Let us first understand what are differential equations. An equation that contains the derivative of an unknown function is called a differential equation. Differential equations are mainly used in the fields of biology, physics, engineering, and many. In mathematics, an exact differential equation is a certain kind of ordinary differential equation which is widely used in physics and engineering.

What is a Differential Equation?

A differential equation is a mathematical equation that shows how a quantity changes with respect to another.

It includes a derivative, which tells us the rate of change of one variable compared to another.

For example, in the equation:

dy/dx = f(x)

x is called the independent variable (the one you change),
y is the dependent variable (the one that changes based on x).

A simple example is:

dy/dx = 7x

This means that the rate at which y changes with respect to x is equal to 7x.

Differential equations can involve:

• Ordinary derivatives (when there's only one independent variable)
• Partial derivatives (when there are two or more independent variables)

In short, differential equations help us describe real-world situations where one quantity depends on how another changes — like speed, growth, or temperature changes over time.

Learn about applications of differential equations

What is Exact Differential Equation?

A differential equation of type \(P(x,y)dx+Q(x,y)dy=0\) is called an exact differential equation if there exists a function of two variables \(u(x,y)\) with continuous partial derivatives such that

\(du(x,y)=P(x,y)dx+Q(x,y)dy\).

The general solution of an exact equation is given by

\(u(x,y)=C\)

where \(C\) is an arbitrary constant.

Let functions \(P(x,y)\) and \(Q(x,y)\) have continuous partial derivatives in a certain domain \(D\). The differential equation \(P(x,y)dx+Q(x,y)dy=0\) is an exact equation if and only if

\(\frac{\partial Q}{\partial x}=\frac{\partial P}{\partial y}\).

Integrating Factor for Exact Differential Equation

If the differential equation \(P(x,y)dx+Q(x,y)dy=0\) is not exact, it can be made precise by multiplying it with a relevant factor \(u(x, y)\) known as the integrating factor for the given differential equation. Let us understand how to find the integrating factor with the help of an example.

For example, consider the equation \(2ydx+xdy=0\).

Now check whether the given differential equation is exact using testing for exactness.

Here \(P=2y\) and \(Q=x\), which gives

\(\frac{\partial P}{\partial y}=\frac{\partial (2y)}{\partial y}=2\), and \(\frac{\partial Q}{\partial x}=\frac{\partial (x)}{\partial x}=1\).

\(\Rightarrow\) \(\frac{\partial P}{\partial y}\neq\frac{\partial Q}{\partial x}\)

So, the given differential equation is not exact.

In order to convert it into the exact differential equation, multiply by the integrating factor \(u(x,y)= x\), the differential equation becomes,

\(2 xy dx + x^{2} dy = 0\).

The above resultant equation is an exact differential equation because \(\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}=2x\).

Note: It is not easy to find the integrating factor every time. But, there are two cases of differential equations whose integrating factor may be found easily. Those equations have the integrating factor having the functions of either \(x\) alone or \(y\) alone.

When you consider the differential equation \(P(x,y)dx + Q(x,y)dy=0\), the two cases involved are:

• Case 1: If \(\left[\frac{1}{Q(x,y)}\right]\left[P_{y}(x,y)-Q_{x}(x,y)\right]=h(x)\), which is a function of \(x\) alone, then \(e^{\int h(x)dx}\) is an integrating factor.
• Case 2: If \(\left[\frac{1}{P(x,y)}\right]\left[Q_{x}(x,y)-P_{y}(x,y)\right]=k(y)\), which is a function of \(y\) alone, then \(e^{\int k(y)dy}\) is an integrating factor.

Learn about Solution of Differential Equations

How to Solve Exact Differential Equations?

The following steps explain how to solve the exact differential equation in a detailed way.

Step 1: Check if the equation is exact.
Suppose you have an equation of the form:
P(x, y) dx + Q(x, y) dy = 0
To check if it's exact, test whether:
∂Q/∂x = ∂P/∂y (partial derivatives are equal)
If they are equal, the equation is exact and you can continue.

Step 2: Set up the system to find a function u(x, y).
We are trying to find a function u(x, y) such that:
∂u/∂x = P(x, y) and ∂u/∂y = Q(x, y)

Step 3: Integrate P(x, y) with respect to x.
Find u(x, y) by integrating P(x, y) with respect to x:
u(x, y) = ∫P(x, y) dx + φ(y)
(Instead of adding a constant, we add an unknown function φ(y) because y is treated like a constant during integration.)

Step 4: Differentiate the result with respect to y.
Now take the partial derivative of your u(x, y) with respect to y, and set it equal to Q(x, y):
∂u/∂y = Q(x, y)
This helps us find φ′(y), the derivative of φ(y).

Step 5: Find φ(y)
Solve for φ′(y) and then integrate it to get φ(y).
Now you can write the full expression for u(x, y).

Step 6: Write the final solution
The general solution to the exact differential equation is:
u(x, y) = C,
where C is an arbitrary constant.

Note: In Step 3, we can integrate the second equation over the variable \(y\) instead of integrating the first equation over \(x\). After integration we need to find the unknown function \(\psi(x)\).

Example of Exact Differential Equation

Some examples of the exact differential equation are listed below:

• \((2xy – 3x^{2})dx + (x^{2} – 2y)dy = 0\).
• \((xy^{2} + x)dx + yx^{2}dy = 0\).
• \(\cos(y) dx + (y^{2} – x \sin(y))dy = 0\).
• \((6x^{2} – y + 3)dx + (3y^{2} – x – 2)dy = 0\).
• \(e^{y}dx + ( 2y + xe^{y} )dy = 0\).
• \(3x(xy -2)dx + (x^{2} + 2y)dy = 0\).

The above equations are exact differential equations because it all satisfies the condition of exactness, which is the differential equation \(P(x,y)dx+Q(x,y)dy=0\) is an exact equation if and only if

\(\frac{\partial Q}{\partial x}=\frac{\partial P}{\partial y}\).

Learn about solutions of differential equations

Important Points of Exact Differential Equation

• 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 \(u(x,y)\).

Properties of Exact Differential Equations

1. Definition-Based Property
   A differential equation of the form:
   P(x, y) dx + Q(x, y) dy = 0
   is called exact if there exists a function u(x, y) such that:
   ∂u/∂x = P(x, y) and ∂u/∂y = Q(x, y)
2. Exactness Condition (Test for Exactness)
   To check if the equation is exact, verify:
   ∂P/∂y = ∂Q/∂x
   If this condition is true, the differential equation is exact.
3. Existence of a Potential Function
   If the equation is exact, then there exists a function u(x, y) such that:
   du = P dx + Q dy
   The general solution is then: u(x, y) = C, where C is a constant.
4. Path Independence
   In an exact differential equation, the solution depends only on the starting and ending points, not on the path between them.
   This property is similar to conservative force fields in physics.
5. Closed Curve Property
   If the path of integration forms a closed curve (starts and ends at the same point), then:
   ∮(P dx + Q dy) = 0
   This means the total change around a loop is zero.
6. Integration Method (Solving an Exact Equation)
   To solve:

• First, integrate P(x, y) with respect to x (treating y as constant)
• Then use the condition ∂u/∂y = Q(x, y) to find the missing part of the function

Exact Differential Equation Solved Examples

Example 1: Is the following differential equation exact or not: \((y^{2} – 2x)dx + (2xy + 1)dy = 0\).

Solution: Check whether the given differential equation is exact using testing for exactness.

Here \(P = (y^{2} – 2x)\) and \(Q = (2xy + 1)\), which gives

\(\frac{\partial P}{\partial y}=\frac{\partial (y^{2}-2x)}{\partial y}=2y\), and

\(\frac{\partial Q}{\partial x}=\frac{\partial (2xy+1)}{\partial x}=2y\).

\(\Rightarrow\) \(\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}\)

So, the given differential equation is exact.

Example 2: Solve the differential equation: (2xy + cos(x)) dx + (x² + sin(y)) dy = 0

Solution:

We are given:
P = 2xy + cos(x)
Q = x² + sin(y)

Step 1: Check for Exactness

Find the partial derivatives:

• ∂P/∂y = ∂/∂y (2xy + cos(x)) = 2x
• ∂Q/∂x = ∂/∂x (x² + sin(y)) = 2x

Since ∂P/∂y = ∂Q/∂x, the equation is exact.

Step 2: Set up and Integrate

We want a function u(x, y) such that:
∂u/∂x = 2xy + cos(x)

Integrate with respect to x:

u(x, y) = ∫(2xy + cos(x)) dx
= x²y + sin(x) + φ(y)
(φ(y) is a function of y)

Step 3: Differentiate u(x, y) with respect to y

u(x, y) = x²y + sin(x) + φ(y)

So, ∂u/∂y = x² + φ′(y)

But from Q, we also have:
∂u/∂y = x² + sin(y)

So:

x² + φ′(y) = x² + sin(y)
⇒ φ′(y) = sin(y)

Step 4: Integrate φ′(y)

φ(y) = ∫sin(y) dy = –cos(y)

Step 5: Final Solution

Substitute back:

u(x, y) = x²y + sin(x) – cos(y)

So, the general solution is:

x²y + sin(x) – cos(y) = C

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