u_1'(x) dx = \int \frac{k}{e^{2x}}dx = \frac{-k}{2e^{2x}} + C_1\]
\[u_2(x) = \int u_2'(x) dx = \int \frac{l}{e^{-2x}}dx = \frac{l}{2e^{-2x}} + C_2\]
Now, substitute `u_1(x)` and `u_2(x)` back into the expression for `y_p(x)`:
\[y_p(x) = \frac{-k}{2e^{x}} + C_1e^x + \frac{l}{2e^{-x}} + C_2e^{-x}\]
As we are only interested in finding the particular solution, we will not consider the constants C_1 and C_2 as they will be absorbed into the complementary solution. Therefore, the particular solution is:
\[y_p(x) = \frac{-k}{2e^{x}} + \frac{l}{2e^{-x}}\]
05
Combine the complementary function and particular solution
Finally, to obtain the general solution, we combine the complementary function and particular solution:
\[y(x) = y_c(x) + y_p(x) = c_1e^x + c_2e^{-x} + \frac{-k}{2e^{x}} + \frac{l}{2e^{-x}}\]
This is the general solution to the given differential equation.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Differential Equations
Differential equations are mathematical equations that relate a function with its derivatives. They are used to describe various physical phenomena, such as heat transfer, wave propagation, and population dynamics. In essence, a differential equation specifies how a quantity changes over time or space. Differential equations are classified based on their order, linearity, and homogeneity.
For instance, the given equation in our problem, \( y'' - y = 2 \tanh x \), is a second-order linear differential equation. Here's why:
For instance, the given equation in our problem, \( y'' - y = 2 \tanh x \), is a second-order linear differential equation. Here's why:
- **Second-order:** The highest derivative in the equation is the second derivative, \( y'' \).
- **Linear:** Each term involving the function \( y \) or its derivatives is linear, meaning it is not raised to a power or multiplied by another derivative.
- **Inhomogeneous:** The equation includes a nonzero term on the right-hand side, \( 2 \tanh x \), making it inhomogeneous.
Complementary Function
The complementary function is a critical part of solving non-homogeneous linear differential equations. It satisfies the corresponding homogeneous equation, where all non-homogeneous terms are set to zero. For the equation \( y'' - y = 0 \), the complementary function accounts for the solution of the associated homogeneous differential equation.
In our exercise, finding the complementary function involves determining the roots of the characteristic equation \( r^2 - 1 = 0 \). Solving this, we obtain roots \( r = 1 \) and \( r = -1 \). The complementary function, denoted as \( y_c(x) \), is then constructed as follows:
In our exercise, finding the complementary function involves determining the roots of the characteristic equation \( r^2 - 1 = 0 \). Solving this, we obtain roots \( r = 1 \) and \( r = -1 \). The complementary function, denoted as \( y_c(x) \), is then constructed as follows:
- \( y_c(x) = c_1e^x + c_2e^{-x} \)
Particular Solution
The particular solution of a non-homogeneous differential equation addresses the specific terms on the right-hand side, independent of arbitrary constants. In our problem, the challenge is to account for the term \( 2 \tanh x \). The variation of parameters method is a technique employed here to determine a particular solution.
This method involves first expressing the particular solution \( y_p(x) \) as a linear combination of the homogeneous solutions with variable coefficients:
Following the procedure means deriving expressions and simplifying to isolate the terms \( u_1'(x) \) and \( u_2'(x) \). These derivatives are then integrated to find \( u_1(x) \) and \( u_2(x) \). The complete particular solution is derived by substituting these functions back into the expression for \( y_p(x) \). Importantly, the particular solution captures the behavior defined solely by the non-homogeneous term \( 2 \tanh x \). This solution will combine appropriately with the complementary function to give a comprehensive depiction of the system described by the differential equation.
This method involves first expressing the particular solution \( y_p(x) \) as a linear combination of the homogeneous solutions with variable coefficients:
- \( y_p(x) = u_1(x)e^x + u_2(x)e^{-x} \)
Following the procedure means deriving expressions and simplifying to isolate the terms \( u_1'(x) \) and \( u_2'(x) \). These derivatives are then integrated to find \( u_1(x) \) and \( u_2(x) \). The complete particular solution is derived by substituting these functions back into the expression for \( y_p(x) \). Importantly, the particular solution captures the behavior defined solely by the non-homogeneous term \( 2 \tanh x \). This solution will combine appropriately with the complementary function to give a comprehensive depiction of the system described by the differential equation.
General Solution
The general solution of a differential equation combines the complementary function and the particular solution, accounting for both the homogeneous and non-homogeneous parts of the equation. It signifies the total set of solutions for the equation.
For the differential equation given, the general solution is composed of:
For the differential equation given, the general solution is composed of:
- The complementary function \( y_c(x) = c_1e^x + c_2e^{-x} \)
- The particular solution derived from the non-homogeneous component \( y_p(x) = \frac{-k}{2e^{x}} + \frac{l}{2e^{-x}} \)