Ordinary Differential Equations/Exact 1

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First Order Differential Equations

This page details a method for trying to find solutions to equations of the form

This is often written as a differential form

.

Subsequently, we will refer to this expression as ODE. Differential forms frequently come up in multivariable calculus while studying line integrals.

Contents

Exact Differential EquationsEdit

Before we begin identifying and solving exact differential equation it helps to make a few observations. We will begin reminding ourselves of the chain rule from multivariable calculus. Which states how to compute the derivative of a composition of two or more functions. Suppose that   is a function of two real variables, and we are given functions   and   which are functions of a single real variable. Then the function   is simply a function of  , with   and   being plugged into   as   and  . The chain rule from multivariable calculus tells us how to calculate the derivative of  . It states that:

 

If we slightly abuse the notation and call the two functions   and   (instead of   and  ) then we can write the chain rule as

 

As an example we could let   and we could let   and  . Then according to the chain rule

 

Of course this would have been seen more directly by substituting for   and   to discover that  , but this simply gives us a verification that we took the derivative correctly.

We will use this theory to evaluate:

 

If we examine this expression carefully it looks equal to the left hand side of our ODE above. Specifically if   and   then our ODE is:

 

This type of equation is especially easy to solve. The only functions whose derivatives are 0 are constant functions or simple constants. Thus the solution to our ODE, i.e. integrating it, will be given by

 

Now consider the following example, applying what we have just figured out.

 

In this example  ,  , and  . Notice that if   then   and  . By the way, if you ever wanted to check for yourself what we are doing here and your calculus is a bit rusty, get maxima (http://maxima.sourceforge.net or prepackaged for your favorite Linux distribution or for Android). The derivation of   we have just made can easily be replayed in maxima, like so,

(%i1) psi:x*y-(x^3/3);

and

(%i2) diff(psi,x);

yielding

(%o1) y - x^2.

Or going from   to  ,

(%i1) psi_prime:y-x^2;

and

(%i2) integrate(psi_prime,x);

yielding

(%o1) x*y-(x^3/3)

Turning back to our problem, by our observations above the solution of this equation should be given by  , or in other words:

 

This particular equation is linear so we may easily verify that the solution obtained in this way is correct. When there is a function   so that   and   then the equation is called exact. Unfortunately not every differential equation of the form   is exact. In order for this to be an effective method for solving differential equation we need a way to distinguish if a differential equation is exact, and what the function   is if the function is exact.

In order to see that   and   could not be arbitrary, remember form multivariable calculus that   (read: the order of the partial derivatives of   is exchangeable) whenever the derivatives exist and are continuous. Since  , then  , similarly  . Hence, if the equation is exact we would definitely need

 

or the same stated differently

 

This can be put into a theorem.

Theorem

Suppose that   and   have continuous partial derivatives and they satisfy the relationship   Then there is a function   so that   and  

Proof. We prove this by giving an explicit construction of  . First notice that if   exists then by integrating the expression   in   we get:

 

Here   just taking an anti-derivative   with respect to x, treating y as a constant. It is necessary to add a function  , because for any function  . So that with the above definition  .

Now we need to determine  . To do this we use that  .

Note that when this is the case, Pdx+Qdy and   must be the same, meaning that   and  . This implies that  . We will now prove that this is also a sufficient condition when the mixed derivative   is continuous.


Proof:

First, take the integral

 

This obviously satisfies the condition that P= .

In order for it to satisfy the other condition, Q(x,y)=  meaning that

 
 
 

Canceling Q(x,y) from both sides, we get  .

This proves that the equation is exact and that

  is an integral of the differential equation.

Note that only C is the arbitrary constant. Changing   only changes the integral by a constant value, which is absorbed by the C. Changing   will also only change it by a constant because of the fact that  .

ExampleEdit

Consider the following DE:

 

Note that:

  and 1 is a continuous function so this equation is exact by what has been proven above.

Therefore, the integral is

 

Take   and  

Which is  

Integrating Factors for an Ordinary Linear Differential Equation of the First OrderEdit

Consider an equation of the form

 

where P(x), Q(x) and y are all functions of x. This is a first-order linear differential equation as discussed previously. For this to work this form must be closely adhered to - the derivative must be by itself.

In general these equations are not exact. They can, however, be made exact by multiplying through by an integrating factor, I(x), another function of x as we have done previously.

Multiply our original equation by I(x):

(1): 

This will be our new, solvable, DE. Now consider the derivative of the product below:

(2): 

Now, if we make the RHS of (1) equal to the LHS of (2), then

(3): 

Which, by the other halves of the equations, makes:

(4): 

Which simplifies to:

(5): 

By equating the equations to get (3) forces multiplying by I(x) to produce a derivative of a product on the RHS of (1), i.e.

(6): 

The new DE is therefore exact, and can be solved more easily. We now find the function I(x) from (5). We will change notation slightly here.

 
 
 
 
 
 

We take this to be our integrating factor. We can ignore the negative factor, because when both sides of the DE are multiplied by it, they will cancel. So, our integrating factor is:

 

To solve the DE, we then multiply by this factor, and solve the equation, given that one side will be able to be turned into a derivative of a product.

General Integrating FactorsEdit

Now we generalize first-order linear differential equations to functions of the sort Pdx+Qdy=0 which is our ODE from the beginning this section. They are sometimes not exact, too. However, when multiplied by a function h(x,y), the product hPdx+hQdy=0 may be exact.

Theorem: An equation of the form Pdx+Qdy=0 which has exactly one integral solution with one arbitrary constant C has infinitely many integrating factors.

Proof: Suppose that the solution is f(x,y)=C. The differential is

 

Since f(x,y)=c is a solution of Pdx+Qdy=0, it must hold true that

 

Which means that a function h exists such that

 

and

 .

Obviously, this is an integrating factor. Furthermore, let S(f) be any function of f.

Then   would equal   so hS(f) is also an integrating factor.