Exact Differential Equation

Exact Differential Equation

Komal MiglaniUpdated on 02 Jul 2025, 06:37 PM IST

Let's begin by understanding what differential equations are. A differential equation is an equation involving one or more terms and the derivatives of one dependent variable with respect to the other independent variable. An exact differential equation is a specific type of ordinary differential equation commonly utilized in physics and engineering. Exact differential equations are widely used in thermodynamics for calculating internal energy and also used in electrostatics for calculating electric potential.

This Story also Contains

  1. What is a Differential Equation?
  2. Exact Differential Equation
  3. Exact Differential Equation Integrating Factor
  4. How to Solve Exact Differential Equations
  5. Solved Examples Based On Exact Differential Equations
Exact Differential Equation
Exact Differential Equation

In this article, we will cover the Exact differential equations. This concept falls under the broader category of differential equations. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination (JEE Main), and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more. Over the last ten years of the JEE Main exam (from 2013 to 2023), a total of five questions have been asked on this concept, including one in 2019, one in 2020, two in 2021, and one in 2022.

What is a Differential Equation?

A differential equation is an equation involving one or more terms and the derivatives of one dependent variable with respect to the other independent variable.

Differential equation: $\frac{d y}{d x}=f(x)$
Where " $x$ " is an independent variable and " $y$ " is a dependent variable

Example of differential equation: $x \frac{d y}{d x}+2 y=0$
The above-written equation involves variables as well as the derivative of the dependent variable $\mathrm{y}$ with respect to the independent variable $\mathrm{x}$. Therefore, it is a differential equation.
The following relations are some of the examples of differential equations:
(i) $\frac{d y}{d x}=\sin 2 x+\cos x$
(ii) $\mathrm{k} \frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}=\left[1+\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^2\right]^{3 / 2}$

Commonly Asked Questions

Q: What are some real-world applications where exact differential equations naturally arise?
A:
Exact differential equations often appear in physics and engineering, particularly in:
Q: How does the concept of exactness extend to higher-order differential equations?
A:
The concept of exactness primarily applies to first-order equations. However, some higher-order equations can be reduced to systems of first-order equations, where exactness might be applicable. In general, exactness is less commonly used for higher-order equations.
Q: Can an exact differential equation have a unique solution without an initial condition?
A:
An exact differential equation without an initial condition will have a general solution f(x,y) = C, representing a family of solutions. A unique solution requires an initial condition to determine the specific value of C.
Q: How does the concept of exactness relate to the existence and uniqueness theorem for differential equations?
A:
Exactness ensures the existence of a solution (the potential function). However, uniqueness still requires initial conditions. The existence and uniqueness theorem provides broader conditions for solution existence and uniqueness, while exactness offers a specific method for finding solutions when applicable.
Q: How does the method for solving exact differential equations differ from the method for separable equations?
A:
While both methods aim to find a general solution, they differ in approach:

Exact Differential Equation

The equation $\mathrm{A}(\mathrm{x}, \mathrm{y}) \mathrm{dx}+\mathrm{B}(\mathrm{x}, \mathrm{y}) \mathrm{dy}=0$ is an exact differential equation if there exists a function of two variables $x$ and $y$ having continuous partial derivatives such that the exact differential equation definition is separated as follows

$
u_x(x, y)=A(x, y) \text { and } u_y(x, y)=B(x, y)
$

General Form

The general form of the exact differential equation is $A(x, y) d x+B(x, y) d y=0$ where A and B are the polynomial functions in terms of x and y.

Sometimes some differential equations can be solved using observation only. In such equations, we can get a differential of a function of x and y.

The equation $A(x, y) d x+B(x, y) d y=0$ is an exact differential equation if there exists a function of two variables $x$ and $y$ having continuous partial derivatives such that the exact differential equation definition is separated as follows $u_x(x, y)=A(x, y)$ and $u_y(x, y)=B(x, y)$

General Form
Sometimes some differential equations can be solved using observation only. In such equations, we can get a differential of a function of x and y .

Exact Differential Equation Integrating Factor

If the differential equation $A(x, y) d x+B(x, y) d y=0$ is not exact, it is possible to make it exact by multiplying using a relevant factor $\mathrm{u}(\mathrm{x}, \mathrm{y})$ which is known as integrating factor for the given differential equation.

Consider an example,

$
2 y d x+x d y=0
$

Now check it whether the given differential equation is exact using testing for exactness.
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 x y d x+x^2 d y=0
$

The above resultant equation is an exact differential equation because the left side of the equation is a total differential of $x^2 y$.

Sometimes it is difficult to find the integrating factor. But, there are two classes 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 $A(x, y) d x+B(x, y) d y=0$, the two cases involved are:

Case 1: If $[1 / B(x, y)]\left[A_y(x, y)-B_x(x, y)\right]=h(x)$, which is a function of $x$ alone, then $e^{\int h(x) d x}$ is an integrating factor.

Case 2: If $[1 / A(x, y)]\left[B_x(x, y)-A_y(x, y)\right]=k(y)$, which is a function of $y$ alone, then $e^{j k(y) d y}$ is an integrating factor.

Commonly Asked Questions

Q: What is an integrating factor and when is it used?
A:
An integrating factor is a function that, when multiplied by a non-exact differential equation, makes it exact. It's used when a differential equation is not initially exact but can be made exact through multiplication by a suitable function.
Q: Can all first-order differential equations be made exact?
A:
No, not all first-order differential equations can be made exact. However, many non-exact equations can be transformed into exact equations using an integrating factor. The existence of an integrating factor depends on the specific form of the equation.
Q: What is the relationship between exact differential equations and the total differential in calculus?
A:
An exact differential equation M(x,y)dx + N(x,y)dy = 0 is equivalent to the total differential df = 0, where f is the potential function. This connection highlights why exact equations are called "exact" - they represent the exact differential of some function.
Q: What is the geometric interpretation of an exact differential equation?
A:
Geometrically, the solution to an exact differential equation f(x,y) = C represents a family of level curves in the xy-plane. Each curve corresponds to a different value of C and represents a solution to the differential equation.
Q: Can a differential equation be exact even if it's not in the standard form M(x,y)dx + N(x,y)dy = 0?
A:
Yes, a differential equation can be exact even if it's not initially written in the standard form. It may require rearrangement or multiplication by a constant to bring it into the form M(x,y)dx + N(x,y)dy = 0 before checking for exactness.

How to Solve Exact Differential Equations

The following steps explain how to solve the exact differential equation:

Step 1: The first step to solving the exact differential equation is to make sure the given differential equation is exact.

$
\frac{\partial Q}{\partial x}=\frac{\partial P}{\partial y}
$

Step 2: Write the system of two differential equations that defines the function $u(x, y)$. That is

$
\begin{aligned}
& \frac{\partial u}{\partial x}=P(x, y) \\
& \frac{\partial u}{\partial y}=Q(x, y)
\end{aligned}
$

Step 3: Integrating the first equation over the variable $x$, we get

$
u(x, y)=\int P(x, y) d x+\phi(y)
$

Step 4: Differentiating concerning $y$, substitute the function $u(x, y)$ in the second equation

$
\frac{\partial u}{\partial x}=\frac{\partial}{\partial x}\left[\int P(x, y) d x+\phi(y)\right]=Q(x, y)
$

Step 5: We can find the function $\varphi(y)$ by integrating the last expression so that the function $\mathrm{u}(\mathrm{x}, \mathrm{y})$ becomes

$
u(x, y)=\int P(x, y) d x+\phi(y)
$

Step 6: Finally, the general solution of the exact differential equation is given by

$
u(x, y)=C
$

Illustration 1 :

Solution of the differential equation $2 x y d x+\left(x^2+3 y^2\right) d y=0$ is Let us first separate terms containing only x with dx and terms containing only y with dy

$
2 x y d x+x^2 d y+3 y^2 d y=0
$

Here first two terms have both $x$ and $y$. We can make an observation that first two terms are the differentiation of $x^2 y$. Hence we can write this equation as

$
d\left(x^2 y\right)+3 y^2 d y=0
$

Integrating this, we get

$
x^2 y+y^3+c=0
$
This is the solution of this equation

The presence of the following exact differentials should be observed in a given differential equation

1. $x d y+y d x=d(x y)$
2. $x d x+y d y=\frac{1}{2} d\left(x^2+y^2\right)$
3. $\frac{x d y-y d x}{x^2}=d\left(\frac{y}{x}\right)$
4. $\frac{y d x-x d y}{y^2}=d\left(\frac{x}{y}\right)$
5. $\frac{x d y-y d x}{x y}=\frac{d y}{y}-\frac{d x}{x}=d\left[\log \left(\frac{y}{x}\right)\right]$
6. $\frac{y d x-x d y}{x y}=d\left[\log \left(\frac{x}{y}\right)\right]$

Illustration 2:

$
\frac{x d y-y d x}{x^2+y^2}+e^x d x=0
$

Observe that

$
\frac{x d y-y d x}{x^2+y^2}=\frac{\frac{x d y-y d x}{x^2}}{1+\frac{y^2}{x^2}}=\frac{d\left(\frac{y}{x}\right)}{1+\left(\frac{y}{x}\right)^2}=d\left[\tan ^{-1}\left(\frac{y}{x}\right)\right]
$

So the equation is

$
d\left[\tan ^{-1} \frac{y}{x}\right]+e^x d x=0
$
Integrating

$\tan ^{-1} \frac{y}{x}+e^x+c=0$

Recommended Video Based on Exact Differential Equations


Solved Examples Based On Exact Differential Equations

Example 1: The solution of the differential equation dy/dx = -1 is

Solution:

$
\begin{aligned}
& d(x+y)=d x+d y \\
& \frac{d y}{d x}=-1 \Rightarrow d x+d y=0 \Rightarrow d(x+y)=0 \\
& \Rightarrow \int d(x+y)=C \Rightarrow(x+y)=C
\end{aligned}
$

Hence, the answer is $\mathrm{x}+\mathrm{y}=\mathrm{C}$.
Example 2: The solution of a differential equation $\frac{d y}{d x}=\frac{-\cos x}{\cos y}$ is
Solution:
As we have learned
The general form of Variable Separation -

$
\begin{aligned}
& d(x+y)=d x+d y \\
& \cos x d x+\cos y d y=0 \text { is also a linear differential eq } \\
& d \sin x+d(\sin y)=0 \Rightarrow d(\sin x+\sin y)=0 \\
& \int d(\sin x+\sin y)=C \Rightarrow \sin x+\sin y=C
\end{aligned}
$


$
\cos x d x+\cos y d y=0 \text { is also a linear differential equation that we can write }
$

Hence, the answer is $\sin x+\sin y=C$..

Example 3: Solution of differential equation $\sin x \cos y d y+\sin y \cos x d x$
Solution:
As we have learned
The general form of Variable Separation -

$
d(x y)=y d x+x d y
$

The equation can be written as

$
\begin{aligned}
& (\sin x) d(\sin y)+(\sin y) d(\sin x)=0 \Rightarrow d(\sin x \cdot \sin y)=0 \\
& \Rightarrow \int d(\sin x \cdot \sin y)=C \Rightarrow \sin x \cdot \sin y=C
\end{aligned}
$

Hence, the answer is $\sin x \cdot \sin y=C$.

Example 4: The solution of the differential equation $(2 x y-\sin x) d x+\left(\left(x^2\right)-\cos y\right) d y=0$ is

Solution:
As we have learned

The general form of Variable Separation -

$
d(x y)=y d x+x d y
$

The equation can be written as

$
\begin{aligned}
& 2 x y d x-\sin x d x+x^2 d y-\cos y d y=0 \\
& 2 x y d x+x^2 d y-\sin x d x-\cos y d y=0 \\
& \Rightarrow y(2 x) d x+x^2 d y-\sin x d x-\cos y d y=0 \\
& \Rightarrow d\left(y \cdot x^2\right)+d(\cos x)-d(\sin y)=0 \\
& \Rightarrow \int d\left(y \cdot x^2\right)+\int d(\cos x)-\int d(\sin y)=C \\
& \Rightarrow x^2 y+\cos x-\sin y=C
\end{aligned}
$

Hence, the answer is $x^2 y+\cos x-\sin y=C$

Example 5: The solution of the differential equation $y d x / d y=x+2 \sqrt{ } y^2-x^2$ is
Solution:
As we have learned
The general form of Variable Separation -

$
d\left(\frac{x}{y}\right)=\frac{y d x-x d y}{y^2}
$

The equation can be written as

$
\begin{aligned}
& y d x=x d y+2 \sqrt{y^2-x^2} d y \\
& \Rightarrow y d x-x d y=2 \sqrt{y^2-x^2} d y \\
& \Rightarrow y d x-x d y=2 y \sqrt{1-x^2 / y^2} d y
\end{aligned}
$

dividing both sides by $y^2$ we get

$
\begin{aligned}
& \frac{y d x-x d y}{y^2}=\frac{2 \sqrt{1-(x / y)^2}}{y} d y \\
& \Rightarrow \frac{d(x / y)}{\sqrt{1-(x / y)^2}}=2 / y d y
\end{aligned}
$

on integrating, we get

$
\sin ^{-1} x / y=2 \ln |y|+C
$
Hence, the answer is $\sin ^{-1} x / y=2 \ln |y|+C$ .

Frequently Asked Questions (FAQs)

Q: How does the process of solving exact differential equations relate to finding orthogonal trajectories?
A:
Orthogonal trajectories are curves that intersect the solution curves of a differential equation at right angles. For an exact equation f(x,y) = C, the orthogonal trajectories satisfy the equation -∂f/∂y dx + ∂f/∂x dy = 0. This relationship allows for finding orthogonal trajectories once the exact solution is known.
Q: What is the significance of the Frobenius theorem in relation to exact differential equations?
A:
The Frobenius theorem provides conditions for when a system of differential forms (which includes exact differential equations as a special case) has a solution. It generalizes the concept of exactness to higher dimensions and is crucial in understanding when a system of equations can be solved by finding a potential function or integrating factor.
Q: How does the method of solving exact differential equations compare to the method of characteristics for PDEs?
A:
While exact differential equations are typically used for ODEs, the method of characteristics is used for first-order PDEs. Both methods aim to find a solution by following certain paths: exact equations use level curves of a potential function, while the method of characteristics follows characteristic curves along which the PDE simplifies.
Q: How does the presence of parameters in an exact differential equation affect the solution process?
A:
Parameters in an exact differential equation are treated as constants during the solution process. The general solution will include these parameters, potentially leading to a family of potential functions. Analyzing how the solution changes with respect to these parameters can provide insights into the system's behavior.
Q: How do you approach an exact differential equation where the functions M and N are defined implicitly?
A:
When M and N are defined implicitly, you may need to use implicit differentiation to verify the exactness condition and to perform the necessary integrations. The solution process might involve solving auxiliary equations or using more advanced techniques from implicit function theory.
Q: Can the method for solving exact differential equations be adapted for complex-valued functions?
A:
Yes, the method can be extended to complex-valued functions. The exactness condition and solution process remain similar, but you'll need to work with complex differentials and ensure that the Cauchy-Riemann equations are satisfied for analytic functions.
Q: How does the concept of exactness relate to Hamiltonian systems in classical mechanics?
A:
In Hamiltonian mechanics, the equations of motion form a system of exact differential equations. The Hamiltonian function serves as the potential function, and the symplectic structure of phase space ensures that the system satisfies the exactness condition. This connection highlights the deep relationship between exactness and conservation laws in physics.
Q: What are some common mistakes students make when solving exact differential equations?
A:
Common mistakes include:
Q: How does the concept of exactness extend to differential forms on manifolds?
A:
On manifolds, exact differential equations correspond to exact differential forms. A form ω is exact if there exists a function f such that ω = df. The concept generalizes to higher dimensions, where an n-form ω is exact if there exists an (n-1)-form α such that ω = dα. This extension is crucial in differential geometry and theoretical physics.
Q: Can numerical methods be effectively applied to solve exact differential equations?
A:
While exact differential equations have analytical solutions, numerical methods can be useful for visualizing solutions or handling cases where the integrals are difficult to evaluate analytically. Methods like Runge-Kutta can be applied, but it's important to remember that these numerical solutions approximate the true exact solution.