How To Solve Exact Differential Equations

Solving exact differential equations is a fundamental skill in the study of differential equations, offering a powerful technique for finding solutions when certain conditions are met. These equations appear in various fields of science and engineering, making their understanding and resolution essential for modeling and analyzing real-world phenomena.

Understanding Exact Differential Equations

An exact differential equation is a first-order ordinary 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 two variables, x and y. The key to identifying an exact differential equation lies in the following condition:

∂M/∂y = ∂N/∂x

This condition states that the partial derivative of M with respect to y must be equal to the partial derivative of N with respect to x. If this condition holds, the differential equation is exact, implying that there exists a function F(x, y) such that:

∂F/∂x = M(x, y) and ∂F/∂y = N(x, y)

Finding this function F(x, y) is the core of solving the exact differential equation, as the general solution is given implicitly by:

F(x, y) = C

where C is an arbitrary constant. This implicit solution represents a family of curves that satisfy the differential equation.

Steps to Solve Exact Differential Equations

Solving an exact differential equation involves a systematic approach that ensures finding the correct solution. Here's a detailed breakdown of the steps:

1. Verify Exactness:

The first and foremost step is to verify that the given differential equation is indeed exact. This is done by checking the condition:

∂M/∂y = ∂N/∂x

Calculate the partial derivatives of M(x, y) and N(x, y) and compare them. If they are equal, proceed to the next step. If not, the equation is not exact, and other methods of solution are required.

2. Find the Potential Function F(x, y):

Since the equation is exact, there exists a function F(x, y) such that:

∂F/∂x = M(x, y) and ∂F/∂y = N(x, y)

To find F(x, y), integrate either M(x, y) with respect to x or N(x, y) with respect to y. Let's say you choose to integrate M(x, y) with respect to x:

F(x, y) = ∫M(x, y) dx + g(y)

Here, g(y) is an arbitrary function of y, which acts as the "constant of integration" since we are performing a partial integration with respect to x.

3. Determine the Function g(y):

Now, differentiate the expression for F(x, y) obtained in the previous step with respect to y:

∂F/∂y = ∂/∂y [∫M(x, y) dx + g(y)] = N(x, y)

This equation allows you to solve for g'(y), the derivative of g(y) with respect to y. Integrate g'(y) with respect to y to find g(y):

g(y) = ∫g'(y) dy

4. Write the General Solution:

Substitute the expression for g(y) back into the expression for F(x, y) obtained in step 2:

F(x, y) = ∫M(x, y) dx + g(y) = C

This is the general solution of the exact differential equation, where C is an arbitrary constant. This solution is implicit, meaning that y is not explicitly expressed as a function of x.

5. Apply Initial Conditions (if given):

If initial conditions are provided (e.g., y(x₀) = y₀), substitute these values into the general solution to find the value of the constant C. This gives the particular solution that satisfies the given initial condition.

Detailed Examples

Let's illustrate the steps involved in solving exact differential equations with a couple of detailed examples:

Example 1:

Solve the following differential equation:

(2x + y²) dx + (2xy + 3y²) dy = 0

1. Verify Exactness:

M(x, y) = 2x + y² N(x, y) = 2xy + 3y²

∂M/∂y = 2y ∂N/∂x = 2y

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

2. Find the Potential Function F(x, y):

F(x, y) = ∫M(x, y) dx + g(y) = ∫(2x + y²) dx + g(y) = x² + xy² + g(y)

3. Determine the Function g(y):

∂F/∂y = 2xy + g'(y) = N(x, y) = 2xy + 3y²

g'(y) = 3y² g(y) = ∫3y² dy = y³ + K (where K is a constant, which we can absorb into the general constant C)

4. Write the General Solution:

F(x, y) = x² + xy² + y³ = C

This is the general solution of the given differential equation.

Example 2:

Solve the following differential equation:

(y cos(xy) + 1) dx + (x cos(xy)) dy = 0

1. Verify Exactness:

M(x, y) = y cos(xy) + 1 N(x, y) = x cos(xy)

∂M/∂y = cos(xy) - xy sin(xy) ∂N/∂x = cos(xy) - xy sin(xy)

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

2. Find the Potential Function F(x, y):

F(x, y) = ∫M(x, y) dx + g(y) = ∫(y cos(xy) + 1) dx + g(y) = sin(xy) + x + g(y)

3. Determine the Function g(y):

∂F/∂y = x cos(xy) + g'(y) = N(x, y) = x cos(xy)

g'(y) = 0 g(y) = ∫0 dy = K (where K is a constant, which we can absorb into the general constant C)

4. Write the General Solution:

F(x, y) = sin(xy) + x = C

This is the general solution of the given differential equation.

Why Does This Method Work? (Theoretical Justification)

The method for solving exact differential equations relies on a fundamental theorem from multivariable calculus. The key idea is that if the differential equation M(x, y) dx + N(x, y) dy = 0 is exact (i.e., ∂M/∂y = ∂N/∂x), then the expression M(x, y) dx + N(x, y) dy is an exact differential. This means it is the total differential of some function F(x, y).

In other words:

dF = (∂F/∂x) dx + (∂F/∂y) dy

If we can find such a function F(x, y), then the differential equation M(x, y) dx + N(x, y) dy = 0 can be rewritten as dF = 0. Integrating both sides gives F(x, y) = C, where C is a constant. This represents the family of solutions to the differential equation.

The condition ∂M/∂y = ∂N/∂x is crucial because it guarantees the existence of the function F(x, y). This condition is a direct consequence of the fact that mixed partial derivatives are equal for sufficiently smooth functions (Clairaut's Theorem). Specifically, if F(x, y) exists such that ∂F/∂x = M and ∂F/∂y = N, then:

∂²F/∂y∂x = ∂M/∂y ∂²F/∂x∂y = ∂N/∂x

By Clairaut's Theorem, ∂²F/∂y∂x = ∂²F/∂x∂y, which implies ∂M/∂y = ∂N/∂x.

Therefore, the exactness condition ensures that the line integral of M(x, y) dx + N(x, y) dy is path-independent, meaning the integral's value depends only on the starting and ending points and not on the path taken. This path independence is what allows us to find a function F(x, y) such that its total differential equals the given expression.

Common Mistakes and How to Avoid Them

Solving exact differential equations can be tricky, and certain common mistakes can lead to incorrect solutions. Here are some pitfalls to watch out for:

• Failing to Verify Exactness: This is the most crucial step. If you skip it and the equation is not exact, the subsequent steps will be meaningless, and the resulting "solution" will be wrong. Always check that ∂M/∂y = ∂N/∂x before proceeding.

• Incorrectly Calculating Partial Derivatives: Errors in calculating ∂M/∂y or ∂N/∂x will lead to an incorrect conclusion about exactness and potentially an incorrect solution. Double-check your partial derivatives carefully.

• Forgetting the Constant of Integration (g(y) or h(x)): When integrating M(x, y) with respect to x (or N(x, y) with respect to y), remember to include an arbitrary function of the other variable (g(y) or h(x)). This is essential because you are performing a partial integration, and the "constant of integration" can depend on the other variable.

• Incorrectly Determining g(y) (or h(x)): After finding the initial expression for F(x, y), carefully differentiate it with respect to the other variable and equate it to N(x, y) (or M(x, y)) to solve for g'(y) (or h'(x)). Make sure to integrate correctly to find g(y) (or h(x)).

• Algebraic Errors: Be meticulous with your algebra throughout the process. Simple errors in simplifying expressions or solving equations can lead to an incorrect solution.

• Confusing Implicit and Explicit Solutions: Remember that the general solution F(x, y) = C is usually an implicit solution. It may not always be possible to solve for y explicitly as a function of x.

Applications of Exact Differential Equations

Exact differential equations have numerous applications in various fields of science and engineering. Here are a few examples:

• Thermodynamics: In thermodynamics, exact differentials are used to describe state functions such as internal energy, enthalpy, and entropy. The change in these functions depends only on the initial and final states of the system and not on the path taken.

• Fluid Dynamics: Exact differential equations can arise in the study of fluid flow, particularly in describing conservative force fields.

• Electromagnetism: In electromagnetism, the concept of potential is closely related to exact differentials. The electric potential and magnetic potential are scalar functions whose gradients give the electric and magnetic fields, respectively.

• Mechanics: Conservative force fields in mechanics, such as gravity, can be described using potential energy functions. The work done by a conservative force depends only on the initial and final positions and can be expressed as an exact differential.

• Economics: In economics, exact differential equations can be used in utility theory to model consumer preferences and derive demand functions.

Beyond the Basics: Integrating Factors

While the method described above works perfectly for exact differential equations, many differential equations are not exact in their original form. However, it is sometimes possible to transform a non-exact differential equation into an exact one by multiplying it by a suitable function called an integrating factor.

An integrating factor, denoted by μ(x, y), is a function that, when multiplied by the original differential equation, makes it exact. That is, if M(x, y) dx + N(x, y) dy = 0 is not exact, then:

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

can be exact for a suitable choice of μ(x, y).

Finding an integrating factor can be challenging, but there are some common cases where it can be determined relatively easily:

• Integrating Factor Depends Only on x: If (∂M/∂y - ∂N/∂x)/N is a function of x only, say f(x), then an integrating factor is given by:

μ(x) = exp(∫f(x) dx)

• Integrating Factor Depends Only on y: If (∂N/∂x - ∂M/∂y)/M is a function of y only, say g(y), then an integrating factor is given by:

μ(y) = exp(∫g(y) dy)

Once you find an integrating factor, multiply the original differential equation by it and verify that the resulting equation is exact. Then, proceed to solve the exact equation as described earlier.

Conclusion

Solving exact differential equations is a valuable technique with applications across various scientific and engineering disciplines. By systematically verifying exactness, finding the potential function, and applying initial conditions, you can effectively solve these equations and gain insights into the behavior of the systems they model. Remember to avoid common mistakes and explore the concept of integrating factors to tackle non-exact differential equations. Mastering this technique will significantly enhance your problem-solving abilities in the field of differential equations.