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Finding Solutions to Exact Equations: A Comprehensive Guide

Finding solutions to exact equations is a crucial skill in differential equations. This article provides a comprehensive guide, explaining the process, offering examples, and addressing common challenges. We'll explore how to identify exact equations and systematically solve them. Understanding exact equations is fundamental to mastering more advanced techniques in differential equations.

What is an Exact Equation?

An exact equation is a first-order differential equation of the form:

M(x, y)dx + N(x, y)dy = 0

where M and N are functions of x and y, and the equation satisfies the condition:

∂M/∂y = ∂N/∂x

This condition signifies that there exists a function F(x, y) such that:

∂F/∂x = M(x, y) and ∂F/∂y = N(x, y)

The solution to the exact equation is then implicitly defined by:

F(x, y) = C

where C is an arbitrary constant.

Identifying Exact Equations

The first step in solving an exact equation is verifying whether the given equation is indeed exact. This involves calculating the partial derivatives ∂M/∂y and ∂N/∂x and checking if they are equal. If they are equal, the equation is exact.

Example 1: Verifying an Exact Equation

Consider the equation:

(2xy + y²)dx + (x² + 2xy)dy = 0

Here, M(x, y) = 2xy + y² and N(x, y) = x² + 2xy. Calculating the partial derivatives:

∂M/∂y = 2x + 2y

∂N/∂x = 2x + 2y

Since ∂M/∂y = ∂N/∂x, the equation is exact.

Solving Exact Equations: A Step-by-Step Approach

Once we confirm an equation is exact, we can proceed to find its solution. The process involves these steps:

1. Integrate M(x, y) with respect to x: This gives us a partial solution, F(x, y), but it will contain an arbitrary function of y, g(y).

2. Differentiate the partial solution with respect to y: This will give us ∂F/∂y.

3. Compare ∂F/∂y with N(x, y): This comparison allows us to determine g'(y).

4. Integrate g'(y) with respect to y: This completes the determination of the function g(y).

5. Substitute g(y) back into F(x, y): This gives the complete solution F(x, y) = C.

Example 2: Solving an Exact Equation

Let's solve the exact equation from Example 1: (2xy + y²)dx + (x² + 2xy)dy = 0

1. Integrate M(x, y) = 2xy + y² with respect to x: ∫(2xy + y²)dx = x²y + xy² + g(y)

2. Differentiate with respect to y: ∂F/∂y = x² + 2xy + g'(y)

3. Compare with N(x, y) = x² + 2xy: We see that g'(y) = 0.

4. Integrate g'(y) = 0 with respect to y: g(y) = K (where K is a constant)

5. Substitute back into F(x, y): F(x, y) = x²y + xy² + K = C Therefore, the solution is x²y + xy² = C (where C is an arbitrary constant).

Dealing with Non-Exact Equations: Integrating Factors

Not all first-order differential equations are exact. However, some non-exact equations can be made exact by multiplying them by an integrating factor, μ(x, y). Finding the appropriate integrating factor can be challenging and often requires specific techniques depending on the structure of the equation.

Conclusion

Solving exact equations is a fundamental technique in differential equations. By understanding the conditions for exactness and following a systematic approach, you can efficiently find solutions to a broad class of differential equations. Remember that while many equations are exact, others require the use of integrating factors, which adds another layer of complexity to the process. Practice is key to mastering this crucial skill.