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
FAQs on Exact Differential Equation
1. What is an exact differential equation in simple terms?
An exact differential equation is a type of first-order differential equation that represents the total differential of some function f(x, y). Think of it as an equation that can be formed by directly differentiating a function. It's typically written in the form M(x, y)dx + N(x, y)dy = 0.
2. How can you tell if a differential equation is exact?
There's a simple test to check if an equation in the form M(x, y)dx + N(x, y)dy = 0 is exact. You need to find the partial derivative of M with respect to y (∂M/∂y) and the partial derivative of N with respect to x (∂N/∂x). If these two partial derivatives are equal (∂M/∂y = ∂N/∂x), the equation is exact. This is known as the condition for exactness.
3. What is the main difference between an exact and a non-exact differential equation?
The key difference lies in whether they satisfy the condition for exactness.
- An exact differential equation satisfies the condition where the partial derivative of M with respect to y equals the partial derivative of N with respect to x. It can be solved directly by integration.
- A non-exact differential equation does not satisfy this condition. It cannot be solved directly and requires an extra step, like finding an 'integrating factor', to make it exact first.
4. What happens if a differential equation isn't exact? Can it still be solved?
Yes, a non-exact differential equation can often be solved. If an equation is not exact, we try to find a special function called an integrating factor (often denoted by μ(x, y)). When we multiply the entire non-exact equation by this integrating factor, it transforms into a new equation that is exact and can then be solved using the standard method.
5. Why is the condition ∂M/∂y = ∂N/∂x so important for checking exactness?
This condition is crucial because it confirms that the equation could have originated from a single potential function, f(x, y). It's based on Clairaut's theorem about the equality of mixed partial derivatives. If an equation Mdx + Ndy = 0 is exact, then M = ∂f/∂x and N = ∂f/∂y. The test ∂M/∂y = ∂N/∂x simply checks if ∂²f/∂y∂x = ∂²f/∂x∂y, which must be true for such a function f(x,y) to exist.
6. Where are exact differential equations used in real-world applications?
Exact differential equations are very useful in science and engineering, especially when dealing with conservative systems. For example, they are used in:
- Thermodynamics: To model relationships between pressure, volume, and temperature where state functions are involved.
- Physics: To describe conservative force fields, like gravitational or electrostatic fields, where the work done between two points is independent of the path taken.
- Economics: In modelling utility functions and understanding economic equilibrium.
