Exact Equation Differential Equations Examples

Exact Equation Differential Equations: A Comprehensive Guide with Examples

Differential equations are the cornerstone of many scientific and engineering disciplines, describing the relationships between variables and their rates of change. Among the various types of differential equations, exact equations hold a special place due to their straightforward solution method. This article will delve into the theory behind exact equations, provide clear step-by-step solutions for several examples, and address frequently asked questions to solidify your understanding. We will cover various techniques and approaches to solving these equations, offering a comprehensive guide for students and professionals alike.

Understanding Exact Differential Equations

An exact differential equation is a first-order differential equation that can be expressed in the form:

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

where M(x, y) and N(x, y) are functions of x and y. The key characteristic of an exact equation is that it represents the total differential of a function, often denoted as F(x, y). In other words, there exists a function F(x, y) such that:

dF = ∂F/∂x dx + ∂F/∂y dy = M(x, y)dx + N(x, y)dy = 0

This implies that:

∂F/∂x = M(x, y) and ∂F/∂y = N(x, y)

The crucial test to determine if a given differential equation is exact is to check the following condition:

∂M/∂y = ∂N/∂x

If this condition holds true, then the equation is exact. If not, it is not an exact equation, and other methods must be employed to solve it.

Solving Exact Differential Equations: A Step-by-Step Approach

Solving an exact differential equation involves finding the function F(x, y) whose total differential is equal to the given equation. Here's a step-by-step method:

1. Verify Exactness:

First, identify M(x, y) and N(x, y) from the given equation. Then, calculate ∂M/∂y and ∂N/∂x. If ∂M/∂y = ∂N/∂x, the equation is exact; otherwise, it's not.

2. Integrate M(x, y) with respect to x:

Integrate M(x, y) with respect to x, treating y as a constant. This will give you a partial solution for F(x, y), which will include an arbitrary function of y, say g(y).

3. Differentiate the partial solution with respect to y:

Differentiate the partial solution obtained in step 2 with respect to y.

4. Compare with N(x, y) and solve for g'(y):

Compare the result from step 3 with N(x, y). This will allow you to determine g'(y).

5. Integrate g'(y) to find g(y):

Integrate g'(y) with respect to y to find g(y). Remember to include the constant of integration, C.

6. Write the final solution:

Substitute the expression for g(y) back into the partial solution obtained in step 2 to obtain the general solution F(x, y) = C.

Examples of Exact Differential Equations and Their Solutions

Let's work through several examples to solidify our understanding.

Example 1:

Solve the differential equation:

(2x + y)dx + (x + 2y)dy = 0

Solution:

1. Verify Exactness:

   M(x, y) = 2x + y => ∂M/∂y = 1 N(x, y) = x + 2y => ∂N/∂x = 1 Since ∂M/∂y = ∂N/∂x, the equation is exact.

2. Integrate M(x, y) with respect to x:

   ∫(2x + y)dx = x² + xy + g(y)

3. Differentiate the partial solution with respect to y:

   ∂/∂y (x² + xy + g(y)) = x + g'(y)

4. Compare with N(x, y) and solve for g'(y):

   x + g'(y) = x + 2y => g'(y) = 2y

5. Integrate g'(y) to find g(y):

   ∫2y dy = y² + C

6. Write the final solution:

   F(x, y) = x² + xy + y² = C

Example 2:

Solve the differential equation:

(y cos x + 2xe^y)dx + (sin x + x²e^y + 2y)dy = 0

Solution:

1. Verify Exactness:

   M(x, y) = y cos x + 2xe^y => ∂M/∂y = cos x + 2xe^y N(x, y) = sin x + x²e^y + 2y => ∂N/∂x = cos x + 2xe^y Since ∂M/∂y = ∂N/∂x, the equation is exact.

2. Integrate M(x, y) with respect to x:

   ∫(y cos x + 2xe^y)dx = y sin x + x²e^y + g(y)

3. Differentiate the partial solution with respect to y:

   ∂/∂y (y sin x + x²e^y + g(y)) = sin x + x²e^y + g'(y)

4. Compare with N(x, y) and solve for g'(y):

   sin x + x²e^y + g'(y) = sin x + x²e^y + 2y => g'(y) = 2y

5. Integrate g'(y) to find g(y):

   ∫2y dy = y² + C

6. Write the final solution:

   F(x, y) = y sin x + x²e^y + y² = C

Example 3: A Non-Exact Equation Demonstrating the Need for an Integrating Factor

Consider the equation: y dx + (x + y²) dy = 0

If we check for exactness:

M(x, y) = y; ∂M/∂y = 1 N(x, y) = x + y²; ∂N/∂x = 1

While ∂M/∂y = ∂N/∂x, this is a deceptively simple case. Trying to solve directly, we get: ∫y dx = xy + g(y). Differentiating with respect to y: x + g'(y) = x + y². This implies g'(y) = y², which integrates to g(y) = (1/3)y³ + C. However, this gives us xy + (1/3)y³ = C, which is not a correct solution.

This case highlights the importance of correctly identifying an exact equation and showcases that even if ∂M/∂y = ∂N/∂x, it doesn't guarantee the direct integration method will work correctly without a proper methodical approach.

Integrating Factors: Handling Non-Exact Equations

Many differential equations that are not initially exact can be made exact by multiplying by an integrating factor. Finding the appropriate integrating factor can be challenging and often requires trial and error or specific techniques depending on the form of the equation. This is a more advanced topic and will not be covered extensively here.

Frequently Asked Questions (FAQ)

Q1: What if the equation is not exact?

A1: If the equation is not exact (∂M/∂y ≠ ∂N/∂x), you might need to find an integrating factor to make it exact. This is a more advanced technique.

Q2: Is there a unique solution to an exact differential equation?

A2: No, there's a family of solutions represented by the general solution F(x, y) = C, where C is an arbitrary constant. A specific solution can be found if an initial condition is given.

Q3: Can this method be applied to higher-order differential equations?

A3: No, this method is specifically designed for first-order differential equations. Higher-order equations require different solution techniques.

Q4: How can I check my solution?

A4: Once you find a solution, you can implicitly differentiate it to check if it satisfies the original differential equation.

Conclusion

Exact differential equations represent a significant class of first-order differential equations solvable through a systematic approach. By understanding the underlying principles and diligently following the steps outlined above, you can confidently tackle a wide range of problems. Remember, verifying exactness is the crucial first step, followed by careful integration and differentiation. While some non-exact equations can be transformed into exact equations using integrating factors, mastering the techniques presented here will equip you with a fundamental tool in your arsenal for solving differential equations. Continued practice with diverse examples will build proficiency and solidify your understanding of this important concept in calculus and differential equations.