Problem 38 - Advanced Engineering Mathematics

Question

We note that \(\left(M_{y}-N_{x}\right) / N=-3 /(1+x),\) so an integrating factor is \(e^{-3 \int d x /(1+x)}=1 /(1+x)^{3}\). Let \(M=\) \(\left(x^{2}+y^{2}-5\right) /(1+x)^{3}\) and \(N=-(y+x y) /(1+x)^{3}=-y /(1+x)^{2},\) so that \(M_{y}=2 y /(1+x)^{3}=N_{x} .\) From \(f_{y}=-y /(1+x)^{2}\) we obtain \(f=-\frac{1}{2} y^{2} /(1+x)^{2}+h(x), h^{\prime}(x)=\left(x^{2}-5\right) /(1+x)^{3},\) and \(h(x)=2 /(1+x)^{2}+\) \(2 /(1+x)+\ln |1+x| .\) A solution of the differential equation is $$-\frac{y^{2}}{2(1+x)^{2}}+\frac{2}{(1+x)^{2}}+\frac{2}{(1+x)}+\ln |1+x|=c$$

Step-by-Step Solution

Answer

The solution is \[-\frac{y^2}{2(1+x)^2} + \frac{2}{(1+x)^2} + \frac{2}{1+x} + \ln|1+x| = c.\]

Step 1: Identify Integrating Factor

The integrating factor is given as \(e^{-3 \int dx /(1+x)} = \frac{1}{(1+x)^3}\). We use this factor to simplify the differential equation.

Step 2: Determine M and N Functions

Given \(M = \frac{x^2 + y^2 - 5}{(1+x)^3}\) and \(N = -\frac{y}{(1+x)^2}\). These are the coefficients of the differential equation in the form \(Mdx + Ndy = 0\).

Step 3: Check Exactness

To verify exactness, compare \(M_y = \frac{2y}{(1+x)^3}\) with \(N_x = \frac{2y}{(1+x)^3}\). Since \(M_y = N_x\), the equation is exact.

Step 4: Solve for Function f

From \(f_y = -\frac{y}{(1+x)^2}\), integrate with respect to \(y\) to find \(f(x,y)\). This yields \(f(x,y) = -\frac{1}{2} \frac{y^2}{(1+x)^2} + h(x)\), where \(h(x)\) is a function of \(x\) alone.

Step 5: Derive h(x) Function

We have \(h'(x) = \frac{x^2 - 5}{(1+x)^3}\). Integrate this expression with respect to \(x\), yielding \(h(x) = \frac{2}{(1+x)^2} + \frac{2}{1+x} + \ln|1+x|\).

Step 6: Combine to Form General Solution

Combine \(f(x,y)\) and \(h(x)\) to form the general solution of the differential equation:\[-\frac{y^2}{2(1+x)^2} + \frac{2}{(1+x)^2} + \frac{2}{1+x} + \ln|1+x| = c\] where \(c\) is the integration constant.

Key Concepts

Integrating Factor

An integrating factor is a critical tool used to make certain types of differential equations easier to solve, particularly non-exact equations. It’s essentially a function that, when multiplied with the original differential equation, transforms it into an exact equation. In our problem, we used an integrating factor defined as \(e^{-3 \int dx /(1+x)} = \frac{1}{(1+x)^3}\). The process involves integrating the term \(-3 dx/(1+x)\), which simplifies to an exponential function that depends on \(x\).

The integrating factor's main purpose is to adjust the coefficients of the differential equation to meet the requirements of exactness. This adjustment allows us to treat the equation as though it were exact, facilitating a more straightforward method of finding a solution. Remember,

• An integrating factor makes calculations and transformations feasible,
• Is particularly helpful when the equation isn't inherently exact,
• Leads the way to solving for a potential function \(f(x,y)\) that satisfies the equation.

Exactness Condition

The exactness condition involves checking whether a given differential equation is exact. For an equation of the form \(Mdx + Ndy = 0\), we say it is exact if the mixed partial derivatives of any function derived from the coefficients \(M\) and \(N\) are equal. Specifically, this means:

• \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\)

This equality means that the differential of the function within the equation can be expressed as a total derivative \(dF\) where \(F(x,y)\) is some potential function.

In the exercise, we confirmed the condition of exactness by calculating:

• \(M_y = \frac{2y}{(1+x)^3}\)
• \(N_x = \frac{2y}{(1+x)^3}\)

Since \(M_y = N_x\), the equation is exact, which allows us to proceed with solving it by seeking a potential function \(f(x,y)\). This potential function enables us to integrate easily and find the solution to the differential equation.

Integration

Integration is a fundamental concept in solving differential equations, used to find the antiderivative of a function. In this context, integration serves multiple purposes. During the exercise, integration is centrally involved in two stages:

• First, integrating \(f_y\) with respect to \(y\) to find part of the potential function \(f(x,y)\). For instance, integrating \(f_y = -\frac{y}{(1+x)^2}\) yields \(f(x,y) = -\frac{1}{2} \frac{y^2}{(1+x)^2} + h(x)\).
• Second, integrating \(h'(x)\) with respect to \(x\) to find \(h(x)\), which is only a function of \(x\). In this exercise, \(h'(x) = \frac{x^2 - 5}{(1+x)^3}\), and integrating gives \(h(x) = \frac{2}{(1+x)^2} + \frac{2}{1+x} + \ln|1+x|\).

The objective is to locate an expression for \(f(x,y)\), and assembling the results from these integrations marks the path to finding the general solution. Integration allows us to backtrack from the derivatives (observed in the differential equation) to the original function, critical for solving these types of problems.

Ordinary Differential Equations

Ordinary Differential Equations (ODEs) involve derivatives of a function with respect to one independent variable. These are critical in mathematical modeling across various fields including physics, engineering, and economics. The focus on ODEs in this exercise showcases their role in describing how quantities change over time or space.

This particular equation falls into the category of first-order ODEs, which involve only the first derivative of the function. The goal is to find the unknown function, which satisfies the equation when substituted back.

To solve ODEs:

• Understand the equation type - here, being an exact differential equation was critical.
• Identify suitable methods for simplification, like using integrating factors.
• Apply specific integration techniques to find solutions.

ODEs form an essential component of understanding mathematical relationships describing dynamic systems, and learning to solve them expands your problem-solving toolkit. This exercise illustrates the process of resolving an ODE through understanding and applying the right scientific mathematics principles.