Find The Solution To The Exact Equation In The Form

Finding Solutions to Exact Equations: A Comprehensive Guide

Finding solutions to exact equations is a fundamental concept in differential equations. An exact equation is a differential equation of the form M(x, y)dx + N(x, y)dy = 0, where the left-hand side is the total differential of some function F(x, y). This means that there exists a function F(x, y) such that ∂F/∂x = M(x, y) and ∂F/∂y = N(x, y). This article provides a comprehensive guide to understanding and solving exact equations, covering various techniques and examples.

Understanding Exact Equations

The key to recognizing an exact equation lies in the condition for exactness: If ∂M/∂y = ∂N/∂x, then the equation M(x, y)dx + N(x, y)dy = 0 is exact. This condition ensures that the mixed partial derivatives of F(x, y) are equal, a requirement for the existence of such a function. Let's break this down:

• M(x, y) and N(x, y): These are functions of x and y, representing the coefficients of dx and dy, respectively.
• ∂M/∂y: This represents the partial derivative of M with respect to y, treating x as a constant.
• ∂N/∂x: This represents the partial derivative of N with respect to x, treating y as a constant.
• Exactness Condition: If ∂M/∂y = ∂N/∂x, it implies that the equation is the total differential of some function F(x, y). This means that dF = ∂F/∂x dx + ∂F/∂y dy = 0. The solution then becomes F(x, y) = C, where C is an arbitrary constant.

Solving Exact Equations: A Step-by-Step Approach

Once we've confirmed that an equation is exact (by verifying ∂M/∂y = ∂N/∂x), we can proceed to find its solution. Here's a step-by-step approach:

1. Identify M(x, y) and N(x, y): Rewrite the equation in the standard form M(x, y)dx + N(x, y)dy = 0.

2. Verify Exactness: Calculate ∂M/∂y and ∂N/∂x. If they are equal, the equation is exact.

3. Find F(x, y): This is the most crucial step. We integrate M(x, y) with respect to x, treating y as a constant:

   ∫M(x, y)dx = F(x, y) + g(y)

   Note the inclusion of g(y), an arbitrary function of y. This accounts for any terms that might vanish when differentiating with respect to x.

4. Determine g(y): Now, differentiate the expression for F(x, y) + g(y) with respect to y:

   ∂/∂y [F(x, y) + g(y)] = ∂F/∂y + g'(y)

   This should be equal to N(x, y). Therefore, we solve for g'(y) and integrate to find g(y).

5. Write the General Solution: Substitute the expression for g(y) back into F(x, y) + g(y) to obtain the general solution F(x, y) = C, where C is an arbitrary constant.

Examples: Illustrating the Process

Let's work through some examples to solidify our understanding:

Example 1: Solve the equation (2xy + y²)dx + (x² + 2xy)dy = 0

1. Identify M and N: M(x, y) = 2xy + y², N(x, y) = x² + 2xy

2. Verify Exactness: ∂M/∂y = 2x + 2y, ∂N/∂x = 2x + 2y. Since ∂M/∂y = ∂N/∂x, the equation is exact.

3. Find F(x, y): ∫(2xy + y²)dx = x²y + xy² + g(y)

4. Determine g(y): ∂/∂y [x²y + xy² + g(y)] = x² + 2xy + g'(y) = x² + 2xy. This implies g'(y) = 0, so g(y) = K (a constant).

5. General Solution: F(x, y) = x²y + xy² + K = C, which simplifies to x²y + xy² = C (where C is an arbitrary constant).

Example 2: Solve the equation (e^x + y)dx + (x + 6y²)dy = 0

1. Identify M and N: M(x, y) = e^x + y, N(x, y) = x + 6y²

2. Verify Exactness: ∂M/∂y = 1, ∂N/∂x = 1. The equation is exact.

3. Find F(x, y): ∫(e^x + y)dx = e^x + xy + g(y)

4. Determine g(y): ∂/∂y [e^x + xy + g(y)] = x + g'(y) = x + 6y². This gives g'(y) = 6y², so g(y) = 2y³ + K.

5. General Solution: F(x, y) = e^x + xy + 2y³ + K = C, or e^x + xy + 2y³ = C.

Dealing with Non-Exact Equations: Integrating Factors

Not all first-order differential equations are exact. However, sometimes we can transform a non-exact equation into an exact one by multiplying it by an integrating factor. This factor, usually a function of x or y (or both), makes the equation exact. Finding the appropriate integrating factor can be challenging and often requires intuition and trial-and-error.

Identifying Potential Integrating Factors:

• Function of x: If [(∂M/∂y) - (∂N/∂x)]/N is a function of x only, say μ(x), then μ(x) = exp(∫[(∂M/∂y) - (∂N/∂x)]/N dx) is an integrating factor.

• Function of y: If [(∂N/∂x) - (∂M/∂y)]/M is a function of y only, say μ(y), then μ(y) = exp(∫[(∂N/∂x) - (∂M/∂y)]/M dy) is an integrating factor.

Example 3 (Non-Exact with Integrating Factor): Solve the equation (y² + 2xy)dx + x²dy = 0

1. Check for Exactness: ∂M/∂y = 2y + 2x, ∂N/∂x = 2x. The equation is not exact.

2. Find an Integrating Factor: [(∂N/∂x) - (∂M/∂y)]/M = (-2y)/ (y² + 2xy) = -2/(y + 2x). This is not a function of y alone. Let's try the other condition: [(∂M/∂y) - (∂N/∂x)]/N = (2y)/(x²) = (2y/x²). This is not a function of x alone.

Let's try an integrating factor of the form 1/x^n

Multiplying by 1/x² gives:

(y²/x² + 2y/x)dx + dy = 0

Then ∂M/∂y = 2y/x² + 2/x and ∂N/∂x = 0. This is still not exact.

Let's consider an integrating factor of the form 1/x:

(y²/x + 2y)dx + xdy = 0

Then ∂M/∂y = 2y/x + 2 and ∂N/∂x = 1. This is still not exact.

This example showcases a situation where a simple integrating factor is not readily apparent, and more advanced techniques might be required. This often involves exploring different forms of integrating factors or utilizing alternative solution methods for non-exact equations. In practice, determining the integrating factor can be challenging and sometimes requires advanced techniques beyond the scope of this introductory guide.

Conclusion

Solving exact equations is a cornerstone of differential equations. Understanding the condition for exactness and the step-by-step solution process is crucial for tackling these types of problems. While not all equations are immediately exact, the concept of integrating factors offers a pathway to transform many non-exact equations into solvable forms. Remember that finding the correct integrating factor can be challenging and may require exploration of different methods and techniques. This comprehensive guide provides a solid foundation for mastering this essential aspect of differential equations.