Problem 27 - Differential Equations and Linear Algebra

Question

Use a Green's function to solve the given initial-value problem. [Hint: Choose \(\left.x_{0}=0 .\right]\) $$y^{\prime \prime}-4 y^{\prime}+4 y=5 x e^{2 x}, \quad y(0)=1, \quad y^{\prime}(0)=0$$

Step-by-Step Solution

Answer

#tag_title# Step 2: Compute Green's function #tag_content# Now, let's find the Green's function. The fundamental solutions based on the homogeneous solution are \(y_1(x) = e^{2x}\) and \(y_2(x) = xe^{2x}\). Using these solutions, we derive the Wronskian \(W(x) = \left|\begin{matrix}y_1 & y_2\\y_1' & y_2'\end{matrix}\right| = e^{4x}\). The Green's function can then be derived using the fundamental solutions and the Wronskian: \(G(x, x_0) = \dfrac{1}{W(x_0)}\left[y_1(x_0)\cdot y_2(x) - y_1(x) \cdot y_2(x_0)\right] = \dfrac{1}{e^{4x_0}}\left[e^{2x_0} \cdot xe^{2x} - e^{2x} \cdot x_0e^{2x_0}\right]\). Since we are given the hint to choose \(x_0 = 0\), we substitute and get: \(G(x, 0) = \dfrac{1}{1}\left[1 \cdot xe^{2x} - e^{2x} \cdot 0\right] = xe^{2x}\). #tag_title# Step 3: Find the particular solution #tag_content# Using the Green's function, we can derive the particular solution for the IVP: \(y_p(x) = y_h(0) + \int_0^x G(x, s) f(s) ds = 1 + \int_0^x se^{2s} \cdot 5se^{2s} ds\), where \(f(s) = 5s e^{2s}\) is the right side of the original inhomogeneous equation. Evaluating the integral, we get \(y_p(x) = 1 + \dfrac{5}{4}x^2 e^{2x}\). #tag_title# Step 4: Combine solutions #tag_content# The general solution for the IVP is the sum of the homogeneous and particular solutions: \(y(x) = y_h(x) + y_p(x) = c_1 e^{2x} + c_2 xe^{2x} + 1 + \dfrac{5}{4}x^2 e^{2x}\). Using the initial conditions \(y(0) = 1\) and \(y'(0) = 0\), we can find the constants as \(c_1 = 0\) and \(c_2 = -\dfrac{5}{4}\). Thus, the solution to the IVP is: \(y(x) = -\dfrac{5}{4}xe^{2x} + 1 + \dfrac{5}{4}x^2 e^{2x}\).

Step 1: Analyze the problem

Identify the type of problem and the appropriate mathematical technique to apply.

Step 2: Apply the technique and solve

#tag_title# Step 2: Compute Green's function #tag_content# Now, let's find the Green's function. The fundamental solutio.

Step 3: Verify the result

Check the answer by substitution or alternative methods to confirm correctness.

Key Concepts

Homogeneous Solution

In the context of solving differential equations, a homogeneous solution is key. It pertains to the solution of the homogeneous form of the differential equation, which is the equation with the right-hand side set to zero.
For example, given the differential equation: \(y'' - 4y' + 4y = 5xe^{2x}\), its homogeneous counterpart is \(y'' - 4y' + 4y = 0\).
In this equation, we are solving for when the output is zero.
The homogeneous solution is often represented as \(y_h(x)\). To find it, we typically solve a characteristic equation derived from the differential equation. This solution only reflects the behavior determined by the system itself, without any external input.
The actual general solution to the original differential equation is the sum of the homogeneous solution and a particular solution that takes the right-hand side into account.

Characteristic Equation

The characteristic equation is a fundamental algebraic tool used in solving linear differential equations. Derived from the differential equation by replacing derivatives with powers of an unknown parameter, generally noted as \(r\), it provides insights into the behavior of the equation's solutions.
The differential equation: \(y'' - 4y' + 4y = 0\) leads to the characteristic equation: \(r^2 - 4r + 4 = 0\).
Solving this equation helps determine the type of solutions we should expect.
If the characteristic equation has distinct roots, the solutions are exponential functions. In cases where roots are repeated, like when \(r=2\) in our example: \(r^2 - 4r + 4 = (r-2)^2 = 0\), the solution involves terms such as \(e^{rx}\) and \(xe^{rx}\). Seeing double roots often indicates a combination of exponential and polynomial functions in the solution.

Initial-Value Problem

Initial-value problems form an essential part of solving differential equations. These problems specify the values of the solution and possibly its derivatives at a given point, usually called the 'initial point.'
In our exercise, we are dealing with an initial-value problem described by: \(y^{\prime\prime}-4y^{\prime}+4y=5xe^{2x}, y(0)=1, y^{\prime}(0)=0\).
This setup dictates that at \(x_0 = 0\), the value of \(y(x)\) should be 1, and the value of its derivative should be 0.
The given initial values help uniquely determine the particular solution of the differential equation. This is crucial because without these initial values, the solution would have arbitrary constants, resulting in infinitely many potential solutions.
Initial-value problems are especially important in modeling real-world occurrences where initial conditions are known.

Differential Equation

Differential equations are mathematical equations that involve functions and their derivatives. They describe how a particular quantity changes in relation to another. In mathematics and engineering, they are vital for modeling the behavior of systems.
The given differential equation for this problem is: \(y^{\prime\prime}-4y^{\prime}+4y=5xe^{2x}\).
This type of equation is a linear second-order differential equation with constant coefficients.
Understanding and solving differential equations can predict how processes evolve over time or space.
Solutions to such equations involve finding a function, or set of functions, that satisfies the relation described by the equation. They can often be broken down into homogeneous and particular parts, especially when dealing with linear equations.
Knowledge of differential equations enables us to solve initial-value problems, hence, their importance in calculus and mathematical modeling.