Question

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

Short answer

Question: Determine if the given equation is exact, and if so, find the solution. Given Equation: $$ \left(2 x y^{2}+2 y\right)+\left(2 x^{2} y+2 x\right) y^{\prime}=0 $$ Answer: The given equation is exact, and its general solution is: $$ x^2y^2 + 2xy = C $$

Step-by-step solution

Calculate partial derivatives

Calculate \(\frac{\partial M}{\partial y}\) and \(\frac{\partial N}{\partial x}\) for the given equation: $$ M(x, y) = 2xy^2 + 2y,\quad N(x, y) = 2x^2y + 2x $$ Calculate \(\frac{\partial M}{\partial y}\): $$ \frac{\partial}{\partial y}(2xy^2 + 2y) = 4xy + 2 $$ Calculate \(\frac{\partial N}{\partial x}\): $$ \frac{\partial}{\partial x}(2x^2y + 2x) = 4xy + 2 $$

Determine if the equation is exact

As per the condition, if \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\), then the equation is exact. We can see that: $$ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} = 4xy + 2 $$ Since the condition is satisfied, the given equation is exact.

Find the solution

To find the solution, we will integrate \(M(x, y)\) with respect to \(x\) and \(N(x, y)\) with respect to \(y\). Then, we will add a constant function for both integrations and compare them. $$ \int M dx = \int (2xy^2 + 2y) dx = x^2y^2 + 2xy + h(y) $$ $$ \int N dy = \int (2x^2y + 2x) dy = x^2y^2 + 2xy + g(x) $$ By comparing the two integrals, we can see that \(h(y)\) and \(g(x)\) are constants. Thus, the general solution of the given exact equation is: $$ x^2y^2 + 2xy = C $$

Key concepts

Partial Derivatives

In the context of exact differential equations, partial derivatives play a crucial role. They involve the differentiation of a function with respect to one variable while holding the other variables constant. For example, for the function \(M(x, y) = 2xy^2 + 2y\), finding the partial derivative with respect to \(y\) involves treating \(x\) as a constant. This gives \(\frac{\partial M}{\partial y} = 4xy + 2\). Similarly, for \(N(x, y) = 2x^2y + 2x\), the partial derivative with respect to \(x\) is \(\frac{\partial N}{\partial x} = 4xy + 2\).

Understanding these derivatives helps us to determine the exactness of a differential equation. If the condition \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\) is satisfied, the equation is exact. In simple terms, matching partial derivatives indicates that the equation can potentially represent a gradient field, meaning an exact solution exists. This determination lays the foundation for solving the equation effectively.

Partial derivatives are a stepping stone for exploring multivariable calculus, offering insight into how functions change around a certain point and how these changes relate to solving complex equations.

Integration of Functions

Once we establish that a differential equation is exact, the next step involves integrating the functions. This means finding the antiderivative or the original function whose derivative would give us the function we're integrating. In our problem, we integrate \(M(x, y)\) with respect to \(x\) and \(N(x, y)\) with respect to \(y\).

Integrating \(M(x, y) = 2xy^2 + 2y\) with respect to \(x\) gives us \(\int M \, dx = x^2y^2 + 2xy + h(y)\). Similarly, integrating \(N(x, y) = 2x^2y + 2x\) with respect to \(y\) results in \(\int N \, dy = x^2y^2 + 2xy + g(x)\). These expressions involve adding arbitrary functions \(h(y)\) or \(g(x)\), representing constants that might come from integration.

Integrals provide a "smoothing out" process, where instead of focusing on the rate of change (as derivatives do), we look at cumulative effects, or the total amounts. Understanding integration helps us construct solutions for exact differential equations, where we equate integrals and adjust these constant terms to find a consistent representation of the solution.

Differential Equation Solutions

Solving exact differential equations relies on both partial derivatives and integration to piece together a coherent solution. After confirming the exactness of an equation, the ultimate goal is to find a solution that satisfies this specific arrangement. Generally, the solution is a relationship involving the variables, expressed as a constant value.

In our example, integration results in the expressions \(x^2y^2 + 2xy + h(y)\) and \(x^2y^2 + 2xy + g(x)\). By comparing these, we realize \(h(y)\) and \(g(x)\) must be constants for the expressions to match. Thus, our solution simplifies to \(x^2y^2 + 2xy = C\), where \(C\) is a constant that might depend on initial conditions or constraints imposed on the problem.

This final equation forms a family of curves where each curve satisfies the original differential equation. Understanding how to derive this through checking exactness with partial derivatives and resolving through integration forms the basis for solving more complex systems involving differential equations. This process underscores the interconnectedness of calculus concepts, demonstrating their practical utility in addressing real-world problems.