Question

Use the method suggested by Exercise 34 to find a particular solution in the form \(y_{p}=\int_{x_{0}}^{x} G(x, t) F(t) d t,\) given the indicated fundamental set of solutions. Assume that \(x\) and \(x_{0}\) are in an interval on which the equation is normal. $$ y^{(4)}-5 y^{\prime \prime}+4 y=F(x) ; \quad\left\\{e^{x}, e^{-x}, e^{2 x}, e^{-2 x}\right\\} $$

Short answer

Question: Given a fourth-order non-homogeneous linear differential equation with forcing function \(F(x)\) and a fundamental set of solutions \(\{e^x, e^{-x}, e^{2x}, e^{-2x}\}\), find a particular solution using the method of integrating factors and Green's function. Answer: The particular solution can be found using the formula $$ y_{p}(x) = \int_{x_0}^{x} G(x, t) F(t) dt $$ where the Green's function, \(G(x, t)\), is $$ G(x, t) = \frac{e^{x}e^{-t} - e^{-x}e^{t}}{96 \cdot (e^{2t} - e^{-2t})(e^{4t} + e^{-4t} + 1)} $$ To find the particular solution, plug the given forcing function \(F(x)\) into the integral and evaluate it.

Step-by-step solution

Identify the fundamental set of solutions

We are given the fundamental set of solutions as \(\{e^x, e^{-x}, e^{2x}, e^{-2x}\}\).

Construct the Wronskian of the given fundamental set

The Wronskian is a determinant that involves the derivatives of the functions in the fundamental set. For the given functions, the Wronskian is: $$ W(x) = \det \begin{pmatrix} e^x & e^{-x} & e^{2x} & e^{-2x} \\ e^x & -e^{-x} & 2e^{2x} & -2e^{-2x} \\ e^x & e^{-x} & 4e^{2x} & 4e^{-2x} \\ e^x & -e^{-x} & 8e^{2x} & -8e^{-2x} \end{pmatrix} $$

Compute the Wronskian

Calculating the determinant, we get: $$ W(x) = 96 \cdot (e^{2x} - e^{-2x})(e^{4x} + e^{-4x} + 1) $$

Construct Green's function

Green's function, \(G(x, t)\), is defined as: $$ G(x, t) = \frac{\varphi_1(x) \varphi_2(t) - \varphi_1(t) \varphi_2(x)}{W(t)} $$ where \(\varphi_1(x)\) and \(\varphi_2(x)\) are the first two functions in the given fundamental set (i.e., \(e^x\) and \(e^{-x}\)).

Compute Green's function

Plugging the functions into the Green's function formula, we get: $$ G(x, t) = \frac{e^{x}e^{-t} - e^{-x}e^{t}}{96 \cdot (e^{2t} - e^{-2t})(e^{4t} + e^{-4t} + 1)} $$

Calculate particular solution using given formula

Finally, the particular solution, \(y_p\), can be computed using the given formula: $$ y_{p}(x) = \int_{x_0}^{x} G(x, t) F(t) dt $$ The fundamental set of solutions already gives the complementary solution, so integrating the Green's function with the forcing function, \(F(x)\), will result in the particular solution. Plug the given \(F(x)\) into this equation to find \(y_p(x)\).

Key concepts

Wronskian

The Wronskian is a crucial concept in differential equations, as it helps us determine whether a set of solutions is linearly independent. In simpler terms, it tells us if these solutions can uniquely solve the differential equation. To compute the Wronskian, we form a matrix using our fundamental set of solutions and their derivatives. For example, given the set \( \{e^x, e^{-x}, e^{2x}, e^{-2x}\} \), we construct a matrix with rows comprising each function and its derivatives.
The determinant of this matrix gives us the Wronskian. If the Wronskian is non-zero at some point in the interval, the solutions are linearly independent. In our example, the Wronskian is:

• \(W(x) = \det\begin{pmatrix}e^x & e^{-x} & e^{2x} & e^{-2x} \e^x & -e^{-x} & 2e^{2x} & -2e^{-2x} \e^x & e^{-x} & 4e^{2x} & 4e^{-2x} \e^x & -e^{-x} & 8e^{2x} & -8e^{-2x} \end{pmatrix}\)

This result ensures that our functions can form the general solution needed for our differential equation. It’s like confirming that our toolbox has all the right tools to fix a complex machine.

Green's Function

Green's functions are a powerful tool in solving nonhomogeneous differential equations. Imagine Green's functions as the "bridge" between the inputs to a system and the resulting outputs.
They help us find particular solutions depending on a given forcing function, \(F(x)\). We define the Green's function for our equation using two functions from our fundamental set. For instance, with \(e^x\) and \(e^{-x}\), we have:

• \(G(x, t) = \frac{e^{x}e^{-t} - e^{-x}e^{t}}{W(t)}\)

Where \(W(t)\) is the Wronskian calculated at \(t\). The function \(G(x, t)\) transforms our equation into an easier computation. By integrating this with \(F(t)\), we find how the system responds to the function \(F\). This results in our particular solution. With this approach, we can tackle difficult problems more efficiently by breaking them down into manageable pieces.

Fundamental Set of Solutions

A fundamental set of solutions is like a family of functions that helps to solve a differential equation completely. These functions are carefully chosen so that any solution of the differential equation can be expressed as a combination of them.
In our example, the set is \( \{e^x, e^{-x}, e^{2x}, e^{-2x}\} \). We use these because they satisfy the homogeneous version of the differential equation. Each function in the set is crucial, and their independence is confirmed using the Wronskian. This independence ensures they can "cover all bases" when we’re piecing together a general solution. By using a fundamental set, we not only solve for the complementary solution but also lay the groundwork for finding particular solutions. Thus, it’s like setting the stage for a complex play—the right actors ensure everything goes as planned and the story unfolds smoothly.

Particular Solution

The particular solution is the key to solving nonhomogeneous differential equations. Unlike the complementary solution that handles just the homogeneous part, the particular solution addresses the entirety of the differential equation, including the forcing function \(F(x)\).
We find it by utilizing Green's function. The solution comes from integrating Green's function with \(F(t)\):

• \(y_{p}(x) = \int_{x_0}^{x} G(x, t) F(t) dt\)

This integral essentially "weighs" the effect of \(F(t)\) over the interval, yielding the particular solution. It’s like tailoring a piece of clothing; it fits the specifics of the problem perfectly, producing the precise outcome needed. Combined with the complementary solution, the particular solution completes the puzzle of the differential equation. It ensures we have the full picture and can describe any behavior modeled by the equation across its domain.