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Magazine Chapter 2: Problem 1
In Problems 1-24 determine whether the given equation is exact. If it is exact, solve it. $$ (2 x-1) d x+(3 y+7) d y=0 $$
Short Answer
Expert verified
The equation is exact and the solution is \(x^2 - x + \frac{3}{2}y^2 + 7y = C\).
Step by step solution
01
Identify and Separate Functions M(x, y) and N(x, y)
Given the differential equation:\[(2x - 1) dx + (3y + 7) dy = 0\]We identify the functions:- \(M(x, y) = 2x - 1\)- \(N(x, y) = 3y + 7\)
02
Check Exactness Condition
A differential equation is exact if:\[\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\]Compute the partial derivatives:- \(\frac{\partial M}{\partial y} = 0\)- \(\frac{\partial N}{\partial x} = 0\)Since \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\), the equation is exact.
03
Determine the Potential Function \(\Psi(x, y)\)
Since the equation is exact, there exists a potential function \(\Psi\) such that:\[\frac{\partial \Psi}{\partial x} = M = 2x - 1\]Integrate \(M\) with respect to \(x\):\[\Psi(x, y) = \int (2x - 1) \, dx = x^2 - x + h(y)\]where \(h(y)\) is an arbitrary function of \(y\).
04
Determine \(h(y)\) by Using \(N(x, y)\)
We also know:\[\frac{\partial \Psi}{\partial y} = N = 3y + 7\]Substitute \(\Psi(x, y) = x^2 - x + h(y)\) into the partial derivative with respect to \(y\):\[\frac{\partial \Psi}{\partial y} = \frac{d}{dy}[x^2 - x + h(y)] = \frac{dh}{dy} = 3y + 7\]Integrate \( \frac{dh}{dy} \):\[h(y) = \int (3y + 7) \, dy = \frac{3}{2}y^2 + 7y + C\]where \(C\) is a constant.
05
Write the Potential Function and Solve
Substitute \(h(y)\) back into \(\Psi(x, y)\):\[\Psi(x, y) = x^2 - x + \frac{3}{2}y^2 + 7y + C\]The solution to the differential equation is:\[x^2 - x + \frac{3}{2}y^2 + 7y = C\]
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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 multivariable calculus. Imagine you have a function with several variables. A partial derivative helps determine how the function changes as only one variable changes, with all other variables held constant.
To calculate a partial derivative, you focus only on the variable you are differentiating with respect to and treat all other variables as constants.
To calculate a partial derivative, you focus only on the variable you are differentiating with respect to and treat all other variables as constants.
- For example, if you have a function \(f(x, y) = 2x - 1\), the partial derivative of \(f\) with respect to \(x\) is \(\frac{\partial f}{\partial x} = 2\) since \(y\) is treated as a constant.
- Similarly, \(\frac{\partial f}{\partial y}\) or any partial derivative with respect to another variable is calculated the same way.
- This is done in the original problem when identifying \(M(x,y)\) and \(N(x,y)\), and computing \(\frac{\partial M}{\partial y} = 0\) and \(\frac{\partial N}{\partial x} = 0\).
Integration
Integration is the process of finding a function given its derivative. It is essentially the reverse operation of differentiation, just like subtraction undoes addition.
In the context of exact differential equations, integration assists us in finding a potential function. After verifying that an equation is exact, you can integrate the function \(M(x, y)\) with respect to \(x\) to begin constructing the potential function.
In the context of exact differential equations, integration assists us in finding a potential function. After verifying that an equation is exact, you can integrate the function \(M(x, y)\) with respect to \(x\) to begin constructing the potential function.
- For the equation \((2x-1) dx + (3y+7) dy=0\), we integrate \(2x - 1\) with respect to \(x\), resulting in \(x^2 - x + h(y)\).
- The integration with respect to \(x\) treats \(y\) as a constant, introducing \(h(y)\) as an undetermined function that may depend on \(y\).
- Subsequently, we use \(N(x, y)\) to find \(h(y)\) by integrating \(3y + 7\) with respect to \(y\), obtaining \(\frac{3}{2}y^2 + 7y + C\).
Potential Function
A potential function is a scalar function from which a vector field can be derived. In the context of exact differential equations, if the equation is exact, it means that there exists a potential function \(\Psi(x, y)\) such that the gradient of \(\Psi\) is the vector field of the differential equation.
This function becomes the solution to the differential equation when derived correctly.
This function becomes the solution to the differential equation when derived correctly.
- The potential function \(\Psi\) in our example is constructed from the integral of \(M\) with respect to \(x\): \(\Psi(x, y) = x^2 - x + h(y)\).
- To ensure completeness, the component \(h(y)\) is determined by making the derivative of \(\Psi\) with respect to \(y\) equal to \(N(x, y)\).
- This leads to \(h(y) = \frac{3}{2}y^2 + 7y + C\), completing \(\Psi(x, y)\). As a result, \(x^2 - x + \frac{3}{2}y^2 + 7y = C\) gives us the solution contour of the differential equation.
Exactness Condition
The exactness condition is a crucial criterion for determining whether a differential equation can be solved using a potential function. For an equation in the form \( M(x, y) \, dx + N(x, y) \, dy = 0 \) to be exact, we must have:
\( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \).
If this condition is satisfied, it indicates the existence of a potential function \(\Psi(x, y)\) whose partial derivatives correspond to \(M\) and \(N\).
\( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \).
If this condition is satisfied, it indicates the existence of a potential function \(\Psi(x, y)\) whose partial derivatives correspond to \(M\) and \(N\).
- In our exercise, both partial derivatives are zeros, thus the exactness condition holds true.
- This implies that there isn鈥檛 any discrepancy in derivatives making it possible to assemble a potential function that consistently aligns with \(M\) and \(N\).
- Recognizing exactness hence simplifies our differential equations problem-solving approach.
