Question

Determine whether or not each of the equations is exact. If it is exact, find the solution. $$ \left(3 x^{2}-2 x y+2\right) d x+\left(6 y^{2}-x^{2}+3\right) d y=0 $$

Short answer

Question: Determine if the given differential equation is exact, and if so, find the implicit solution to the equation: $$ (3x^2 - 2xy + 2)dx + (6y^2 - x^2 + 3)dy = 0 $$ Answer: The given differential equation is exact, and the implicit solution to the equation is: $$ x^3 - x^2y + 2x + 2y^3 + 3y = C $$

Step-by-step solution

Verify if the equation is exact

To check if the given differential equation is exact, we need to verify if the mixed partial derivatives of the potential function (M and N) are equal. That is, we need to check if: \(M_{y} = N_{x}\) where \(M = 3x^2 - 2xy + 2, N = 6y^2 - x^2 + 3\) Proceed to find \(M_{y}\) and \(N_{x}\):

Calculate M_y and N_x

Calculate the partial derivatives: \(M_{y} = \frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(3x^2 - 2xy + 2) = -2x\) \(N_{x} = \frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(6y^2 - x^2 + 3) = -2x\)

Check if M_y = N_x

Compare the two partial derivatives: Since \(M_{y} = -2x\) and \(N_{x} = -2x\), we can conclude that the given differential equation is exact.

Find the potential function (ψ)

To find the potential function, integrate M with respect to x, and N with respect to y: \(ψ_{x} = M\) so, \(\int (3x^2 - 2xy + 2) dx = x^3 - x^2y + 2x + h(y)\) \(ψ_{y} = N\) so, \(\int (6y^2 - x^2 + 3) dy = 2y^3 - x^2y + 3y + g(x)\)

Match the cross terms and find ψ

Now, we have to compare both expressions for ψ and eliminate any arbitrary functions h(y) and g(x) by matching the cross terms: From \(\int M dx\) and \(\int N dy\), we have: \(ψ(x, y) = x^3 - x^2y + 2x + 2y^3 + 3y + C\) Where C is a constant.

Write the solution as an equation

Finally, the solution to the given exact differential equation can be found by setting ψ(x, y) equal to a constant: $$x^3 - x^2y + 2x + 2y^3 + 3y = C$$ This equation represents the implicit solution to the given exact differential equation.

Key concepts

Partial Derivatives

Partial derivatives play a crucial role in understanding and solving exact differential equations. They are used to determine how a function changes as each variable changes while keeping the other variables constant. When given an equation like \( M dx + N dy = 0 \), where \( M \) and \( N \) are functions of \( x \) and \( y \), respectively, one must calculate the partial derivative of \( M \) with respect to \( y \) and \( N \) with respect to \( x \).

An exact differential equation signifies that there exists a potential function \( \psi \) where \( M \) is its partial derivative with respect to \( x \) and \( N \) with respect to \( y \)—symbolically, \( M = \psi_x \) and \( N = \psi_y \). If \( M_y = N_x \) after taking these partial derivatives, the equation is exact, indicating the potential function's existence. For students, mastering partial derivative calculations is an essential step toward solving these types of problems effectively.

Potential Function

The notion of a potential function is at the heart of exact differential equations. It serves as a unifying factor for the functions \( M \) and \( N \) in exact equations of the form \( M dx + N dy = 0 \). The potential function, denoted as \( \psi \) here, is essentially the function whose partial derivatives generate \( M \) and \( N \)—in mathematical terms, \( \psi_x = M \) and \( \psi_y = N \).

To find the potential function for an exact equation, one integrates \( M \) with respect to \( x \) and \( N \) with respect to \( y \) independently. It's necessary to match cross terms and to consider arbitrary functions which are functions solely in terms of \( y \) and \( x \) respectively to ensure the potential function aligns with both \( M \) and \( N \) correctly. This result is paramount, as the potential function encapsulates the solution to the differential equation in an implicit form.

Implicit Solutions

In the realm of differential equations, implicit solutions are those that define the dependent variable in terms of the independent variables not by a direct formula, but through an equation that relates them. For exact differential equations, the solution is typically presented implicitly as \(\psi(x, y) = C\), where \( C \) is a constant, and \( \psi \) is the potential function discovered from the integration process.

Implicit solutions require no further algebraic manipulation to solve for a dependent variable explicitly. They're extremely useful because they may often express complex relationships more succinctly than explicit equations. Furthermore, implicit solutions can describe relationships between variables that may not be easily expressible in an explicit form. Training in recognizing and working with implicit solutions allows students to tackle a wider array of problems in calculus and applied mathematics disciplines.

Boundary Value Problems

Boundary value problems (BVPs) are a category of differential equations where the solution must satisfy certain fixed conditions, referred to as boundary conditions, at the endpoints of the domain. In the context of an exact differential equation, the solution \( \psi(x, y) = C \) is often part of a larger BVP where values of \( x \) and \( y \) may be specified along the boundary of the region of interest.

BVPs are ubiquitous in physics and engineering because they model phenomena where conditions are known at the boundaries rather than at an initial point, which is the case for initial value problems (IVPs). The correct implementation of boundary conditions is crucial for obtaining a meaningful solution that corresponds to the physical situation being modeled. Mastering BVPs provides students with the skills needed to solve practical problems they will encounter in their academic and professional careers.