Question

In Exercises \(1-17\) determine which equations are exact and solve them. $$ \left(\frac{1}{x}+2 x\right) d x+\left(\frac{1}{y}+2 y\right) d y=0 $$

Short answer

Question: Determine if the given equation is exact, and if it is, find its solution: $$ \left(\frac{1}{x}+2 x\right) d x+\left(\frac{1}{y}+2 y\right) d y=0 $$ Answer: The given equation is exact. The solution is: $$ \ln\left(\frac{x}{y}\right) + x^2 - y^2 = c $$

Step-by-step solution

Check if the given equation is exact

To check the exactness of the given equation, we need to find the partial derivative of its coefficients. Let \(M(x, y) = \frac{1}{x} + 2x\) and \(N(x, y) = \frac{1}{y} + 2y\). Now, we need to find the partial derivatives \(\frac{\partial M}{\partial y}\) and \(\frac{\partial N}{\partial x}\) to check if they are equal. Computing the partial derivatives: $$ \frac{\partial M}{\partial y} = \frac{\partial}{\partial y} \left(\frac{1}{x} + 2x\right) = 0, $$ and $$ \frac{\partial N}{\partial x} = \frac{\partial}{\partial x} \left(\frac{1}{y} + 2y\right) = 0. $$ Since \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\), the given equation is exact.

Integrate M and N to obtain the solution

As the given equation is exact, we will now integrate both M and N to obtain the solution. We start by integrating M with respect to x and N with respect to y. Integrating \(M(x,y)\) with respect to x: $$ \int M(x, y) dx = \int \left(\frac{1}{x} + 2x\right) dx = \ln|x| + x^2 + f(y) $$ Here, \(f(y)\) is an arbitrary function of y since we integrated with respect to x. Integrating \(N(x,y)\) with respect to y: $$ \int N(x, y) dy = \int \left(\frac{1}{y} + 2y\right) dy = \ln|y| + y^2 + g(x) $$ Here, \(g(x)\) is an arbitrary function of x since we integrated with respect to y.

Find the solution from the integrals

Now, we must combine the two integrals we got from step 2. We have: $$ \ln|x| + x^2 + f(y) = \ln|y| + y^2 + g(x) $$ As there is no common term to find \(f(y)\) and \(g(x)\), we can rewrite the equation as: $$ \ln|x| - \ln|y| + x^2 - y^2 = g(x) - f(y) $$ Let \(c = g(x) - f(y)\), which is a constant. Therefore, the solution is: $$ \ln|x| - \ln|y| + x^2 - y^2 = c \implies \ln\left(\frac{x}{y}\right) + x^2 - y^2 = c $$

Key concepts

Partial Derivatives

Partial derivatives are a fundamental concept in differential equations used to explore the way a multi-variable function changes with respect to one of those variables. When you have a function like \( M(x, y) = \frac{1}{x} + 2x \) and \( N(x, y) = \frac{1}{y} + 2y \), partial derivatives allow us to focus on one variable at a time.

• To determine whether an equation is exact, we first need to calculate \( \frac{\partial M}{\partial y} \) and \( \frac{\partial N}{\partial x} \).
• For example, when we compute \( \frac{\partial M}{\partial y} \), we treat \( x \) as a constant, which results in 0 because there is no \( y \) term in \( M(x, y) \).
• Similarly, \( \frac{\partial N}{\partial x} \) is computed by treating \( y \) as constant, also giving 0.

By comparing these derivatives, we can check if the condition of exactness holds, which simplifies solving the equation.

Integration Techniques

Integration techniques play a crucial role when solving exact differential equations. Once exactness is established, we need to integrate the partial derivatives back to their original potential function.

• First, integrate \( M(x, y) \) with respect to \( x \), treating \( y \) as a constant. This usually results in a general expression like \( \int (\frac{1}{x} + 2x) \, dx = \ln|x| + x^2 + f(y) \), where \( f(y) \) is an arbitrary function of \( y \).
• Similarly, integrate \( N(x, y) \) with respect to \( y \), yielding \( \int (\frac{1}{y} + 2y) \, dy = \ln|y| + y^2 + g(x) \), where \( g(x) \) is arbitrary regarding \( x \).

These steps consolidate our solutions by compensating for the functions lost during differentiation, ultimately leading to integrating each part of the equation accurately to retrieve the potential function.

Exactness Condition

The exactness condition is the key criterion for verifying if a differential equation can be categorized as exact. An exact differential equation implies that it comes from a single potential function whose total derivative vanishes.

• The condition for exactness is that the mixed partial derivatives \( \frac{\partial M}{\partial y} \) and \( \frac{\partial N}{\partial x} \) must be equal across continuous functions. In this exercise, both are zero, fulfilling the condition.
• When an equation is exact, it signifies a hidden relationship, hinting at a conserved quantity or an underlying potential function \( \Psi(x, y) \) that is rearranged by the differential terms \( M(x, y) \) and \( N(x, y) \).

By confirming the exactness, differential equations can be simplified significantly, easing the path to finding explicit solutions.

Solving Differential Equations

Solving differential equations involves piecing together all the concepts learned to find a solution that fulfills the given conditions. For exact equations, once exactness is determined and integrals are obtained, the next step is to form a coherent solution.

• You combine the results of the partial integrations performed on \( M(x, y) \) and \( N(x, y) \) to construct the original function \( \Psi(x, y) \).
• The expression \( \ln|x| + x^2 + f(y) = \ln|y| + y^2 + g(x) \) gives us the path to equate both sides by determining the constants and analyzing any shared explicit terms or exhibiting a constant \( c \).
• Ultimately, the solution takes a form like \( \ln\left(\frac{x}{y}\right) + x^2 - y^2 = c \), unraveling the abstract equation into a quantifiable relationship between \( x \) and \( y \).

This solution embodies the final answer and serves as a major accomplishment in solving such differential equations.