Solving Exact Differential Equations: A Comprehensive Guide

Exact differential equations are a significant part of mathematical analysis. They are encountered in various fields, such as physics, engineering, and economics. An exact differential equation is one that can be solved by finding an integrating factor or by utilizing the method of partial derivatives.

Introduction to Exact Differential Equations

An exact differential equation is of the form M(x, y)dx N(x, y)dy 0, where there exists a function f(x, y) such that f_x(x, y) M(x, y) and f_y(x, y) N(x, y). This equation can be solved by finding the potential function f(x, y) directly from the given differentials.

Case Study: Solving the Given Problem

Given Equations

Consider the differential equation where:

N_{xy} x - 2e^y M_{xy} yxsinx

The equation dM/dy 1 dN/dx confirms that the given differential equation is exact. This means there exists a function f(x, y) such that:

f_{xy} N(x, y) and f_{yx} M(x, y).

Step-by-Step Solution

To find the potential function f(x, y), we can follow these steps:

Integrate N(x, y) dy xy - 2e^y g(x) Find g'(x) xsinx Integrate g'(x) by parts to get g(x) sinx - xcosx Combine the results to get f(x,y) xy - 2e^y sinx - xcosx c

This function f(x, y) is the potential function that satisfies the given differential equation, and its level curves will represent the solution to the equation.

General Method for Solving Exact Equations

Method Summary

To solve an exact differential equation, follow these steps:

Check if the equation is exact by ensuring dM/dy dN/dx. Integrate N(x, y) dy g(x) to find f(x, y). Differentiate with respect to x to find g(x). Integrate g'(x) to find g(x). Combine the results to get the potential function f(x, y) c.

Hints and Tips

Hints for Solving Exact Equations

xdy - ydx d(xy) xsinxdx d(sinx - xcosx) The process of solving exact equations can be simplified by recognizing standard integrals and using integration by parts.

Solving a More Complex Exact Equation

Consider the differential equation:

yx - x sinx dy/dx - 2e^{yx} x 0.

Identification of M and N

Let P_x(y) yx - x sinx and Q_x(y) -2e^{yx} x. The equation is exact because dP_x(y)/dy 1 dQ_x(y)/dx.

Defining f(x, y)

Define a function f(x, y) such that df(x, y)/dx P_x(y) and df(x, y)/dy Q_x(y). The solution will be given by f(x, y) c_1, where c_1 is an arbitrary constant.

Potential Function Integration

Integrate P_x(y) dy xy - x cosx g(y) Find g'(y) -2e^{yx} Integrate g'(y) to get g(y) -2e^{yx} Combine the results to get f(x, y) x y - x cosx c

The solution to the differential equation is provided by the potential function f(x, y) c_1, where c_1 const.

From the potential function, the solution can be back-solved to find the expression for y(x)

yx x cosx - x W(-2 e^{cosx - sinx/x}) c_1/x

Conclusion

Exact differential equations have wide-ranging applications in various fields of science and engineering. By mastering the solution techniques, such as the method of partial derivatives, you can handle more complex problems in these areas. Understanding the steps involved in solving exact equations not only enhances your problem-solving skills but also provides a solid foundation for more advanced mathematical concepts.