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The differential equation $$ P(x, y) d x+Q(x, y) d y=0 $$ is exact. Construct a solution $$ \varphi(x, y)=\int_{x_{0}}^{x} P(x, y) d x+\int_{y_{0}}^{y} Q\left(x_{0}, y\right) d y=\text { constant. } $$

Short Answer

Expert verified
The solution is \(\varphi(x, y) = \int_{x_0}^x P(x, y) \, dx + \int_{y_0}^y Q(x_0, y) \, dy = \text{constant}.\)

Step by step solution

01

Definition of Exact Equation

An exact differential equation of the form \(P(x, y) \, dx + Q(x, y) \, dy = 0\) satisfies \(\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} \). This condition means there exists a potential function \(\varphi(x, y)\) such that \(\frac{\partial \varphi}{\partial x} = P\) and \(\frac{\partial \varphi}{\partial y} = Q\).
02

Construct the Potential Function \(\varphi\)

Begin with the first integral: \(\int_{x_0}^x P(x, y) \, dx\). This results in a function that can vary with \(y\). Add a function \(g(y)\) to account for integration with respect to \(x\). So, it becomes \(\varphi(x, y) = \int_{x_0}^x P(x, y) \, dx + g(y)\).
03

Differentiate with Respect to \(y\)

Differentiate \(\varphi(x, y) = \int_{x_0}^x P(x, y) \, dx + g(y)\) with respect to \(y\): \(\frac{\partial \varphi}{\partial y} = \int_{x_0}^x \frac{\partial P}{\partial y} \, dx + \frac{dg}{dy}\). By the exactness condition, this equals \(Q(x, y)\).
04

Solve for \(g'(y)\)

Set \(\frac{\partial \varphi}{\partial y} = Q(x, y)\). Therefore, \(\frac{d g}{dy} = Q(x_0, y)\). This implies \(g(y) = \int_{y_0}^y Q(x_0, y) \, dy + C\), where \(C\) is a constant of integration.
05

Combine to Form the Solution

Substituting \(g(y)\) back, the potential function \(\varphi(x, y)\) becomes, \(\varphi(x, y) = \int_{x_0}^x P(x, y) \, dx + \int_{y_0}^y Q(x_0, y) \, dy + C\). Thus, the solution is \(\varphi(x, y) = \text{constant}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Potential Function
An exact differential equation refers to a type of equation where a potential function exists. This potential function, often denoted as \(\varphi(x, y)\), is a scalar function from which the vector field components can be derived through partial derivatives. The function \(\varphi(x, y)\) satisfies the conditions \(\frac{\partial \varphi}{\partial x} = P(x, y)\) and \(\frac{\partial \varphi}{\partial y} = Q(x, y)\). Understanding these conditions is critical because they ensure the path-independence of the line integral, confirming that the differential equation is exact.
In simpler terms, a potential function \(\varphi\) acts like an underlying map through which both functions \(P\) and \(Q\) can be drawn. This explains why the differential equations relate to conservative fields, where any closed-loop path integration yields zero. Constructing \(\varphi\) involves combining the integrals of \(P\) and \(Q\) across the respective variables.
Partial Derivatives
Partial derivatives are the backbone of dealing with functions of several variables. They measure how functions change as only one variable varies, keeping others constant. For an exact differential equation, these partial derivatives play a crucial role. The condition \(\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}\) ensures the equation is exact, meaning the partial derivatives must be equal.
This mutual relationship between \(P\) and \(Q\), via their partial derivatives, allows us to find a potential function. When calculating the partial derivative of the potential function \(\varphi\) to verify the exactness or derive expressions for \(P\) and \(Q\), the partial derivative becomes a tool to transition smoothly between expressions.
Mastering partial derivatives helps unravel the complexity of multivariable functions, making it easier to solve, verify, or even visualize the behavior of functions involved in exact differential equations.
Integration Techniques
To solve exact differential equations, familiarity with integration techniques becomes indispensable. Starting with one variable, we integrate function \(P(x, y)\) with respect to \(x\), assuming \(y\) is constant, to begin constructing the potential function \(\varphi(x, y)\). This often results in a function of \(x\) plus an unknown function of \(y\).
After determining the potential function's first part, integrating \(Q(x_0, y)\) with respect to \(y\) completes the solution. Here, the critical aspect is to recognize and properly account for the added constant of integration or functions \(g(y)\), to match the exact form of the differential.
Integration techniques, thus, require a strategic step-by-step approach, ensuring each component fits perfectly into the potential function \(\varphi(x, y)\). This process transforms both integrals into constants, cementing \(\varphi = \text{constant}\) as the final solution.
Mathematical Methods for Physicists
The concept of exact differential equations and their solutions through potential functions are not just mathematical exercises; they are ingrained in physics and engineering. They model various phenomena, such as gravitational and electromagnetic fields, where potential functions describe energies or forces that have no directional dependencies.
In physics, solving such equations using integration and recognizing exactness can help in determining energy conservation or formulating conservative systems. These methods often intertwine with principles from calculus and linear algebra, providing tools for analysis and simplification.
By applying mathematical methods for physicists to exact differential equations, problems become more tractable. Key mathematical techniques—like recognizing conservative forces using potential function mapping—empower physicists to simplify, understand, and calculate complex systems efficiently. This alignment signifies the interplay between precise mathematics and practical physical applications.

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Most popular questions from this chapter

In the development of a series solution of the simple harmonic oscillator equation the second series coefficient \(a_{1}\) was neglected except to set it equal to zero. From the coefficient of the next to the lowest power of \(x, x^{k-1}\), develop a second indicial type equation. (a) (SHO equation with \(k=0 .\) ) Show that \(a_{1}\), may be assigned any finite value (including zero). (b) (SHO equation with \(k=1\).) Show that \(a_{1}\) must be set equal to zero.

In the equation $$ y_{2}(x)=y_{1}(x) \int^{x} \frac{\exp \left[-\int^{s} P(t) d t\right]}{\left[y_{1}(s)\right]^{2}} d s $$ \(y_{1}(x)\) satisfies $$ y_{1}^{\prime \prime}+P(x) y_{1}^{\prime}+Q(x) y_{1}=0 . $$ The function \(y_{2}(x)\) is a linearly independent second solution of the same equation. Show that the inclusion of lower limits on the two integrals leads to nothing new; that is, it throws in only overall factors and/or a multiple of the known solution \(y_{1}(x)\).

In the linear homogeneous differential equation $$ \frac{d v}{d t}=-a v $$ the variables are separable. When the variables are separated the equation is exact. Solve this differential equation subject to \(v(0)=v_{0}\) by the following three methods: (a) separating variables and integrating; (b) treating the separated variable equation as exact; and (c) using the result for a linear homogeneous differential equation. $$ \text { ANS. } v(t)=v_{0} e^{-a t} $$

An atomic (quantum mechanical) particle is confined inside a rectangular box of sides \(a, b\), and \(c .\) The particle is described by a wavefunction \(\psi\) which satisfies the Schrödinger wave equation $$ -\frac{\hbar^{2}}{2 m} \nabla^{2} \psi=E \psi. $$ The wavefunction is required to vanish at each surface of the box (but not to be identically zero). This condition imposes constraints on the separation constants and therefore on the energy \(E .\) What is the smallest value of \(E\) for which such a solution can be obtained?

To a good approximation, the interaction of two nucleons may be described by a mesonic potential $$ V=\frac{A e^{-e x}}{x} $$ attractive for \(A\) negative. Develop a series solution of the resultant Schrödinger wave equation $$ \frac{\hbar^{2}}{2 m} \frac{d^{2} \psi}{d x^{2}}+(E-V) \psi=0 $$ Through the first three nonvanishing coefficients: $$ \psi_{k=1}=a_{0}\left\\{x+\frac{1}{2} A^{\prime} x^{2}+\frac{1}{6}\left[\frac{1}{2} A^{2}-E^{\prime}-a A^{\prime}\right] x^{3}+\cdots\right\\} $$ where the prime indicates multiplication by \(2 m / \hbar^{2}\).

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