Exact Differential Equation

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

ux(x,y) = p(x,y) and uy (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

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

Procedures 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 – 3x2 ) dx + ( x2 – 2y ) dy = 0
  • ( xy2 + x ) dx + yx2 dy = 0
  • Cos y dx + ( y2 – x sin y ) dy = 0
  • ( 6x2 – y +3 ) dx + (3y2 -x – 2) dy =0
  • ey dx + ( 2y + xey ) dy = 0

Exact Differential Equation Solved Problem


Find the solution for the differential equation ( 2xy – sin x ) dx + ( x2 – cos y) dy = 0


Given, ( 2xy – sin x ) dx + ( x2 – 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 )

\(\frac{\partial u}{\partial x}=2xy-\sin x\) ….(1)

\(\frac{\partial u}{\partial y}=x^{2}-\cos y\) ….(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 ) = x2y + cos x – sin y

Therefore the general solution for the given differential equation is

x2y + cos x – sin y = C

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