Solving Exact Differential Equations: A Step-by-Step Guide to the Equation y cos x dx sin x dy - 2 x e^y dx - x^2 e^y dy dy 0

In this article, we will explore the process of solving an exact differential equation of the form A(x,y)dx B(x,y)dy 0. Specifically, we will delve into the following equation:

Introduction

The given differential equation is:

y cos x dx sin x dy - 2 x e^y dx - x^2 e^y dy dy 0

Verifying Exactness

For a differential equation of the form M(x,y)dx N(x,y)dy 0 to be exact, the condition ?M/?y ?N/?x must be satisfied.

First, we define:

M(x,y) y cos x - 2 x e^y

N(x,y) sin x - x^2 e^y - 1

We need to compute the partial derivatives:

?M/?y cos x - 2 x e^y

?N/?x cos x - 2 x e^y

Since ?M/?y ?N/?x, the given differential equation is exact.

Finding the Potential Function

To find the potential function ψ(x,y), we integrate M(x,y)dx with respect to x:

ψ(x,y) ∫ y cos x - 2 x e^y dx

This integral can be solved as follows:

ψ(x,y) y sin x - x^2 e^y g(y)

where g(y) is an arbitrary function of y.

Next, we differentiate ψ(x,y) with respect to y:

?ψ/?y sin x - x^2 e^y g'(y)

Setting ?ψ/?y N(x,y), we get:

sin x - x^2 e^y g'(y) sin x - x^2 e^y - 1

This simplifies to:

g'(y) -1

Solving for g(y), we find:

g(y) -y - C

Thus, the potential function is:

ψ(x,y) y sin x - x^2 e^y - y - C

Solution to the Differential Equation

The solution to the differential equation is given by setting the potential function equal to a constant:

y sin x - x^2 e^y - y C

This implicit equation represents the solution to the original exact differential equation.

Conclusion

Understanding and solving exact differential equations involves several key steps: verifying exactness, finding the potential function, and combining the results to form the final solution. By following these steps, you can handle a variety of differential equations and find their solutions efficiently.

Keywords: exact differential equations, differential equations, solution methods