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Chapter 2: Problem 5

Determine whether the given differential equation is exact. If it is exact, solve it. $$ \left(2 x y^{2}-3\right) d x+\left(2 x^{2} y+4\right) d y=0 $$

Short Answer

Expert verified
The differential equation is exact, and the solution is \( x^2y^2 - 3x + 4y = C \).

Step by step solution

01

Identify the Functions M(x, y) and N(x, y)

Given a differential equation of the form \( M(x, y)\, dx + N(x, y)\, dy = 0 \), we identify \( M(x, y) = 2xy^2 - 3 \) and \( N(x, y) = 2x^2y + 4 \).
02

Compute Partial Derivatives

To check for exactness, we compute the partial derivatives \( \frac{\partial M}{\partial y} \) and \( \frac{\partial N}{\partial x} \). Compute \( \frac{\partial M}{\partial y} = \frac{\partial}{\partial y} (2xy^2 - 3) = 4xy \).Compute \( \frac{\partial N}{\partial x} = \frac{\partial}{\partial x} (2x^2y + 4) = 4xy \).
03

Check If the Equation is Exact

For the differential equation to be exact, the condition \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \) must hold.Here, both partial derivatives are equal to \( 4xy \), thus the equation is exact.
04

Solve the Exact Differential Equation

Since the equation is exact, there exists a function \( \Phi(x, y) \) such that:- \( \frac{\partial \Phi}{\partial x} = M(x, y) = 2xy^2 - 3 \)- \( \frac{\partial \Phi}{\partial y} = N(x, y) = 2x^2y + 4 \)Integrate \( M(x, y) \) with respect to \( x \):\[ \Phi(x, y) = \int (2xy^2 - 3) \, dx = x^2y^2 - 3x + g(y) \]Differentiate this result with respect to \( y \):\[ \frac{\partial}{\partial y}(x^2y^2 - 3x + g(y)) = 2x^2y + \frac{dg}{dy} \]Set this equal to \( N(x, y) \):\[ 2x^2y + \frac{dg}{dy} = 2x^2y + 4 \]Thus, \( \frac{dg}{dy} = 4 \) implying \( g(y) = 4y + C \).
05

Write Out the Solution

Combine components to write the complete solution:\[ \Phi(x, y) = x^2y^2 - 3x + 4y + C \]Thus, the solution to the differential equation is:\[ x^2y^2 - 3x + 4y = C \] where \( C \) is an arbitrary constant.

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Key Concepts

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

Partial Derivatives
Partial derivatives are a fundamental concept in calculus and are crucial for understanding exact differential equations. In mathematics, when dealing with functions of several variables, a partial derivative measures how a function changes as one of its variables changes, while keeping the other variables constant.
  • For a function \( z = f(x, y) \), the partial derivative with respect to \( x \), denoted as \( \frac{\partial f}{\partial x} \), captures the rate of change of \( z \) as \( x \) varies, holding \( y \) constant.
  • Similarly, \( \frac{\partial f}{\partial y} \) shows how \( z \) changes with \( y \) while \( x \) is fixed.
Partial derivatives are essential for checking the exactness condition in differential equations, where comparing partial derivatives helps determine if an equation is exact.
Integration
Integration is the process of finding a function given its derivative. It is the inverse operation of differentiation and helps to construct solutions to differential equations. In exact differential equations, integration is used to find the potential function \( \Phi(x, y) \) that satisfies the given differential equation.
  • To integrate a function with respect to \( x \), you sum up all the tiny changes in the function as \( x \) changes, which model the area under the curve of the function.
  • When solving exact differential equations, one typically performs integration on \( M(x, y) \) with respect to \( x \), followed by ensuring consistency using \( N(x, y) \) and integration with respect to \( y \), if needed.
This step is essential as it allows us to find the potential function that reveals the solution to the differential equation.
Differential Equations
Differential equations are equations that involve the derivatives of a function. They play a significant role in describing various physical phenomena through mathematical models. Understanding differential equations involves learning how they are structured and how to solve them.
  • An example of a differential equation is \( M(x, y) \, dx + N(x, y) \, dy = 0 \), which may represent a scenario in physics, chemistry, or engineering.
  • The goal often is to find the functions that make this equation true; these functions describe the behavior of the system modeled by the equation.
Finding solutions to differential equations, like exact differential equations, requires a strategy of checking for exactness and possibly integrating to find the system's underlying potential function.
Exactness Condition
The exactness condition is a criterion used to determine if a differential equation can be classified as exact. This is crucial as exact differential equations have a special structure that allows them to be solved more directly.
  • For an equation \( M(x, y) \, dx + N(x, y) \, dy = 0 \) to be exact, it's required that the condition \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \) holds.
  • This means that the mixed partial derivatives are equal, indicating that the differential equation derives from a single potential function \( \Phi(x, y) \).
When the exactness condition is satisfied, it suggests the presence of a function whose derivatives correspond to the components \( M(x, y) \) and \( N(x, y) \), leading us towards the function鈥檚 solution.

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Most popular questions from this chapter

Use a numerical solver and Euler's method to obtain a four-decimal approximation of the indicated value. First use \(h=0.1\) and then use \(h=0.05\). \(-y\) $$ y^{\prime}=x^{2}+y^{2}, \quad y(0)=1 ; y(0.5) $$

Solve the given initial-value problem. Give the largest interval \(I\) over which the solution is defined. $$ x y^{\prime}+y=e^{x}, \quad y(1)=2 $$

Two chemicals \(A\) and \(B\) are combined to form a chemical \(C\). The rate, or velocity, of the reaction is proportional to the product of the instantaneous amounts of \(A\) and \(B\) not converted to chemical \(C\). Initially there are 40 grams of \(A\) and 50 grams of \(B\), and for each gram of \(B\), 2 grams of \(A\) is used. It is observed that 10 grams of \(C\) is formed in 5 minutes. How much is formed in 20 minutes? What is the limiting amount of \(C\) after a long time? How much of chemicals \(A\) and \(B\) remains after a long time?

A differential equation governing the velocity \(v\) of a falling mass \(m\) subjected to air resistance proportional to the square of the instantaneous velocity is $$ m \frac{d v}{d t}=m g-k v^{2}, $$ where \(k>0\) is the drag coefficient. The positive direction is downward. (a) Solve this equation subject to the initial condition \(v(0)=v_{0}\) (b) Use the solution in part (a) to determine the limiting, or terminal, velocity of the mass. We saw how to determine the terminal velocity without solving the \(\mathrm{DE}\) in Problem 39 in Exercises \(2.1\). (c) If distance \(s\), measured from the point where the mass was released above ground, is related to velocity \(v\) by \(d s / d t=v(t)\), find an explicit expression for \(s(t)\) if \(s(0)=0\).

Use Euler's method to obtain a four-decimal approximation of the indicated value. Carry out the recursion of (3) by hand, first using \(h=0.1\) and then using \(h=0.05\). $$ y^{\prime}=2 x-3 y+1, \quad y(1)=5 ; y(1.2) $$
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