Exact Differential Equation

An equation with one or more terms is known as a differential equation. It entails calculating the derivative of one variable (dependent variable) in relation to another one (independent variable). The differential equation for a given function is f(x) = dy/dx, where “x” is the independent variable and “y” is the dependent variable. We will go through what an exact differential equation is, standard form, integrating factor, and how to solve exact differential equations in detail with examples and solved problems in this post.

P (x,y) dx + Q (x,y) dy=0 P (x,y) dx + Q (x,y) dy=0 If a function f of two variables x and y with continuous partial derivatives exists, the exact differential equation definition is split as follows. ux(x, y) = p(x, y) and uy (x, y) = Q(x, y); ux(x, y) = p(x, y); As a result, u(x, y) = C is the general solution of the equation. “C” stands for an arbitrary constant.

EXACTNESS CHECKING:

Assume that the functions P(x, y) and Q(x, y) have continuous partial derivatives in a certain domain D, and that the differential equation is exact if and only if the condition is met, ∂Q/∂x = ∂P/∂y.

INTEGRATING FACTOR FOR EXACT DIFFERENTIAL EQUATIONS:

If the differential equation P (x, y) dx + Q (x, y) dy = 0 is not exact, it can be made precise by multiplying it with a relevant factor u(x, y), also known as the integrating factor for the given differential equation.

Consider the following scenario:

0 = 2ydx + xdy

Now use exactness checking to see if the given differential equation is correct.

The differential equation provided is not accurate.

Multiply the differential equation by the integrating factor u(x,y)= x to get the precise differential equation.

0 = 2 xy dx + x2 dy

Because the left side of the equation is a complete differential of x2y, the resultant equation is an exact differential equation.

EXACT DIFFERENTIAL EQUATIONS: HOW TO SOLVE THEM

The procedures that follow will show you how to solve the exact differential equation in detail.

Step 1: To solve an exact differential equation, the first step is to ensure that the provided differential equation is exact by performing an exactness test,

∂Q/∂x = ∂P/∂y. 

Step 2: Define the function u using a system of two differential equations (x,y). That is 

∂u/∂x = P(x,y)

∂u/∂y = Q(x,y)

Step 3: Taking the first equation and integrating it over the variable x, we get 

u(x,y) = ∫ P(x,y)dx + ϕ(y)

Write an unknown function of y instead of an arbitrary constant C.

Step 4: Substitute the function u(x,y) in the second equation, differentiating with respect to y.

∂u/∂y =∂/∂y [∫ P(x,y)dx + ϕ(y)] = Q(x,y)

The derivative of the unknown function ϕ(y) is obtained from the above statement and it’s provided by

ϕ(y) = Q(x,y) – ∂/∂y [∫ P(x,y)dx]

Step 5: We’ll look for the function ϕ(y) by integrating the last expression, resulting in the function 

u(x,y) = ∫ P(x,y)dx + ϕ(y)

Step 6: Finally, the exact differential equation’s general solution is provided by C = u (x,y).

REAL LIFE EXAMPLES OF EXACT DIFFERENTIAL EQUATION:

Some examples of the exact differential equations are:

Q1. ( 2xy – 3x2 ) dx + ( x2 – 2y ) dy = 0

Q2. ( xy2 + x ) dx + yx2 dy = 0

Q3. Cos y dx + ( y2 – x sin y ) dy = 0

Q4. ( 6x2 – y +3 ) dx + (3y2 -x – 2) dy =0

Q5. ey dx + ( 2y + xey ) dy = 0

Some solved examples are given below:

Q1. ( 2xy – sin x ) dx + ( x2 – cos y) dy = 0

Solution:

Given, ( 2xy – sin x ) dx + ( x2 – cos y) dy = 0

∂Q/∂x = 2x

∂P/∂y = 2x

The equation is thus exact because it satisfies the condition, ∂Q/∂x = ∂P/∂y.

Now, let a function u ( x, y )

∂u/∂x = 2xy – sin x   ….(1)

∂u/∂y = x2 – cos y  ….(2)

By integrating equation (1) with respect to x, we get

u(x,y) = ∫( 2xy – sin x )dx = x2 + cos x + ϕ(y)

Substituting the above equation in equation (2) , it becomes

∂u/∂y = ∂/∂y [x2y + cos x + ϕ(y)] = x2 – cos y

x2 + ϕ(y) = x2 – cos y 

So we get,

ϕ(y) = – cos y

Hence,

ϕ(y) = ∫( – cos y )dy = -siny

Therefore the function, u ( x , y ) = x2y + cos x – sin y

Thus the general solution for the given differential equation is

x2y + cos x – sin y = C

Q2: Solve: (3x2y3 – 5x4)dx + (y + 3x3y2)dy = 0

Solution: 

In this question M(x, y) = 3x2y3 – 5x4 & N(x, y) = y + 3x3y2 

∂M/∂y = 6x2y2

∂N/∂x =6x2y2

Since both the values are same, the equation is exact. 

Now, 

I(x,y) = ∫ M(x,y)dx

          = ∫(3x2y3 – 5x4)dx

          = x3y3 – x5 + f(y)

We know, ∂I/∂y = N(x,y)

  3x3y2 +df/dy = y + 3x3y2

  df/dy = y

  df = ydy

               ∫df = ∫ydy

  f = y2/2 + C

put value of f(y) in the general solution, 

I(x,y) =x3y2 –  x5+y2/2  + C

Q3: Solve the differential equation: (x2+3y2)dy + 2xydx = 0

Solution:

M(x, y) = 2xy, N(x, y) =x2+3y2

dm/dy = 2x

dn/dx = 2x

so, the equation is exact.

Now, I(x, y)= ∫M(x,y)dx = ∫(2xy)dx = x2y + C

dI/dy = N(x,y)

Putting I(x, y) we get:

d(x2y + C)/dy = x2+3y2

x2 + dC/dy = x2+3y2

dC/dy = 3y2

dC = ∫ 3y2dy

C = y3 + c

Put the value of C in I(x, y):

I(x, y)= x2y+y3+c

This is the general solution for the given exact differential equation.

CONCLUSION:

A differential equation is a mathematical equation that connects one or more unknown functions with their derivatives. In many applications, the functions are used to represent many physical quantities and their derivatives are used to describe their rates of change with respect to some variable, and the differential equation is used to define the relationship between them and how they vary.