Problem 7
Question
Determine whether the given differential equation is exact. If it is exact, solve it. $$ \left(x^{2}-y^{2}\right) d x+\left(x^{2}-2 x y\right) d y=0 $$
Step-by-Step Solution
Verified Answer
The differential equation is not exact.
1Step 1: Identify M(x,y) and N(x,y)
The given differential equation is \((x^{2}-y^{2}) \, dx + (x^{2}-2xy) \, dy=0.\) Identifying the functions, we have:- \( M(x, y) = x^2 - y^2 \)- \( N(x, y) = x^2 - 2xy \)
2Step 2: Check for Exactness
For the equation to be exact, the partial derivatives \( \frac{\partial M}{\partial y} \) and \( \frac{\partial N}{\partial x} \) must be equal.Calculate \( \frac{\partial M}{\partial y} = -2y \) and\( \frac{\partial N}{\partial x} = 2x \). Since \( \frac{\partial M}{\partial y} eq \frac{\partial N}{\partial x} \) (\(-2y eq 2x\)), the equation is not exact.
Key Concepts
Partial DerivativesDifferential EquationsExactness Condition
Partial Derivatives
Partial derivatives are a fundamental concept in calculus used to analyze how functions change when several variables are involved. In the context of functions with two variables, such as \( f(x, y) \), partial derivatives measure the rate of change of the function with respect to one variable while keeping the other variable constant. This can be visualized as taking a slice of the function's surface along a particular direction and observing how the slice changes.
To compute a partial derivative, we differentiate with respect to the variable of interest, while treating all other variables as constants. For instance, for a function \( M(x, y) = x^2 - y^2 \), the partial derivative with respect to \( y \), denoted as \( \frac{\partial M}{\partial y} \), is calculated by treating \( x \) as a constant, resulting in \( -2y \). Similarly, for a function \( N(x, y) = x^2 - 2xy \), the partial derivative with respect to \( x \), denoted as \( \frac{\partial N}{\partial x} \), results in \( 2x - 2y \).
Understanding partial derivatives is crucial because they are used to determine if a differential equation is exact, which leads us to our next topic.
To compute a partial derivative, we differentiate with respect to the variable of interest, while treating all other variables as constants. For instance, for a function \( M(x, y) = x^2 - y^2 \), the partial derivative with respect to \( y \), denoted as \( \frac{\partial M}{\partial y} \), is calculated by treating \( x \) as a constant, resulting in \( -2y \). Similarly, for a function \( N(x, y) = x^2 - 2xy \), the partial derivative with respect to \( x \), denoted as \( \frac{\partial N}{\partial x} \), results in \( 2x - 2y \).
Understanding partial derivatives is crucial because they are used to determine if a differential equation is exact, which leads us to our next topic.
Differential Equations
Differential equations involve equations that relate a function with its derivatives. These are ubiquitous in mathematical modeling of physical systems, ranging from simple applications like population growth to complex ones such as climate modeling. They describe how a certain quantity changes over time or space and are central to understanding dynamic systems. In simpler terms, differential equations help us understand how things evolve.
A differential equation can be expressed in terms of differentials \( dx \) and \( dy \), which represent infinitesimally small changes in \( x \) and \( y \) respectively. When dealing with a two-variable system, as in our original exercise, we encounter equations of the type \( M(x, y)\, dx + N(x, y)\, dy = 0 \). Here, \( M(x, y) \) and \( N(x, y) \) are functions in terms of \( x \) and \( y \). The solution of a differential equation is a function (or a family of functions) that satisfies the relationship described by the equation.
Understanding differential equations requires knowledge of their exactness, which leads us to the next crucial concept.
A differential equation can be expressed in terms of differentials \( dx \) and \( dy \), which represent infinitesimally small changes in \( x \) and \( y \) respectively. When dealing with a two-variable system, as in our original exercise, we encounter equations of the type \( M(x, y)\, dx + N(x, y)\, dy = 0 \). Here, \( M(x, y) \) and \( N(x, y) \) are functions in terms of \( x \) and \( y \). The solution of a differential equation is a function (or a family of functions) that satisfies the relationship described by the equation.
Understanding differential equations requires knowledge of their exactness, which leads us to the next crucial concept.
Exactness Condition
A differential equation in two variables is termed *exact* if it can be expressed in the form \( M(x, y)\, dx + N(x, y)\, dy = 0 \) and satisfies a specific condition: the partial derivative of \( M \) with respect to \( y \) is equal to the partial derivative of \( N \) with respect to \( x \).
Mathematically, the condition for exactness is \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \). This condition guarantees that the differential equation is the total differential of some function \( \psi(x, y) \), which means there exists a function such that \( d\psi = M(x, y)\, dx + N(x, y)\, dy \).
In our exercise, the equation \((x^2-y^2)\, dx + (x^2-2xy)\, dy=0\) is not exact because \( \frac{\partial M}{\partial y} \) results in \(-2y\), while \( \frac{\partial N}{\partial x} \) results in \( 2x - 2y \). Since these partial derivatives are not equal (except for the specific case when \( y = x \)), the equation does not meet the exactness condition and thus, cannot be solved as an exact equation.
When an equation is not exact, alternative methods, such as finding an integrating factor or using approximation techniques, may be needed to find a solution.
Mathematically, the condition for exactness is \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \). This condition guarantees that the differential equation is the total differential of some function \( \psi(x, y) \), which means there exists a function such that \( d\psi = M(x, y)\, dx + N(x, y)\, dy \).
In our exercise, the equation \((x^2-y^2)\, dx + (x^2-2xy)\, dy=0\) is not exact because \( \frac{\partial M}{\partial y} \) results in \(-2y\), while \( \frac{\partial N}{\partial x} \) results in \( 2x - 2y \). Since these partial derivatives are not equal (except for the specific case when \( y = x \)), the equation does not meet the exactness condition and thus, cannot be solved as an exact equation.
When an equation is not exact, alternative methods, such as finding an integrating factor or using approximation techniques, may be needed to find a solution.
Other exercises in this chapter
Problem 7
Find the general solution of the given differential equation. Give the largest interval over which the general solution is defined. Determine whether there are
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\(\frac{d y}{d x}=e^{3 x+2 y}\)
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In Problems \(1-10\), solve the given differential equation by using an appropriate substitution. $$ \frac{d y}{d x}=\frac{y-x}{y+x} $$
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