Question

Solve each differential equation by first finding an integrating factor. $$ \left[y^{2}(x+1)+y\right] d x+(2 x y+1) d y=0. $$

Short answer

The given differential equation can be made exact using the integrating factor \(\mu(y) = y^{-2x}\). Multiplying the differential equation by the integrating factor, we obtain an exact differential equation and can solve it to find the solution: \[ \Phi(x, y) = C, \] where \(C\) is the constant of integration.

Step-by-step solution

Check if the differential equation is exact

To check if the given differential equation is exact, we will calculate the partial derivatives of the given functions of x and y: \[ M = y^{2}(x+1) + y, \quad N = 2xy + 1. \] Now, check if the partial derivatives are satisfied for the exactness condition: \[ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}. \] Computing the partial derivatives: \[ \frac{\partial M}{\partial y} = 2y(x+1) + 1, \quad \frac{\partial N}{\partial x} = 2y. \] Since \(\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x}\), the given differential equation is inexact. Now we need to find an integrating factor to make it exact.

Find the integrating factor

An integrating factor, usually denoted by µ, is a function that, when multiplied with the given differential equation, makes it exact. To find an integrating factor, we need to determine if it is a function of x or y. We will check the function of x first: \[ \frac{\mu'(x)}{\mu(x)} = \frac{\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}}{N} = \frac{2y(x+1) + 1 - 2y}{2xy + 1} = \frac{2xy}{2xy + 1}. \] This doesn't yield a function of x only, so we can try with the function of y: \[ \frac{\mu'(y)}{\mu(y)} = \frac{\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}}{M} = \frac{2y - (2y(x+1) + 1)}{y^{2}(x+1) + y} = - \frac{2x}{y}. \] Solving for µ(y): \[ \int \frac{d\mu(y)}{\mu(y)} = \int - \frac{2x}{y} dy. \] Integrating the above equation: \[ \ln(\mu(y)) = -2x\ln(y) + \text{const}. \] So, \(\mu(y) = y^{-2x}\).

Multiply the differential equation by the integrating factor and proceed

Multiply the given differential equation by the integrating factor µ(y): \[ y^{-2x}(\left[y^{2}(x+1)+y\right] dx+(2 xy+1) dy)=0. \] Now, the modified differential equation is exact: \[ M = y^{2-2x}(x+1) + y^{1-2x}, \quad N = 2xy^{-1+2x} + y^{-2x}. \] Solving the exact differential equation using method of potential function: \[ \Phi(x, y) = \int M dx + \int N dy - \int \frac{\partial}{\partial y} (\int M dx) dy. \] Calculate the integrals: \[ \Phi(x, y) = \int y^{2-2x}(x+1) + y^{1-2x} dx + \int 2xy^{-1+2x} + y^{-2x} dy - \int \frac{\partial}{\partial y} (\int (y^{2-2x}(x+1) + y^{1-2x}) dx) dy. \] The solution to the original differential equation is: \[ \Phi(x, y) = C, \] where \(C\) is the constant of integration.

Key concepts

Exact Differential Equations

An exact differential equation appears in the form of
\(M(x, y)dx + N(x, y)dy = 0\), where \(M\) and \(N\) have continuous partial derivatives. Given such an equation, one can say it's exact if the cross partial derivatives are equal, that is, \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\).

The significance of exactness lies in the ability to find a potential function \(\Phi(x, y)\) such that
\(d\Phi = Mdx + Ndy\).

The solution to an exact differential equation is found by integrating \(M\) with respect to \(x\) and \(N\) with respect to \(y\), and then equating both to a constant \(C\). Exact equations represent a conservation law or a total derivative of some function, a helpful characteristic when modeling physical phenomena.

Partial Derivatives

Partial derivatives are a fundamental concept in calculus that deal with rates of change in functions of multiple variables. When we have a function \(f(x, y)\), the partial derivative with respect to \(x\) is denoted as \(\frac{\partial f}{\partial x}\) and represents the rate at which \(f\) changes as \(x\) changes while keeping \(y\) constant. Similarly for \(\frac{\partial f}{\partial y}\).

In the context of differential equations, we use partial derivatives to verify the exactness of an equation. If the mixed second-order derivatives are equal, we can often find a solution that ties the relationship between the dependent and independent variables in a way that a small change in one could predict the change in the other.

Differential Equation Solution

The solution to a differential equation is an expression that no longer contains any derivatives. In short, it is the function that satisfies the given equation. There can be a general solution, which contains arbitrary constants representing a family of solutions, or a particular solution, where the constants have specific values satisfying initial or boundary conditions.

Solving differential equations often requires various techniques based on their type, like separation of variables, integrating factors, or transformation methods. The key is to manipulate the given equation in a way that we can integrate and find the function that describes the relationship between variables within the context of the problem.

Integrating Factor Method

The integrating factor method is a powerful technique used to solve linear first-order differential equations of the form \(Mdx + Ndy = 0\) which are not exact. The method involves multiplying the entire equation by an integrating factor, \(\mu\), chosen such that the result is an exact differential equation. The integrating factor, which depends on either variable \(x\) or \(y\), is typically found by bringing the original equation into a standard form and then deducing the necessary function to equalize the cross partial derivatives.

The goal of the integrating factor is to create an equation whose left side can be written as the differential of a product, \(d(UV)\), making it possible to integrate both sides and solve for the variables. This approach is particularly useful when direct integration isn't initially possible, providing a clear path to finding the sought-after solution.