Exact Differential Equation: Example, Solving Equation

Exact differential equation is a type of equation in which we have to find the derivatives exactly by using a formula of მQ / მx = მP / მy. An inexact differential equation is one for which the derivatives cannot be found exactly but can be approximated using various methods. Exact differential equation is a differential equation for which there exists a function F(x, y) such that its partial derivatives with respect to x and y are equal to the respective functions in the differential equation. In the following article, we will get to know about solving the exact differential equations.

Key Takeaways: Exact Differential Equation, differential equation, differentiation, derivatives, functions, equation

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Exact Differential Equation Definition

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An exact differential equation is a differential equation for which there exists a function F(x, y) such that its partial derivatives with respect to x and y are equal to the respective functions in the differential equation. In other words, an exact differential equation is one where the derivative of a function with respect to one variable is equal to the derivative of the same function with respect to the other variable. The equation P(x, y) dx + Q(x, y) dy = 0 can be considered to represent an exact differential equation if there is a function f of two variables, i.e., x and y having constant partial derivatives such as mentioned below:

u_x(x, y) = P(x, y) and u_y (x, y) = Q(x, y);

Hence, the general equation's solution is u(x, y) = C.

Note: Here, "C" is an arbitrary constant.

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Exact Differential Equation Integrating Factor

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The above equation is said to be an exact differential equation if a function u(x,y) exists such that its partial derivative with respect to x is equal to p(x,y), and its partial derivative with respect to y is equal to Q(x,y).

If the equation P (x, y) dx + Q (x, y) dy = 0 is an exact differential equation, then there exists a function u(x,y) such that- P(x,y) = u_x(x, y) = p(x, y) and Q(x, y) = u_y(x, y) = q(x, y).

Hence, the general solution of the equation is u(x, y) = C, where C is an arbitrary constant.

dy/dx = -P(x,y)/Q(x,y).

This can be easily integrated to get the desired function u(x, y).

\(u(x, y) = \int \frac{P(x,y)}{Q(x,y)} dy = \int \frac{p(x,y)}{q(x,y)} dy + C\)

The above equation is the general solution of the differential equation.

The integrating factor method can be used to solve the differential equation if it is not possible to find the function u(x,y) directly.

The integrating factor is a function of the independent variable, which, when multiplied by the terms of a given differential equation, results in an equivalent differential equation whose right-hand side is equal to zero.

This fact can be stated mathematically as follows:

If the equation P(x,y)dx + Q(x,y)dy = 0 is transformed to P(x,y)dx + Q(x,y)dy + R(x,y) = 0, then R(x,y)=0.

The integrating factor R(x,y) can be found by solving the following equation:

\(\frac{\partial R}{\partial x}\) = \(\frac{P}{Q}\), which can be easily integrated to get R(x,y). After finding the integrating factor, it can be multiplied to both sides of the original equation to get the equivalent equation with the right-hand side equal to zero. This equation can be easily solved to get the desired function u(x,y).

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Testing for Exactness

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To test whether a differential equation is exact, we need to check if its total differential is equal to zero.

The total differential of a function is given by:

dF(x,y) = F_x(x,y)dx + F_y(x,y)dy

where F_x and F_y are the partial derivatives of F with respect to x and y respectively.

Thus, for a differential equation application to be exact, we need:

F_x(x,y)dx + F_y(x,y)dy = 0

If this is true, then the equation is exact. Otherwise, it is inexact.

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Example of Exact Differential Equation

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Let us consider the following differential equation:

3x²y dx + (x^3​ + y²) dy = 0.

This equation is an exact differential equation since the derivatives of a function with respect to x and y are equal. The integrating factor can be easily found by solving the following equation:

\(\frac{\partial R}{\partial x}\) = 3x²,

which gives R(x,y) = x³y.

After multiplying this factor to both sides of the original equation, we get 3x²y dx + (x^3​ + y²) dy = 0.

This is the equivalent equation with the right-hand side equal to zero and can be easily solved to get the desired function u(x,y). If it is not a linear equation, you may need a numerical method to find a solution.

Hence, the general solution of the differential equation is u(x,y) = x³y + C, where C is an arbitrary constant.

This is a brief overview of the exact differential equation and how it can be solved using the integrating factor method. For more detailed information on this topic, please refer to a calculus textbook.

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How to Solve Exact Differential Equations?

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Here are some steps to solve the exact differential equation:

Step 1: Firstly, to solve the exact differential equation is to ensure that the provided differential equation is exact utilising testing for exactness. These are all statistical methods.

∂Q∂x = ∂P∂y

Step 2: Now, write the system of two differential equations defining the function u(x,y), i.e.,

∂u∂x = P(x,y)

∂u∂y = Q(x,y)

Step 3: Combining the first equation with the variable x, we get

u(x,y) = ∫P(x,y)dx + φ(y)

Rather than an arbitrary constant C, write an anonymous function of y.

Step 4: The function u(x,y) can be differentiated with respect 

to y in the second equation.

∂u∂y = ∂∂y⌊∫P(x,y)dx + φ(y)⌋ = Q(x,y)

We get the derivative of the anonymous function φ(y) from the above sample, and it is provided by 

φ(y) = Q(x,y) − ∂∂y(∫P(x,y)dx)

Step 5: We can discover the function φ(y) by combining the last expression to turn function u(x,y) to

u(x,y) = ∫P(x,y)dx + φ(y)

Step 6: In the last step, the general solution of the exact differential equation is provided by

u(x, y) = C.

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Points to Remember

Following are some important points:

• Exact differential equation is a differential equation in which the function's derivatives with respect to x and y are equal.
• The integrating factor method can be used to solve the exact differential equation if it is impossible to find the function u(x,y) directly.
• The general solution of the exact differential equation is u(x, y) = C.
• To solve an exact differential equation, it is necessary to test for exactness and write the system of two differential equations defining the function u(x,y).

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