How To Solve Exact Differential Equations

How to Solve Exact Differential Equations: A Comprehensive Guide

Exact differential equations are a specific type of first-order differential equation that can be solved using a straightforward method. Understanding these equations and their solution techniques is crucial for anyone studying differential equations, as they form the foundation for more complex methods. This comprehensive guide will walk you through the theory, practical steps, and examples of solving exact differential equations.

Understanding Exact Differential Equations

A first-order differential equation is said to be exact if it 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, and the following condition holds:

∂M/∂y = ∂N/∂x

This condition ensures that the equation is the total differential of some 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 means that the solution to the exact differential equation is implicitly given by:

F(x, y) = C

where C is an arbitrary constant.

Identifying Exact Differential Equations

Before attempting to solve an equation, it's crucial to determine if it's indeed exact. This is done by checking the condition:

∂M/∂y = ∂N/∂x

Let's look at some examples:

Example 1: (2xy + 3)dx + (x² + 1)dy = 0

Here, M(x, y) = 2xy + 3 and N(x, y) = x² + 1. Therefore:

∂M/∂y = 2x ∂N/∂x = 2x

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

Example 2: (y² + 2x)dx + (2xy + 1)dy = 0

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

∂M/∂y = 2y ∂N/∂x = 2y

Again, ∂M/∂y = ∂N/∂x, indicating this is an exact equation.

Example 3: (2x + y)dx + (x + 2y)dy = 0

Here, M(x, y) = 2x + y and N(x, y) = x + 2y. We calculate:

∂M/∂y = 1 ∂N/∂x = 1

This equation is also exact.

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

Once you've confirmed the equation is exact, you can proceed to find the solution F(x, y) using the following steps:

Step 1: Integrate M(x, y) with respect to x, treating y as a constant.

This gives you a partial solution for F(x, y):

F(x, y) = ∫M(x, y)dx + g(y)

Notice the addition of g(y). This is because the integration with respect to x might omit a function of y that could be present in F(x, y).

Step 2: Differentiate the result from Step 1 with respect to y.

∂F/∂y = ∂/∂y [∫M(x, y)dx + g(y)]

Step 3: Equate the result from Step 2 to N(x, y).

∂F/∂y = N(x, y)

This equation allows you to solve for g'(y).

Step 4: Integrate g'(y) with respect to y to find g(y).

g(y) = ∫g'(y)dy + C₁

Where C₁ is another constant of integration. Note this constant is not relevant to the final solution because it will combine with the arbitrary constant from the final solution.

Step 5: Substitute g(y) back into the expression for F(x, y) from Step 1.

This gives you the complete solution for F(x, y).

Step 6: Write the implicit solution:

F(x, y) = C, where C is an arbitrary constant.

Worked Examples

Let's solve the examples from the previous section:

Example 1: (2xy + 3)dx + (x² + 1)dy = 0

1. Step 1: ∫(2xy + 3)dx = x²y + 3x + g(y)

2. Step 2: ∂/∂y(x²y + 3x + g(y)) = x² + g'(y)

3. Step 3: x² + g'(y) = x² + 1 => g'(y) = 1

4. Step 4: g(y) = ∫1dy = y

5. Step 5: F(x, y) = x²y + 3x + y

6. Step 6: x²y + 3x + y = C

Example 2: (y² + 2x)dx + (2xy + 1)dy = 0

1. Step 1: ∫(y² + 2x)dx = xy² + x² + g(y)

2. Step 2: ∂/∂y(xy² + x² + g(y)) = 2xy + g'(y)

3. Step 3: 2xy + g'(y) = 2xy + 1 => g'(y) = 1

4. Step 4: g(y) = ∫1dy = y

5. Step 5: F(x, y) = xy² + x² + y

6. Step 6: xy² + x² + y = C

Example 3: (2x + y)dx + (x + 2y)dy = 0

1. Step 1: ∫(2x + y)dx = x² + xy + g(y)

2. Step 2: ∂/∂y(x² + xy + g(y)) = x + g'(y)

3. Step 3: x + g'(y) = x + 2y => g'(y) = 2y

4. Step 4: g(y) = ∫2ydy = y²

5. Step 5: F(x, y) = x² + xy + y²

6. Step 6: x² + xy + y² = C

Integrating Factors

Not all first-order differential equations are exact. However, some non-exact equations can be made exact by multiplying them by an appropriate function called an integrating factor. Finding the integrating factor can be challenging, and it’s often a trial and error process or involves solving another differential equation.

Finding Integrating Factors

The most common approach is to look for integrating factors that are functions of either x alone or y alone. If:

• (∂M/∂y - ∂N/∂x)/N is a function of x only, say μ(x), then the integrating factor is e^(∫μ(x)dx).

• (∂N/∂x - ∂M/∂y)/M is a function of y only, say μ(y), then the integrating factor is e^(∫μ(y)dy).

Applying the integrating factor to the original equation transforms it into an exact equation, which can then be solved using the steps outlined previously. It's important to note that finding an integrating factor isn't always possible, and there is no guaranteed method to find one in all cases.

Conclusion

Solving exact differential equations is a fundamental skill in the study of differential equations. By understanding the conditions for exactness and following the systematic steps outlined in this guide, you can effectively solve a wide range of these equations. Remember to always check for exactness before proceeding with the solution, and consider using integrating factors when dealing with non-exact equations. Mastering this technique will build a strong foundation for tackling more complex differential equation problems. Consistent practice with various examples will solidify your understanding and ability to confidently solve these equations.