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:

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

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

4.0Solved Examples on Exact Differential Equation 

Example 1: Solve the equation:

(2x + 3y) dx + (3x + 4y) dy = 0 

Solution:

1. Check if exact:

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

Since they are equal, the equation is exact.

2. Find F(x, y):

$F(x,y)=\int Mdx=\int (2x+3y)dx=x^2+3xy+g(y)$

Differentiate w.r.t y:

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

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

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

Integrating:

$g(y)=2y^2$

3. Final solution:

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

Example 2: Solve the exact differential equation:

$(3x^2+2y)dx+2xdy=0$

Solution:

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

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

Step 2 – Check Exactness:

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

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

Since they are equal, the equation is exact.

Step 3 – Find F(x, y):

$F(x,y)=\int M(x,y)dx=\int (3x^2+2y)dx=x^3+2xy+g(y)$

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

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

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

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

Integrating:

g(y) = C 

Step 5 – General Solution:

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

Example 3: Solve: $(2xy+\cos y)dx+(x^2-y\sin y)dy=0$

Solution:

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

$M(x,y)=2xy+\cos y,N(x,y)=x^2-y\sin y$

Step 2 – Check Exactness:

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

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

They are not equal:

$2x-\sin y\ne 2x$

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: $(2x+3y^2)dx+(6xy+\cos y)dy=0$

Solution:

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

$M(x,y)=2x+3y^2,N(x,y)=6xy+\cos y$

Step 2 – Check Exactness:

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

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

They are equal, so the equation is exact.

Step 3 – Find F(x, y):

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

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

$\frac{\partial F}{\partial y}=6xy+g^{\prime }(y)$

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

So:

$6xy+g^{\prime }(y)=6xy+\cos y⟹g^{\prime }(y)=\cos y$

Integrating:

$g(y)=\sin y$

Step 5 – Final solution:

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

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

Solution:

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

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

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):

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

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

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

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

So:

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

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: $(3x^{2}y+2y^3)dx+(x^3+6x{y}^2)dy=0$ 

Solution:

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

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

Step 2 – Check Exactness:

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

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

They are equal ⇒ Exact equation.

Step 3 – Find F(x, y):

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

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

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

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

Thus:

$g^{\prime }(y)=0⟹g(y)=C$

Step 5 – General solution:

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

Example 7: Solve: $(e^x\sin y+1)dx+(e^x\cos y+2y)dy=0$

Solution:

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

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

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):

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

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

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

But

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

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

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

1. Solve the following exact differential equation:

$(y+x^2)dx+xdy=0$

2. Determine if the following differential equation is exact or not:

$(x\sin y+2)dx+x\cos ydy=0$

3. Solve the exact differential equation:

$(e^x+y)dx+(x+e^y)dy=0$

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: