Exact Differential Equation

An exact equation is a ordinary differential equation that can be expressed in the form

M (x, y) d x + N (x, y) d y = 0

where there exists some function $ψ (x, y) = C$ such that

\frac{\partial ψ}{\partial x} = M (x, y) and \frac{\partial ψ}{\partial y} = N (x, y)

We also call $ψ (x, y)$ the potential function of the exact equation.

Requirement

An equation of the form $M (x, y) d x + N (x, y) d y = 0$ is an exact equation if and only if

\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}

Consider the function $ψ (x, y) = C$ .

Applying the Gradient Operator yields

\nabla ψ = \frac{\partial ψ}{\partial x} \hat{i} + \frac{\partial ψ}{\partial y} \hat{j}

We can write this function in the form

\frac{\partial ψ}{\partial x} d x + \frac{\partial ψ}{\partial y} d y = 0

If we let $M (x, y) = \frac{\partial ψ}{\partial x}$ and $N (x, y) = \frac{\partial ψ}{\partial y}$ , we have

M (x, y) d x + N (x, y) d y = 0

Find the family of solutions to the differential equation

4 x y + 1 + (2 x^{2} + cos y) y^{'} = 0

We then define:

Verify that this is an Exact Equation

\frac{\partial}{\partial y} (4 x y + 1) 4 x = \frac{\partial}{\partial x} (2 x^{2} + cos y) = 4 x

Now we may proceed.

Integrate $M (x, y)$ with respect to $x$ to find $f (x, y)$

f (x, y) = \int 4 x y + 1 d x = 2 y x^{2} + x + C (y)

Note: as $y$ is treated as a constant, the constant of the integral is treated as $C (y)$ .

Now take the partial derivative with respect to $y$ to find an equation for $N (x, y)$

N = \frac{\partial}{\partial y} (2 y x^{2} + x + C (y)) = 2 x^{2} + C^{'} (y)

With two expressions for $N$ , we can equate them

2 x^{2} + C^{'} (y) C^{'} (y) C (y) = 2 x^{2 + c o s y} = cos y = sin y

Therefore, we can write

f (x, y) = 2 x^{2} y + x + sin y

And the solution to the differential equation is

2 x^{2} y + x + sin y = C

The solution is a family of curves due to the arbitrary constant $C$ .