Exact Differential Equation

Exact Differential Equation Definition:

Exact differential equation definition 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). We can represent the differential equation for a given function 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 basically discuss what is the exact differential equation, the standard form,what is the integrating factor, and how to solve exact differential equations in detail with various examples and solved problems.

What is Differential Equation?

A differential equation is an equation which contains one or more terms which involve the derivatives of one variable (i.e., dependent variable) with respect to the other variable (i.e., independent variable)

dy/dx = f(x)

Here “x” is referred to as an independent variable and “y” is known as an dependent variable

For example, dy/dx = 5x

A differential equation contains derivatives which can either be partial derivatives or can be ordinary derivatives. The derivatives represent a rate of change, and the differential equation describes a relationship between the quantity that is continuously varying and the speed of change. There are a lot of differential equation formulas to find the solution of the derivatives.

Exact differential equations are not generally linear. In other words, this can be defined as a method for solving the first-order nonlinear differential equations. 

• The exact differential equation solution can be in the implicit form F(x, y) which is equal to C. 

• Although this is a distinct class of differential equations, it will share many similarities with first-order linear differential equations. Importantly, we will discover that there is often (although not always!) an integration factor required to make a differential equation in the “exact” form. The integration factor we have discussed will take a different form than that of first-order linear equations . 

Exact Differential Equation:

Let us consider the equation P(x, y)dx + Q(x, y)dy equal to 0. Suppose that there exists a function v(x, y) such that dv = Mdx + Ndy, then the differential equation is said to be an exact differential equation solution is given by v(x, y) = c.

Theorem

If P, Q, \[\frac{∂P}{∂y}\] , \[\frac{∂Q}{∂x}\] are continuous functions of x and y, then a necessary and sufficient exact differential equation condition that

Pdx + Qdy = 0 ----🡪(1)

is exact, is that \[\frac{∂P}{∂y}\] = \[\frac{∂Q}{∂x}\]

Proof of the Theorem

Suppose that (1) is exact.

Then by definition there exists a v such that

dv = Pdx + Qdy

But from total derivative formula,

dv = \[\frac{∂v}{∂x}\] dx + \[\frac{∂v}{∂y}\] dy

comparing the above equations we get P = \[\frac{∂v}{∂x}\] ,Q = \[\frac{∂v}{∂y}\]

These two equations lead to \[\frac{∂P}{∂y}\] = \[\frac{∂²v}{∂y∂x}\] and  \[\frac{∂P}{∂y}\] = \[\frac{∂²v}{∂x∂y}\] 

Since \[\frac{∂²v}{∂y∂x}\] = \[\frac{∂²v}{∂x∂y}\] we get  \[\frac{∂P}{∂y}\] =  \[\frac{∂Q}{∂y}\] 

Let Ø(x,y)  be a function for which \[\frac{∂Ø}{∂x}\] = P(x,y)

Then \[\frac{∂²Ø}{∂y∂x}\] = \[\frac{∂P}{∂y}\] ; => \[\frac{∂²Ø}{∂x∂y}\] = \[\frac{∂²Ø}{∂y∂x}\] = \[\frac{∂P}{∂y}\] = \[\frac{∂Q}{∂y}\]

On integrating both side of this equation with respect to x holding y fixed we get,

\[\frac{∂Ø}{∂y}\] = Q + A(y), where A(y) is an arbitrary function of y

Now define a function v as v(x,y) = Ø(x,y) - ∫A(y)dy

Then dv = \[\frac{∂v}{∂x}\]dx + \[\frac{∂v}{∂y}\]dy = \[\frac{∂Ø}{∂x}\]dx + (\[\frac{∂Ø}{∂y}\] - A(y))dy = Pdx + Qdy 

Hence the given equation is exact.

Testing for Exactness (Exact Differential Equation Condition)

Let’s assume function P(x,y) and function Q(x,y) having the continuous partial derivatives in a particular domain named D, the differential equation is an exact differential equation condition if and only if it satisfies the following condition:

 \[\frac{∂P}{∂y}\] =  \[\frac{∂Q}{∂x}\]

Exact Differential Equation Examples

Some of the examples of the exact differential equations are given below:

• ( 2xy – 3x²) dx + ( x² – 2y ) dy equals 0

• ( xy² + x ) dx + yx² dy equals 0

• Cos y dx + ( y² – x sin y ) dy equals 0

• ( 6x² – y +3 ) dx + (3y² – x – 2) dy equals 0

• e\[^{y}\] dx + ( 2y + xe\[^{y}\] ) dy equals 0

Exact Differential Equation Questions:

Question 1) Solve 3x(xy -2)dx + (x2 + 2y)dy = 0 

exact differential equation solution) 

P = 3x(xy -2), Q = (x2 + 2y)

First \[\frac{∂P}{∂y}\] = 3x²  and   \[\frac{∂Q}{∂x}\] = 3x² 

i.e. ,\[\frac{∂P}{∂y}\] = \[\frac{∂Q}{∂x}\]. Also the functions and derivatives are continuous. Hence the equation is exact.

Therefore the solution is given by

∫P(x,y)dx + ∫(terms in Q(x,y) not containing x)dy = C

∫(3x²y - 6x)dx + ∫2ydy = C

Or, x³y - 3x² + y² = C