What makes a differential equation exact?

An equation is exact if . ∂y/∂M = ∂x/∂N

What to do if the differential equation is not exact?

Use an integrating factor to transform the equation into an exact one, then solve.

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Exact Differential Equation

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Exact Differential Equation

1.0What Is an Exact Differential Equation?

An Exact Differential Equation is a type of first-order differential equation that can be written in the form:

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

where M(x, y) and N(x, y) are co ntinuous functions of x and y, and the equation is called exact if there exists a function F(x, y) such that:

$\frac{\partial F}{\partial x} = M (x, y) and \frac{\partial F}{\partial y} = N (x, y)$

Then, the general solution is:

F(x, y) = C

where C is the constant of integration.

2.0Definition of Exact Differential Equation

An equation of the form:

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

is said to be exact if the following condition holds:

$\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$

This condition ensures that M and N are the partial derivatives of some function F(x, y).

3.0Exact vs Non-Exact Differential Equations

Exact Differential Equation	Non-Exact Differential Equation
Satisfies $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$	$\frac{\partial M}{\partial y} \neq = \frac{\partial N}{\partial x}$
Can be solved by finding F(x, y) = C	Requires an integrating factor to make it exact
Example: (2x + 3y) dx + (3x + 4y) dy = 0	Example: $(x + y^{2}) d x + (y - x^{2}) d y = 0$ (non-exact)

4.0Solved Examples on Exact Differential Equation

Example 1: Solve the equation:

(2x + 3y) dx + (3x + 4y) dy = 0

Solution:

Check if exact:

$\frac{\partial N}{\partial x} = \frac{\partial ( 3 x + 4 y )}{\partial x} = 3$

Since they are equal, the equation is exact.

Find F(x, y):

$F (x, y) = \int M d x = \int (2 x + 3 y) d x = x^{2} + 3 x y + g (y)$

Differentiate w.r.t y:

$\frac{\partial F}{\partial y} = 3 x + g^{'} (y)$

But $\frac{\partial F}{\partial y} = N = 3 x + 4 y$
So,

$3 x + g^{'} (y) = 3 x + 4 y ⟹ g^{'} (y) = 4 y$

Integrating:

$g (y) = 2 y^{2}$

Final solution:

$F (x, y) = x^{2} + 3 x y + 2 y^{2} = C$

Example 2: Solve the exact differential equation:

$(3 x^{2} + 2 y) d x + 2 x d y = 0$

Solution:

Step 1 – Identify M(x, y) and N(x, y):

$M (x, y) = 3 x^{2} + 2 y, N (x, y) = 2 x$

Step 2 – Check Exactness:

$\frac{\partial M}{\partial y} = \frac{\partial ( 3 x ^{2} + 2 y )}{\partial y} = 2$

$\frac{\partial N}{\partial x} = \frac{\partial ( 2 x )}{\partial x} = 2$

Since they are equal, the equation is exact.

Step 3 – Find F(x, y):

$F (x, y) = \int M (x, y) d x = \int (3 x^{2} + 2 y) d x = x^{3} + 2 x y + g (y)$

Step 4 – Compute $\frac{\partial F}{\partial y}$ :

$\frac{\partial F}{\partial y} = 2 x + g^{'} (y)$

But $\frac{\partial F}{\partial y} = N (x, y) = 2 x$
Thus:

$2 x + g^{'} (y) = 2 x ⟹ g^{'} (y) = 0$

Integrating:

g(y) = C

Step 5 – General Solution:

$F (x, y) = x^{3} + 2 x y = C$

Example 3: Solve: $(2 x y + cos y) d x + (x^{2} - y sin y) d y = 0$

Solution:

Step 1 – Identify M(x, y) and N(x, y):

$M (x, y) = 2 x y + cos y, N (x, y) = x^{2} - y sin y$

Step 2 – Check Exactness:

$\frac{\partial M}{\partial y} = 2 x - sin y$

$\frac{\partial N}{\partial x} = 2 x$

They are not equal:

$2 x - sin y \neq = 2 x$

This is a non-exact equation.

To solve, we would need an integrating factor, which can be a bit advanced.

Example 4: Solve the equation: $(2 x + 3 y^{2}) d x + (6 x y + cos y) d y = 0$

Solution:

Step 1 – Identify M(x, y) and N(x, y):

$M (x, y) = 2 x + 3 y^{2}, N (x, y) = 6 x y + cos y$

Step 2 – Check Exactness:

$\frac{\partial M}{\partial y} = 6 y$

$\frac{\partial N}{\partial x} = 6 y$

They are equal, so the equation is exact.

Step 3 – Find F(x, y):

$F (x, y) = \int M (x, y) d x = \int (2 x + 3 y^{2}) d x = x^{2} + 3 x y^{2} + g (y)$

Step 4 – Compute $\frac{\partial F}{\partial y}$ :

$\frac{\partial F}{\partial y} = 6 x y + g^{'} (y)$

But $\frac{\partial F}{\partial y} = N (x, y) = 6 x y + cos y$

So:

$6 x y + g^{'} (y) = 6 x y + cos y ⟹ g^{'} (y) = cos y$

Integrating:

$g (y) = sin y$

Step 5 – Final solution:

$F (x, y) = x^{2} + 3 x y^{2} + sin y = C$

Example 5: Solve: $(y e^{x} + 2 x) d x + (e^{x} + 3 y^{2}) d y = 0$

Solution:

Step 1 – Identify M(x, y) and N(x, y):

$M (x, y) = y e^{x} + 2 x, N (x, y) = e^{x} + 3 y^{2}$

Step 2 – Check Exactness:

$\frac{\partial M}{\partial y} = e^{x}$

$\frac{\partial N}{\partial x} = e^{x}$

Since they are equal, the equation is exact.

Step 3 – Find F(x, y):

$F (x, y) = \int M (x, y) d x = \int (y e^{x} + 2 x) d x = y e^{x} + x^{2} + g (y)$

Step 4 – Compute $\frac{\partial F}{\partial y}$ :

$\frac{\partial F}{\partial y} = e^{x} + g^{'} (y)$

But $\frac{\partial F}{\partial y} = N (x, y) = e^{x} + g ’ y^{2}$

So:

$e^{x} + g^{'} (y) = e^{x} + 3 y^{2} ⟹ g^{'} (y) = 3 y^{2}$

Integrating:

$g (y) = y^{3}$

Step 5 – Final solution:

$F (x, y) = y e^{x} + x^{2} + y^{3} = C$

Example 6: Solve the equation: $(3 x^{2} y + 2 y^{3}) d x + (x^{3} + 6 x y^{2}) d y = 0$

Solution:

Step 1 – Identify M(x, y) and N(x, y):

$M (x, y) = 3 x^{2} y + 2 y^{3}, N (x, y) = x^{3} + 6 x y^{2}$

Step 2 – Check Exactness:

$\frac{\partial M}{\partial y} = 3 x^{2} + 6 y^{2}$

$\frac{\partial N}{\partial x} = 3 x^{2} + 6 y^{2}$

They are equal ⇒ Exact equation.

Step 3 – Find F(x, y):

$F (x, y) = \int M (x, y) d x = \int (3 x^{2} y + 2 y^{3}) d x = x^{3} y + 2 x y^{3} + g (y)$

Step 4 – Compute $\frac{\partial F}{\partial y}$ :

$\frac{\partial F}{\partial y} = x^{3} + 6 x y^{2} + g^{'} (y)$

But $\frac{\partial F}{\partial y} = N (x, y) = x^{3} + 6 x y^{2}$

Thus:

$g^{'} (y) = 0 ⟹ g (y) = C$

Step 5 – General solution:

$F (x, y) = x^{3} y + 2 x y^{3} = C$

Example 7: Solve: $(e^{x} sin y + 1) d x + (e^{x} cos y + 2 y) d y = 0$

Solution:

Step 1 – Identify M(x, y) and N(x, y):

$M (x, y) = e^{x} sin y + 1, N (x, y) = e^{x} cos y + 2 y$

Step 2 – Check Exactness:

$\frac{\partial M}{\partial y} = e^{x} cos y$

$\frac{\partial N}{\partial x} = e^{x} cos y$

They are equal ⇒ Exact equation.

Step 3 – Find F(x, y):

$F (x, y) = \int M (x, y) d x = \int (e^{x} sin y + 1) d x = e^{x} sin y + x + g (y)$

Step 4 – Compute $\frac{\partial F}{\partial y}$ :

$\frac{\partial F}{\partial y} = e^{x} cos y + g^{'} (y)$

But

So: $\frac{\partial F}{\partial y} = N (x, y) = e^{x} cos y + 2 y .$

$e^{x} cos y + g^{'} (y) = e^{x} cos y + 2 y ⟹ g^{'} (y) = 2 y$

Integrating:

$g (y) = y^{2}$

Step 5 – Final solution:

$F (x, y) = e^{x} sin y + x + y^{2} = C$

5.0Practice Questions on Exact Differential Equations

Solve the following exact differential equation:

$(y + x^{2}) d x + x d y = 0$

Determine if the following differential equation is exact or not:

$(x sin y + 2) d x + x cos y d y = 0$

Solve the exact differential equation:

$(e^{x} + y) d x + (x + e^{y}) d y = 0$

6.0Why Is Exact Differential Equation Important?

Simplifies solving differential equations in physics, engineering, and mathematics.
Helps in solving problems involving conservative forces and potential functions.
Vital in JEE and advanced-level problems requiring step-by-step methodical solutions.

Also Read:

Differential Equations	Derivative Function Calculus	Differentiability
First Derivative Test	Continuity and Differentiability	Integrals of Particular Functions
Differentiation and Integration	First order Differential Equation	Logarithmic Functions

Join ALLEN!

(Session 2026 - 27)

Name

Mobile Number

Class

Choose class

Goal

Choose your goal

Preferred Programs

Preferred Mode

State

Choose State

I agree to

I authorise ALLEN Career Institute Pvt Ltd to send me regular updates via Phone calls, Whatsapp, SMS, Robocalls (Automated Calls), Emails, or on postal address.

Exact Differential Equation

1.0What Is an Exact Differential Equation?

An Exact Differential Equation is a type of first-order differential equation that can be written in the form:

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

where M(x, y) and N(x, y) are co ntinuous functions of x and y, and the equation is called exact if there exists a function F(x, y) such that:

$\frac{\partial F}{\partial x} = M (x, y) and \frac{\partial F}{\partial y} = N (x, y)$

Then, the general solution is:

F(x, y) = C

where C is the constant of integration.

2.0Definition of Exact Differential Equation

An equation of the form:

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

is said to be exact if the following condition holds:

$\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$

This condition ensures that M and N are the partial derivatives of some function F(x, y).

3.0Exact vs Non-Exact Differential Equations

Exact Differential Equation	Non-Exact Differential Equation
Satisfies $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$	$\frac{\partial M}{\partial y} \neq = \frac{\partial N}{\partial x}$
Can be solved by finding F(x, y) = C	Requires an integrating factor to make it exact
Example: (2x + 3y) dx + (3x + 4y) dy = 0	Example: $(x + y^{2}) d x + (y - x^{2}) d y = 0$ (non-exact)

4.0Solved Examples on Exact Differential Equation

Example 1: Solve the equation:

(2x + 3y) dx + (3x + 4y) dy = 0

Solution:

Check if exact:

$\frac{\partial N}{\partial x} = \frac{\partial ( 3 x + 4 y )}{\partial x} = 3$

Since they are equal, the equation is exact.

Find F(x, y):

$F (x, y) = \int M d x = \int (2 x + 3 y) d x = x^{2} + 3 x y + g (y)$

Differentiate w.r.t y:

$\frac{\partial F}{\partial y} = 3 x + g^{'} (y)$

But $\frac{\partial F}{\partial y} = N = 3 x + 4 y$
So,

$3 x + g^{'} (y) = 3 x + 4 y ⟹ g^{'} (y) = 4 y$

Integrating:

$g (y) = 2 y^{2}$

Final solution:

$F (x, y) = x^{2} + 3 x y + 2 y^{2} = C$

Example 2: Solve the exact differential equation:

$(3 x^{2} + 2 y) d x + 2 x d y = 0$

Solution:

Step 1 – Identify M(x, y) and N(x, y):

$M (x, y) = 3 x^{2} + 2 y, N (x, y) = 2 x$

Step 2 – Check Exactness:

$\frac{\partial M}{\partial y} = \frac{\partial ( 3 x ^{2} + 2 y )}{\partial y} = 2$

$\frac{\partial N}{\partial x} = \frac{\partial ( 2 x )}{\partial x} = 2$

Since they are equal, the equation is exact.

Step 3 – Find F(x, y):

$F (x, y) = \int M (x, y) d x = \int (3 x^{2} + 2 y) d x = x^{3} + 2 x y + g (y)$

Step 4 – Compute $\frac{\partial F}{\partial y}$ :

$\frac{\partial F}{\partial y} = 2 x + g^{'} (y)$

But $\frac{\partial F}{\partial y} = N (x, y) = 2 x$
Thus:

$2 x + g^{'} (y) = 2 x ⟹ g^{'} (y) = 0$

Integrating:

g(y) = C

Step 5 – General Solution:

$F (x, y) = x^{3} + 2 x y = C$

Example 3: Solve: $(2 x y + cos y) d x + (x^{2} - y sin y) d y = 0$

Solution:

Step 1 – Identify M(x, y) and N(x, y):

$M (x, y) = 2 x y + cos y, N (x, y) = x^{2} - y sin y$

Step 2 – Check Exactness:

$\frac{\partial M}{\partial y} = 2 x - sin y$

$\frac{\partial N}{\partial x} = 2 x$

They are not equal:

$2 x - sin y \neq = 2 x$

This is a non-exact equation.

To solve, we would need an integrating factor, which can be a bit advanced.

Example 4: Solve the equation: $(2 x + 3 y^{2}) d x + (6 x y + cos y) d y = 0$

Solution:

Step 1 – Identify M(x, y) and N(x, y):

$M (x, y) = 2 x + 3 y^{2}, N (x, y) = 6 x y + cos y$

Step 2 – Check Exactness:

$\frac{\partial M}{\partial y} = 6 y$

$\frac{\partial N}{\partial x} = 6 y$

They are equal, so the equation is exact.

Step 3 – Find F(x, y):

$F (x, y) = \int M (x, y) d x = \int (2 x + 3 y^{2}) d x = x^{2} + 3 x y^{2} + g (y)$

Step 4 – Compute $\frac{\partial F}{\partial y}$ :

$\frac{\partial F}{\partial y} = 6 x y + g^{'} (y)$

But $\frac{\partial F}{\partial y} = N (x, y) = 6 x y + cos y$

So:

$6 x y + g^{'} (y) = 6 x y + cos y ⟹ g^{'} (y) = cos y$

Integrating:

$g (y) = sin y$

Step 5 – Final solution:

$F (x, y) = x^{2} + 3 x y^{2} + sin y = C$

Example 5: Solve: $(y e^{x} + 2 x) d x + (e^{x} + 3 y^{2}) d y = 0$

Solution:

Step 1 – Identify M(x, y) and N(x, y):

$M (x, y) = y e^{x} + 2 x, N (x, y) = e^{x} + 3 y^{2}$

Step 2 – Check Exactness:

$\frac{\partial M}{\partial y} = e^{x}$

$\frac{\partial N}{\partial x} = e^{x}$

Since they are equal, the equation is exact.

Step 3 – Find F(x, y):

$F (x, y) = \int M (x, y) d x = \int (y e^{x} + 2 x) d x = y e^{x} + x^{2} + g (y)$

Step 4 – Compute $\frac{\partial F}{\partial y}$ :

$\frac{\partial F}{\partial y} = e^{x} + g^{'} (y)$

But $\frac{\partial F}{\partial y} = N (x, y) = e^{x} + g ’ y^{2}$

So:

$e^{x} + g^{'} (y) = e^{x} + 3 y^{2} ⟹ g^{'} (y) = 3 y^{2}$

Integrating:

$g (y) = y^{3}$

Step 5 – Final solution:

$F (x, y) = y e^{x} + x^{2} + y^{3} = C$

Example 6: Solve the equation: $(3 x^{2} y + 2 y^{3}) d x + (x^{3} + 6 x y^{2}) d y = 0$

Solution:

Step 1 – Identify M(x, y) and N(x, y):

$M (x, y) = 3 x^{2} y + 2 y^{3}, N (x, y) = x^{3} + 6 x y^{2}$

Step 2 – Check Exactness:

$\frac{\partial M}{\partial y} = 3 x^{2} + 6 y^{2}$

$\frac{\partial N}{\partial x} = 3 x^{2} + 6 y^{2}$

They are equal ⇒ Exact equation.

Step 3 – Find F(x, y):

$F (x, y) = \int M (x, y) d x = \int (3 x^{2} y + 2 y^{3}) d x = x^{3} y + 2 x y^{3} + g (y)$

Step 4 – Compute $\frac{\partial F}{\partial y}$ :

$\frac{\partial F}{\partial y} = x^{3} + 6 x y^{2} + g^{'} (y)$

But $\frac{\partial F}{\partial y} = N (x, y) = x^{3} + 6 x y^{2}$

Thus:

$g^{'} (y) = 0 ⟹ g (y) = C$

Step 5 – General solution:

$F (x, y) = x^{3} y + 2 x y^{3} = C$

Example 7: Solve: $(e^{x} sin y + 1) d x + (e^{x} cos y + 2 y) d y = 0$

Solution:

Step 1 – Identify M(x, y) and N(x, y):

$M (x, y) = e^{x} sin y + 1, N (x, y) = e^{x} cos y + 2 y$

Step 2 – Check Exactness:

$\frac{\partial M}{\partial y} = e^{x} cos y$

$\frac{\partial N}{\partial x} = e^{x} cos y$

They are equal ⇒ Exact equation.

Step 3 – Find F(x, y):

$F (x, y) = \int M (x, y) d x = \int (e^{x} sin y + 1) d x = e^{x} sin y + x + g (y)$

Step 4 – Compute $\frac{\partial F}{\partial y}$ :

$\frac{\partial F}{\partial y} = e^{x} cos y + g^{'} (y)$

But

So: $\frac{\partial F}{\partial y} = N (x, y) = e^{x} cos y + 2 y .$

$e^{x} cos y + g^{'} (y) = e^{x} cos y + 2 y ⟹ g^{'} (y) = 2 y$

Integrating:

$g (y) = y^{2}$

Step 5 – Final solution:

$F (x, y) = e^{x} sin y + x + y^{2} = C$

5.0Practice Questions on Exact Differential Equations

Solve the following exact differential equation:

$(y + x^{2}) d x + x d y = 0$

Determine if the following differential equation is exact or not:

$(x sin y + 2) d x + x cos y d y = 0$

Solve the exact differential equation:

$(e^{x} + y) d x + (x + e^{y}) d y = 0$

6.0Why Is Exact Differential Equation Important?

Simplifies solving differential equations in physics, engineering, and mathematics.
Helps in solving problems involving conservative forces and potential functions.
Vital in JEE and advanced-level problems requiring step-by-step methodical solutions.

Also Read:

Differential Equations	Derivative Function Calculus	Differentiability
First Derivative Test	Continuity and Differentiability	Integrals of Particular Functions
Differentiation and Integration	First order Differential Equation	Logarithmic Functions

Frequently Asked Questions

What makes a differential equation exact?

What to do if the differential equation is not exact?

Give an example of an exact differential equation.

Frequently Asked Questions

What makes a differential equation exact?

What to do if the differential equation is not exact?

Give an example of an exact differential equation.

Join ALLEN!

Exact Differential Equation

1.0What Is an Exact Differential Equation?

2.0Definition of Exact Differential Equation

3.0Exact vs Non-Exact Differential Equations

4.0Solved Examples on Exact Differential Equation

5.0Practice Questions on Exact Differential Equations

6.0Why Is Exact Differential Equation Important?

Table of Contents

Join ALLEN!

Exact Differential Equation

1.0What Is an Exact Differential Equation?

2.0Definition of Exact Differential Equation

3.0Exact vs Non-Exact Differential Equations

4.0Solved Examples on Exact Differential Equation

5.0Practice Questions on Exact Differential Equations

6.0Why Is Exact Differential Equation Important?

Table of Contents