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Exact Differential Equations

April 5th, 2009 |

Author: Merrin

Exact differential equations are a class of first order differential equations that can be solved, provided some integrations can be carried out. Their name is inherited from the concept of an exact differential of a two dimensional function. Suppose we have a function f(x,y) which has a differential expansion of the following form.

$df=\cfrac{\partial f}{\partial x} dx+\cfrac{\partial f}{\partial y} dy$

When

$df = 0$

it forms a differential equation.

$\cfrac{\partial f}{\partial x} +\cfrac{\partial f}{\partial y} \cfrac{dy}{dx} =0$

It is simpler to write this as

$M(x,y)+N(x,y)\cfrac{dy}{dx} =0$

Hopefully we can reversibly calculate f(x,y)=C which is the solution relating x and y. The condition to identify a random equation from this differential equation of the above form as exact is that the mixed partial derivatives of the solution f(x,y) are equal. We see that this means

$M(x,y)=\cfrac{\partial f}{\partial x} \quad N(x,y)=\cfrac{\partial f}{\partial y}$
$\cfrac{\partial^{2}f}{\partial x\partial y}=\cfrac{\partial^{2} f}{\partial y\partial x}$
$\cfrac{\partial M(x,y)}{\partial y}=\cfrac{\partial N(x,y)}{\partial x}$

We can take the integrals of both of these partial derivatives and try to match the constants of integration such that we can determine f(x,y)=C. An exact equation generally will be nonlinear but we can still find a solution which is implicit.

Example 1. Determine whether the following differential equation is exact. If it is exact find a solution to the differential equation.

$2x+y+(x+3y^{2} )\cfrac{dy}{dx} =0$

Solution 1. We can write this equation in the standard form.

$M(x,y)dx+N(x,y)dy=0$
$(2x+y)dx+(x+3y^{2} )dy=0$

The condition for the equation to be exact is that

$\cfrac{\partial M(x,y)}{\partial y} =\cfrac{\partial N(x,y)}{\partial x}$

We can find the partial derivative of M with respect to y.

$\cfrac{\partial M(x,y)}{\partial y} =\cfrac{\partial }{\partial y} (2x+y)=1$

Next we find the partial derivative of N with respect to x.

$\cfrac{\partial N(x,y)}{\partial x} =\cfrac{\partial }{\partial x} (x+3y^{2} )=1$

Since the partial derivative calculations match the equation is exact. In order to solve the equation, we must find f(x,y). We know that the partial derivative of f(x,y) with respect to x is M so we can start there.

$\cfrac{\partial f(x,y)}{\partial x} =M(x,y)=2x+y$