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Magazine 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 its general solution is \( x^2y^2 - 3x + 4y = C \).
Step by step solution
01
Identify Functions M(x, y) and N(x, y)
The differential equation is given in the form \( M(x, y) dx + N(x, y) dy = 0 \) where \( M(x, y) = 2xy^2 - 3 \) and \( N(x, y) = 2x^2y + 4 \).
02
Find Partial Derivative \( \frac{\partial M}{\partial y} \)
Calculate the partial derivative of \( M \) with respect to \( y \). \[ \frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(2xy^2 - 3) = 4xy \]
03
Find Partial Derivative \( \frac{\partial N}{\partial x} \)
Calculate the partial derivative of \( N \) with respect to \( x \). \[ \frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(2x^2y + 4) = 4xy \]
04
Check Exactness of the Differential Equation
The differential equation is exact if \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \). From the previous steps, we see that \( 4xy = 4xy \), so the differential equation is exact.
05
Integrate \( M \) with Respect to \( x \)
To find the solution, integrate \( M(x, y) = 2xy^2 - 3 \) with respect to \( x \). \[ \int (2xy^2 - 3) \, dx = x^2y^2 - 3x + h(y) \] where \( h(y) \) is an arbitrary function of \( y \).
06
Differentiate the Expression with Respect to \( y \)
Differentiate the result from Step 5 with respect to \( y \). \[ \frac{\partial}{\partial y}(x^2y^2 - 3x + h(y)) = 2x^2y + h'(y) \]
07
Equate and Solve for \( h'(y) \)
Set the derivative from Step 6 equal to \( N(x, y) \) from Step 1. \[ 2x^2y + h'(y) = 2x^2y + 4 \] Solving gives \( h'(y) = 4 \), so \( h(y) = 4y + C \) where \( C \) is a constant.
08
Write the General Solution
Combine the results to get the general solution of the differential equation. \[ x^2y^2 - 3x + 4y = C \] where \( C \) is an integration 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 tool used to examine functions of more than one variable. In the context of differential equations, they help us explore how a function changes in response to changes in one of its variables, keeping others constant. Here, we have a function expressed as a sum of two components: \( M(x, y) = 2xy^2 - 3 \) and \( N(x, y) = 2x^2y + 4 \). To assess whether our differential equation is exact, we need to calculate partial derivatives.
- Compute \( \frac{\partial M}{\partial y} \): This involves taking the derivative of \( M(x, y) \) with respect to \( y \) while treating \( x \) as constant. Simplified, it becomes \( 4xy \).
- Next, compute \( \frac{\partial N}{\partial x} \): Here, the derivative of \( N(x, y) \) with respect to \( x \) is calculated, treating \( y \) as constant. This also results in \( 4xy \).
Integration
Integration is the mathematical process of finding the integral of a function, which can be thought of as the inverse process of differentiation. In the context of exact differential equations, integration acts as a pivotal step towards determining the overall solution.
- We start by integrating \( M(x, y) = 2xy^2 - 3 \) with respect to \( x \). This means we treat \( y \) as a constant and focus on the variable \( x \), leading to \( \int (2xy^2 - 3) \, dx = x^2y^2 - 3x + h(y) \), where \( h(y) \) is an arbitrary function.
- Next, it requires integrating \( N(x, y) \) in a similar context, but due to our specific problem constraints, we further hone in on matching conditions with our solution from \( M(x, y) \).
General Solution
The general solution of a differential equation is a function that includes one or more arbitrary constants and provides a family of solutions to the equation. After determining exactness and conducting necessary integrations, we arrive at the general solution that satisfies the original differential expression.
- Once the integration of \( M(x, y) \) is performed, the expression contains \( x^2y^2 - 3x + h(y) \).
- By equating the derivative of this solution with \( N(x, y) \) and solving for \( h'(y) \), we find \( h(y) = 4y + C \), where \( C \) is an integration constant.
- The general solution thus becomes \( x^2y^2 - 3x + 4y = C \), encapsulating the complete range of solutions for various values of \( C \).
Exactness Condition
The exactness condition is crucial in confirming whether a differential equation can be directly integrated to find a solution. It hinges on the relationship between the partial derivatives of the functions involved.
- The differential equation is exact if \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \).
- For the given differential equation, after calculating \( \frac{\partial M}{\partial y} \) as \( 4xy \) and \( \frac{\partial N}{\partial x} \) also as \( 4xy \), we confirm exactness since both derivatives are equal.
- This equality implies that there exists a common solution function whose total differential corresponds with the original equation.
