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Reduce each DE to self-adjoint form. (a) \(x^{2} y^{\prime \prime}+x y^{\prime}+y=0\), (b) \(y^{\prime \prime}+y^{\prime} \tan x=0\).

Short Answer

Expert verified
The function (a) \(x^{2} y^{\prime \prime}+x y^{\prime}+y=0\) is already in the self-adjoint form. The function (b) \(y^{\prime \prime}+y^{\prime} \tan x=0\), when multiplied by the integrating factor, becomes self-adjoint and represented by \((e^{-ln|cosx|}y^{\prime})^{\prime}=0\).

Step by step solution

01

Analyze the First Function

The given function in (a) is \(x^{2} y^{\prime \prime}+x y^{\prime}+y=0\). As it can be seen, the function is already in the self-adjoint form with \(p(x) = x^2\), \(p'(x) = 2x\) and \(q(x) = 1\). Therefore, no modification is needed for this equation.
02

Analyze the Second Function

For the second function (b), \(y^{\prime \prime}+y^{\prime} \tan x=0\), an extra term is present which makes it non-self-adjoint. To transform the equation into the self-adjoint form, an integrating factor will be used. First, recognize that the given equation is of the form \(y^{\prime \prime}+s(x) y^{\prime}=0\), where \(s(x) = tan(x)\). Thus, an integrating factor is \(e^{\int s(x) dx} = e^{\int tan(x) dx} = e^{-ln|cosx|}\)
03

Transform the Equation

Multiply the equation \(y^{\prime \prime}+y^{\prime} \tan x=0\) by the integrating factor just found to convert the equation into self-adjoint form. By doing this, the equation becomes \(e^{-ln|cosx|}(y^{\prime \prime}+y^{\prime} \tan x)=0\), which simplifies to \((e^{-ln|cosx|}y^{\prime})^{\prime}=0\). This is a self-adjoint form.

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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 involve variables and their derivatives. They are fundamental in describing various physical phenomena, from simple motion to complex fluid dynamics. A differential equation expresses the relationship between a function and its derivatives, and they are broadly categorized into ordinary differential equations (ODEs) and partial differential equations (PDEs). Ordinary differential equations involve derivatives with respect to a single variable, whereas partial differential equations involve derivatives with respect to multiple variables.

In solving differential equations, the aim is to find a function that satisfies the given equation. These functions are crucial in physics and engineering because they often model real-world systems. Knowing how to solve differential equations allows one to predict system behaviors over time. It's a stepping stone to understanding more advanced topics, such as dynamics and control systems.

Different methods exist to solve these equations, depending on their complexity and form. Some common approaches include separation of variables, method of undetermined coefficients, and variations of parameters. In our exercise, we focus on reducing a given differential equation to self-adjoint form, which simplifies the solving process.
Self-Adjoint Form
The self-adjoint form of a differential equation is significant because it simplifies analysis and solutions. An equation is self-adjoint if it can be written in a specific symmetric form that allows the use of certain mathematical properties and techniques. These properties often lead to simplifications in finding solutions or determining the behavior of the systems modeled by the differential equations.

A second-order linear differential equation is in a self-adjoint form if it can be expressed as \[ \frac{d}{dx}\left[ p(x) \frac{dy}{dx} \right] + q(x) y = 0. \]
For self-adjoint form, the function \( p(x) \) is the leading coefficient and plays a critical role in transforming other forms into self-adjoint. In our exercise, equation (a) was already in self-adjoint form with \( p(x) = x^2 \). However, equation (b) required transformation using an integrating factor to achieve this form.

By transforming a differential equation into self-adjoint form, it becomes easier to apply various mathematical theorems and techniques, such as Euler's theorem or solving boundary value problems. Therefore, recognizing or converting equations into this form can significantly aid in solving complex mathematical and physical problems.
Integrating Factor
An integrating factor is a mathematical tool used to simplify differential equations and solve them more easily. This technique is particularly useful in handling linear first-order differential equations, and it helps in converting non-self-adjoint equations into self-adjoint form.

The general idea involves multiplying a differential equation by a function, called an integrating factor, that depends on one of the variables. This multiplication simplifies the equation, often making it easier to solve. In the provided exercise, the technique involved finding an integrating factor for equation (b) to convert it into self-adjoint form.

To determine the integrating factor, consider the general form of a first-order linear differential equation:\[ y' + s(x)y = 0. \]
The integrating factor is given by:\[ \mu(x) = e^{\int s(x) dx}. \]
Applying the integrating factor transforms the equation into a more manageable form. In the exercise provided, the integrating factor was found by evaluating \( e^{-\ln|\cos x|} \), which helped transform the equation into a self-adjoint form that could be simplified further. This powerful technique is pivotal in solving various differential equations encountered in mathematical physics.

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Most popular questions from this chapter

Let \(f_{1}, f_{2}\), and \(f_{3}\) be three solutions of the third-order linear differential equation \(y^{\prime \prime \prime}+p_{2}(x) y^{\prime \prime}+p_{1}(x) y^{\prime}+p_{0}(x) y=0 .\) Derive a FODE satisfied by the (generalized \(3 \times 3\) ) Wronskian of these solutions.

The function defined by $$ f(x)=\sum_{n=0}^{\infty} c_{n} x^{n}, \quad \text { where } \quad c_{n+2}=\frac{2 n-\lambda}{(n+1)(n+2)} c_{n}, $$ can be written as \(f(x)=c_{0} g(x)+c_{1} h(x)\), where \(g\) is even and \(h\) is odd in \(x .\) Show that \(f(x)\) goes to infinity at least as fast as \(e^{x^{2}}\) does, i.e., \(\lim _{x \rightarrow \infty} f(x) e^{-x^{2}} \neq 0 .\) Hint: Consider \(g(x)\) and \(h(x)\) separately and show that $$ g(x)=\sum_{n=0}^{\infty} b_{n} x^{n}, \quad \text { where } \quad b_{n+1}=\frac{4 n-\lambda}{(2 n+1)(2 n+2)} b_{n} . $$ Then concentrate on the ratio \(g(x) / e^{x^{2}}\), where \(g\) and \(e^{x^{2}}\) are approximated by polynomials of very high degrees. Take the limit of this ratio as \(x \rightarrow \infty\), and use recursion relations for \(g\) and \(e^{x^{2}}\). The odd case follows similarly.

Solve \(y^{\prime \prime}-2 y^{\prime}+y=x e^{x}\) subject to the initial conditions \(y(0)=0\), \(y^{\prime}(0)=1 .\)

Show that if \(u(x)\) and \(v(x)\) are solutions of the self-adjoint DE \(\left(p u^{\prime}\right)^{\prime}+q u=0\), then Abel's identity, \(p\left(u v^{\prime}-v u^{\prime}\right)=\) constant, holds.

Substitute \(v=y^{\prime} / y\) in the homogeneous SOLDE $$ y^{\prime \prime}+p(x) y^{\prime}+q(x) y=0 $$ and: (a) Show that it turns into \(v^{\prime}+v^{2}+p(x) v+q(x)=0\), which is a firstorder nonlinear equation called the Riccati equation. Would the same Riccati equation substitution work if the DE were inhomogeneous? (b) Show that by an appropriate transformation, the Riccati equation can be directly cast in the form \(u^{\prime}+u^{2}+S(x)=0\).

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