Question

Suppose \(f\) is continuous on an open interval that contains \(x_{0}=0 .\) Use variation of parameters to find a formula for the solution of the initial value problem $$ y^{\prime \prime}-y=f(x), \quad y(0)=k_{0}, \quad y^{\prime}(0)=k_{1}. $$

Short answer

Question: Use the method of variation of parameters to find a solution to the initial value problem \(y^{\prime\prime} - y = f(x)\) with the given initial conditions, \(y(0) = k_0\) and \(y'(0) = k_1\). Answer: The final solution is given in the form of \(y(x) = (C_1 e^{x} + C_2 e^{-x}) + A e^{x} + B e^{-x}\), where the constants \(A\), \(B\), \(C_1\), and \(C_2\) can be determined using the initial conditions \(y(0) = k_0\) and \(y'(0) = k_1\).

Step-by-step solution

Find the complementary solution

To find the complementary solution, we will solve the homogeneous version of the given differential equation: $$ y^{\prime\prime} - y = 0. $$ The characteristic equation is: $$ r^2 - 1 = 0. $$ Factoring, we get \((r-1)(r+1) = 0\), with roots \(r=1\) and \(r=-1\). Consequently, the complementary solution is: $$ y_c(x) = C_1 e^{x} + C_2 e^{-x}. $$

Apply variation of parameters

Using the method of variation of parameters, we introduce two functions, \(u_1(x)\) and \(u_2(x)\), such that the particular solution is given by: $$ y_p(x) = u_1(x)e^{x} + u_2(x)e^{-x}. $$ To find \(u_1(x)\) and \(u_2(x)\), we need to solve two first-order linear differential equations. We start by computing the derivatives of \(y_p(x)\): $$ y_p^{\prime}(x) = u_1^{\prime}(x)e^{x} - u_2^{\prime}(x)e^{-x} + u_1(x)e^{x} + u_2(x)e^{-x}, $$ and $$ y_p^{\prime\prime}(x) = u_1^{\prime\prime}(x)e^{x} + u_2^{\prime\prime}(x)e^{-x} + 2u_1^{\prime}(x)e^{x} - 2u_2^{\prime}(x)e^{-x} + u_1(x)e^{x} + u_2(x)e^{-x}. $$ Next, we substitute these expressions into the given differential equation, resulting in: $$ u_1^{\prime\prime}(x)e^{x} + u_2^{\prime\prime}(x)e^{-x} = f(x). $$ Now, we solve the following two first-order linear differential equations simultaneously: $$ u_1^{\prime}(x)e^{x} - u_2^{\prime}(x)e^{-x} = 0, $$ and $$ u_1^{\prime\prime}(x)e^{x} + u_2^{\prime\prime}(x)e^{-x} = f(x). $$ We can write the system of equations as: $$ \begin{cases} u_1'(x) = u_2'(x)e^{-2x}, \\ u_1''(x) + u_2''(x)e^{2x} = f(x)e^{2x}. \end{cases} $$ Integrating the first equation with respect to \(x\), we get: $$ u_1(x) = \int u_2'(x)e^{-2x}\,dx + A, $$ where \(A\) is a constant of integration. Substituting this expression for \(u_1(x)\) into the second equation, we obtain: $$ \frac{d}{dx}\left(\int u_2'(x)e^{-2x}\,dx + A\right) + u_2''(x)e^{2x} = f(x)e^{2x}. $$ Simplifying, we have: $$ u_2'(x)e^{-2x} + u_2''(x)e^{2x} = f(x)e^{2x}. $$ Integrating this equation with respect to \(x\) yields: $$ u_2(x) = \int\left(f(x)e^{2x} - u_2'(x)e^{-2x}\right)\,dx + B, $$ where \(B\) is another constant of integration. Now, we have expressions for \(u_1(x)\) and \(u_2(x)\). Therefore, the particular solution is: $$ y_p(x) = \left(\int u_2'(x)e^{-2x}\,dx + A\right)e^{x} + \left(\int\left(f(x)e^{2x} - u_2'(x)e^{-2x}\right)\,dx + B\right)e^{-x}. $$

Apply initial conditions to find the final solution

To find the final solution, we combine the complementary solution and particular solutions, and apply the given initial conditions: $$ y(x) = y_c(x) + y_p(x) = (C_1 e^{x} + C_2 e^{-x}) + \left(\int u_2'(x)e^{-2x}\,dx + A\right)e^{x} + \left(\int\left(f(x)e^{2x} - u_2'(x)e^{-2x}\right)\,dx + B\right)e^{-x}. $$ Applying the initial conditions \(y(0) = k_0\) and \(y'(0) = k_1\), we can solve for the constants \(A\), \(B\), \(C_1\), and \(C_2\). The final solution will then be in the form of: $$ y(x) = (C_1 e^{x} + C_2 e^{-x}) + A e^{x} + B e^{-x}. $$

Key concepts

Variation of Parameters

Variation of parameters is a method used to find the particular solution of a nonhomogeneous differential equation. This technique builds on the complementary solution (which solves the corresponding homogeneous equation) and accounts for the non-zero part that varies according to the function on the other side of the equation.

To apply variation of parameters, introduce new functions, known as parameter functions, that will multiply the terms of the complementary solution. These functions are designed to absorb the nonhomogeneous part of the equation. Typically, you'll end up with a system of first-order differential equations to find these functions.

Exercise Improvement Advice

For an educational resource, it is beneficial to incorporate step-by-step visualizations when explaining variation of parameters. Diagrams showing the transition from the complementary solution to the introduction of the parameter functions help in improving student comprehension. Demonstrating how parameter functions are derived through integration can also ease students' understanding.

Differential Equations

Differential equations are mathematical equations that relate some function with its derivatives. They describe a wide variety of phenomena such as physics, engineering, and even biology. In the context of our problem, a second-order differential equation is presented, which includes both the function y and its second derivative y''.

These equations can be classified as either homogeneous or nonhomogeneous. A homogeneous differential equation is set equal to zero, whereas a nonhomogeneous equation equals a non-zero function, often representing an external force or input.

Exercise Improvement Advice

To make differential equations more accessible, it's important to focus on clear examples and contextual applications. By showing how these equations model real-world situations, students can grasp their importance and get motivated to understand the underlying mathematics. Additionally, analogy can be a powerful tool, comparing the solving process to more familiar tasks or ideas to build confidence.

Complementary Solution

The complementary solution, often denoted as y_c, is a general solution to the homogeneous part of a differential equation (i.e., where the equation is set to zero). This solution incorporates all possible solutions to the homogeneous equation and includes arbitrary constants that later will be determined by initial conditions.

In the step-by-step solution, the complementary solution is found by solving the characteristic equation of the differential equation's homogeneous part. Here, it is expressed as a combination of exponential functions with constants C1 and C2.

Exercise Improvement Advice

An important aspect of conveying the complementary solution to students is to emphasize its role in the overall solution of the differential equation. It's useful to articulate that this solution forms the basis upon which the particular solution is added. Visual aids can also illustrate how different parameters can adjust the shape of the complementary solution.

Particular Solution

In contrast with the complementary solution, the particular solution, denoted as y_p, specifically addresses the nonhomogeneous part of our differential equation. This solution does not contain arbitrary constants; it is constructed to ensure that, when added to the complementary solution, it satisfies the entire nonhomogeneous equation.

The particular solution usually arises from a specific form that is guessed based on the nonhomogeneous part and is then tweaked through variation of parameters or another method until it fits the equation. Once the particular solution is found, it is often specific to the nonhomogeneous equation's unique characteristics.

Exercise Improvement Advice

When explaining the particular solution to students, it's effective to show how this solution, combined with the complementary solution, forms the complete solution to the nonhomogeneous equation. It can also help to work through multiple examples with different nonhomogeneous parts to illustrate the process of finding y_p. Real-life applications of these solutions can be particularly motivating, as students see the practical application of their mathematical skills.