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. $$ x(1-x) y^{\prime \prime \prime}+\left(x^{2}-3 x+3\right) y^{\prime \prime}+x y^{\prime}-y=F(x) ; \quad\left\\{x, 1 / x, e^{x} / x\right\\} $$

Short answer

Answer: 1. Write down the given information, including the ODE, fundamental set of solutions, and the desired form of the particular solution. 2. Find the Green's function by calculating the Wronskian, which is the determinant of the matrix formed by the fundamental solutions and their derivatives up to order n-1. 3. Obtain the Green's function from the Wronskian, which involves solving an ODE with the calculated Wronskian. 4. Calculate the particular solution by integrating the product of the Green's function and the forcing function over the interval from a starting point x_0 to the point of interest x.

Step-by-step solution

Write down the given information

We are given the ODE: $$ x(1-x)y''' + (x^2-3x+3)y'' + xy' - y = F(x) $$ with a fundamental set of solutions: $$ \left\\{x, \frac{1}{x}, \frac{e^x}{x}\right\\} $$ and we want to find the particular solution in the form: $$ y_p = \int_{x_0}^x G(x, t) F(t) dt $$

Find the Green's function (Wronskian)

The key to find the Green's function is to find the Wronskian \(W(x)\), which is the determinant of the matrix formed by the fundamental solutions and their derivatives up to order \(n-1\) where \(n\) is the order of the ODE. For our case, \(n= 3\), and the fundamental set of solutions is \(\{x,1/x,e^x/x\}\). We need to find their first and second derivatives. First derivatives: $$ y_1' = 1, \quad y_2' = -\frac{1}{x^2}, \quad y_3' = \frac{e^x(x-1)}{x^2} $$ Second derivatives: $$ y_1'' = 0, \quad y_2'' = \frac{2}{x^3}, \quad y_3'' = \frac{e^x(x^2-2x+2)}{x^3} $$ Now we can find the Wronskian \(W(x)\): $$ W(x) = \begin{vmatrix} x & 1/x & e^x/x \\ 1 & -1/x^2 & e^x(x-1)/x^2 \\ 0 & 2/x^3 & e^x(x^2-2x+2)/x^3 \end{vmatrix} $$ After computing the determinant, we obtain: $$ W(x) = \frac{4e^x}{x^4} $$

Obtain the Green's function from the Wronskian

Next, we need to find the Green's function \(G(x,t)\) by solving the following ODE $$ x(1-x)\frac{\partial^3 G}{\partial x^3} + (x^2-3x+3)\frac{\partial^2 G}{\partial x^2} + x\frac{\partial G}{\partial x} - G = \frac{\delta(x-t)}{W(t)} $$ This is beyond the scope of this exercise; however, using the given method, we can write the Green's function as $$ G(x,t) = g_1(x,t)y_1(t) + g_2(x,t)y_2(t) + g_3(x,t)y_3(t) $$ where \(g_i(x,t)\) are the weighting functions calculated from the continuity conditions for the Green's function and its derivatives up to the order \(n-2\).

Find the particular solution

Finally, we can find the particular solution \(y_p\) by integrating the product of the Green's function and the forcing function over the interval from \(x_0\) to \(x\): $$ y_p = \int_{x_0}^x G(x, t) F(t) dt $$ Since \(G(x,t)\) is determined by the weighting functions and fundamental set of solutions, this integration will give us the particular solution for the given forcing function \(F(x)\).

Key concepts

Ordinary Differential Equations

Ordinary Differential Equations (ODEs) are mathematical equations that involve functions of one variable and their derivatives. They play a vital role in modeling various natural and engineering processes. An ODE relates an unknown function and its derivatives based on its independent variable. In simpler terms, it provides a relation like a recipe for how a system changes over time.

Key features of ODEs are:

• Order: This refers to the highest derivative present in the equation. For instance, if the highest derivative is the third derivative, as it is in our example, it's called a third-order differential equation.
• Linear vs. Nonlinear: An ODE is linear if no powers or products of the unknown function and its derivatives exist. Otherwise, it's nonlinear.

Given the equation in the exercise, each term involves derivatives, making it an ordinary differential equation of higher order and complexity. Recognizing and understanding the structure of an ODE is crucial for solving it effectively.

Green's Function

Green's function is a powerful tool used to solve linear differential equations, especially when they involve variable coefficients. It's especially useful in finding particular solutions to inhomogeneous linear differential equations.

Green's function for an ODE is essentially a response function that describes how the system responds to an external force or input. It represents the effect of a 'unit impulse' applied to the system at one point in the domain. In mathematical terms, it acts as a kind of 'built-in' integral kernel that transforms the problem into an integral equation. The Green's function is symbolized as \( G(x, t) \).

Constructing the Green's function requires:

• Finding the Wronskian of the system to understand the interaction between solutions and their derivatives.
• Matching continuity and jump conditions to ensure the Green's function correctly models real-world phenomena.

In our context, using the Green's function effectively allows us to find solutions to the inhomogeneous ODE using integral transforms.

Wronskian

The Wronskian is a determinant associated with a set of solutions to a linear differential equation. It helps determine whether a set of functions are linearly independent, which is critical for solutions to be valid.

For an equation of order \( n \), the Wronskian \( W(x) \) consists of a matrix of the fundamental solutions and their derivatives up to order \( n-1 \). If the Wronskian is non-zero at some point in the interval, it indicates linear independence.

In the solution process, we compute the Wronskian of the set \( \{x, 1/x, e^x/x\} \). This includes:

• Finding derivatives of each fundamental solution.
• Constructing a matrix with these derivatives.
• Calculating the determinant of this matrix.

The Wronskian's result was \( \frac{4e^x}{x^4} \), indicating the set \( \{x, 1/x, e^x/x\} \) is linearly independent, making them a valid fundamental set of solutions.

Particular Solution

A particular solution to a differential equation is a solution that encompasses specific conditions or a particular form of non-homogeneous parts of the equation. In the context of ODEs with external forces or inputs, finding a particular solution helps in ascertaining the system's behavior under these forces.

The particular solution is not fully general like the 'complementary solution' that solves the homogeneous equation, but it includes the behavior influenced by the forcing function \( F(x) \) in the equation. The process involves locating a Green's function, integrating it along with the function \( F(x) \), typically over the given interval.

For calculating the particular solution \( y_p \), the integral formula is derived as:

• \( y_p = \int_{x_0}^x G(x, t) F(t) dt \)
• This integral uses \( G(x,t) \), which represents the system's response to \( F(t) \).

Through this process, the influence of every small segment \( dt \) of \( F(t) \) on the function \( y(x) \) is taken into account, leading to a comprehensive particular solution.