A differential equation is an equation which contains one or more terms. It involves the derivative of one variable (dependent variable) with respect to the other variable (independent variable). The differential equation for a given function can be represented in a form: f(x) = dy/dx where “x” is an independent variable and “y” is a dependent variable. In this article, we are going to discuss what is exact differential equation, standard form, integrating factor, and how to solve exact differential equation in detail with examples and solved problems.

## Exact Differential Equation Definition

The equation P (x,y) dx + Q (x,y) dy=0 is an exact differential equation if there exists a function f 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) = p(x,y) and u_{y }(x,y) = Q(x,y);

Therefore, the general solution of the equation is u(x,y) = C.

Where C is an arbitrary constant.

## Testing for Exactness

Assume the functions P(x,y) and Q(x,y) having the continuous partial derivatives in a particular domain D, the differential equation is an exact differential equation if and only if it satisfies the condition

\(\frac{\partial Q}{\partial x}=\frac{\partial P}{\partial y}\)## Integrating Factor

If the differential equation P (x,y) dx + Q (x,y) dy = 0 Â is not exact, it is possible to make it exact by multiplying using an appropriate factor u(x,y) which is known as integrating factor for the given differential equation.

Consider an example,

2ydx + x dy = 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,

Â 2 xy dx + x^{2}Â dy = 0

The above resultant equation is exact differential equation because the left side of the equation is a total differential of Â x^{2}y.

Sometimes it is difficult to find the integrating factor. But, there are two classes of differential equation 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 Â P (x,y) dx + Q (x,y) dy=0, the two cases involved are:

**Case 1: **If Â \(\frac{1}{Q (x,y)}[P_{y}(x,y)-Q_{x}(x,y)]= h(x)\), which is a function of x alone, then \(e^{\int h(x)dx}\) Â is an integrating factor

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

## Procedure to Solve Exact Differential Equation

Step 1: The first step to solve exact differential equation is that to make sure with the given differential equation is exact using testing for exactness.

\(\frac{\partial Q}{\partial x}=\frac{\partial P}{\partial y}\)Step 2: Write the system of two differential equations that defines the function u(x,y). That is

\(\frac{\partial u}{\partial x}=P(x,y)\) \(\frac{\partial u}{\partial y}=Q(x,y)\)Step 3: Integrating the first equation over the variable x , we get

\(u(x,y)= \int P(x,y)dx + \varphi (y)\)Instead of an arbitrary constant C, write an unknown function of y.

Step 4: Differentiating with respect to y, substitute the function u(x,y) in the second equation

\(\frac{\partial u}{\partial y}= \frac{\partial }{\partial y}\left \lfloor \int P(x,y)dx+\varphi (y) \right \rfloor=Q(x,y)\)From the above expression we get the derivative of the unknown function \(\varphi (y)\) and it is given by

\(\varphi (y)=Q(x,y)-\frac{\partial }{\partial y}\left ( \int P(x,y)dx \right )\)Step 5: We can find the function \(\varphi (y)\) by integrating the last expression ,so that the function u(x,y) becomes

\(u(x,y)=\int P(x,y)dx+\varphi (y)\)Step 6: Finally, the general solution of the exact differential equation is given by

u (x,y) = C.

## Exact Differential Equation Examples

Some of the examples of the exact differential equations are as follows :

- ( 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

### Exact Differential Equation Problems

Go through the below problem to solve the exact differential equation.

**Question**: Find the solution for the differential equation ( 2xy – sin x ) dx + ( x^{2} – cos y) dy = 0

**Solution:**

Given, ( 2xy – sin x ) dx + ( x^{2} – cos y) dy = 0

First check this equation for exactness,

\(\frac{\partial Q}{\partial x}=\frac{\partial }{\partial x}(x^{2}-\cos y)=2x\) \(\frac{\partial P}{\partial y}=\frac{\partial }{\partial y}(2xy-\sin x)=2x\)The equation is exact because it satisfies the condition

\(\frac{\partial Q}{\partial x}=\frac{\partial P}{\partial y}\)From the system of two equations, find the functions u ( x, y )

\(\frac{\partial u}{\partial x}=2xy-\sin x\) â€¦.(1) \(\frac{\partial u}{\partial y}=x^{2}-\cos y\) â€¦.(2)By integrating the first equation with respect to the variable x, we get

\(u(x,y)= \int (2xy-\sin x)dx=x^{2}+\cos x+\varphi (y)\)Substituting the above equation in equation (2) , it becomes

\(\frac{\partial u}{\partial y}=\frac{\partial }{\partial y}[x^{2}y+\cos x+\varphi (y)]=x^{2}-\cos y\) \(\Rightarrow ^{2}+\varphi (y)=x^{2}-cos y\)We get, \(\Rightarrow \varphi (y)=-cos y\)

Hence, \(\varphi (y)=\int (-\cos y)dy=-\sin y\)

So the function u ( x , y ) becomes

u ( x , y ) = x^{2}y + cos x – sin y

Therefore the general solution for the given differential equation is

x^{2}y + cos x – sin y = C

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