In the study of partial differentiation, recall that a function of two variables that equals a constant, describes the points in the 3-D plane with the same potential; . The curves that connect the points with the same potential are called level curves and have the value of c. A contour map is a level curve graph where common elevations are connected giving a 2-D representation of a 3-D reality.
Using a LiveMath 3-D graph theory you can plot such a function along with the level curves describing the contours associated with that function. The picture below uses the following function to demonstrate this (here is a LiveMath plug-in animation of the graph below).
In general terms, this type of equation is represented by the following:
It describes the level curves and is the solution to the following differential equation. The equation below is just the total derivative of the function above.
Because it is the total derivative of some function z(x,y) it is called an Exact Differential Equation.
To help understand how to solve these types of equations you will look at the solution first and then analyze how to back into that solution.
Expl #11 Exact Equations
In this example you will take the total derivative of a function and analyze its parts. Then you will take this new equation (a differential equation now) and, knowing the answer, describe the method used to solve it.
After setting the RHS equal to zero you will have a differential equation to solve. Notice how the coefficients are neither separable, homogeneous, nor are they linear.
To help analyze this equation, label the coefficient of dx as M and the coefficient of dy as N. Both are functions of x and y so place the equation in the following form.
The total derivative of a function is obtained by adding the partial derivatives of the coefficients. This is done with the equation below.
NOTE: As discussed earlier, using LiveMath's partial derivative Op to find partial derivatives requires an Independence Declaration to be placed in the notebook. To control the scope of these declarations, it is wise to study partial derivatives inside their own case theory. These declarations only affect operations within a case theory or those nested inside that case theory. They will not affect operations outside of the case theory they are in, unless they are at the root, or main theory level, where they affect all operations in the notebook.
You can see that differential equations of this type are Exact. They are immediate derivatives of another function. You know this is true in this example because you developed the equation below by taking the derivative of the original function.
You can describe an Exact Differential Equation as an equation whose dx coefficient is the partial derivative with respect to x of some function f(x,y) and whose dy coefficient is the partial derivative with respect to y of the SAME function. Since you dry-labed the last example you know what this equation, z=f(x,y), is. This will not be the case as you look to solve these problems though, so you need to find a way of determining that an equation is exact, then you will know that M and N are related to the solution equation in this way!
Using the fact that these partial derivatives are of the same function will be the key to the method used to solve these equations. To test a differential equation for exactness, follow the method described in the next example.
Expl #12 Test for Exactness
To test for exactness, equate the partial derivative with respect to y of M and the partial derivative with respect to x of N. Notice that these partials are with respect to the exact opposite variables as those used to determine the total derivative in the last example. The reason for this will become clear to you later when you derive this test.
The fact that they are equal means that the differential equation is exact!
To solve these types of equations you will need to take one or the other coefficient and "go backwards" to determine the solution. You can take either M or N to do this, it is up to you. For this example try using M.
Expl #13 Exact Solution Method
This is very close to the answer, and with a little twist, will lead to a method that you will use to obtain the solution.
Now you have a potential function (), that represents the solution to the problem. To solve, the function u must be determined. If you take the partial derivative of the unknown function with respect to y this time, you will get N. By setting up the equation this way, you can then isolate u.
You already know what N is, so:
The question remains, why do you take partial derivatives of M and N to determine if an equation is exact? M and N have been defined as the partial derivatives of z with respect to x and y respectively.
By taking the second partial derivative of each coefficient WITH RESPECT TO THE OPPOSITE VARIABLE, the LHS of both of these equations is equal and therefore the RHS are equal too.
This is verified below:
Now that the method has been explained, the following example puts it all together to solve a problem.
Expl #14 Solving an Exact Equation
The previous four examples demonstrate a method of finding the solution to an exact differential equation by confirming that the partial derivatives with respect to y of M and x of N are equal. When this test fails, the equation can sometimes be transformed from a non-exact into an exact equation, by applying an integrating factor.
Although there is not a general rule for finding integrating factors for exact equations, there are a few special cases where the method, described below, can sometimes help you find one.
As mentioned eariler, an integrating factor is a function that, when multiplied on to a differential equation, yields an equation that is recognized as a derivative. Once in this form, you can use the solution methods just discussed to solve the equation.
When an equation fails the test for exactness there are five types of integrating factors that, if applicable, you can use to transform an equation into one that you can use with the methods discussed in the last section. You will derive this method in general terms and learn to use LiveMath to find one of these factors. The remaining cases will be listed for reference.
First make the assumption that the equation in question cannot be solved by other methods discussed to this point.
For computational efficiency declare both the unknown factor and the coefficients of the differentials as variables.
If you can isolate from this equation, you can apply the result to the differential equation to make it exact. The following example derives the first case where the potential integrating factor is a function of x alone.
Expl #15 Deriving an Integrating Factor
As mentioned above , M and N are declared as variables. You know that each is a function, but by declaring them user defined variables in LiveMath, the derivation is much easier to follow.
You now have the equations necessary to give an integrating factor for an equation where the factor is a function of x. The Prop defining u gives you a method of testing whether it is a function of x for the problem at hand. Using the same notation, the next example demonstrates how to find this factor using LiveMath.
Expl #16 Finding an Integrating Factor
The equation in this example is first tested for exactness using the methods described in prior examples. Set up your notebook making sure the Case Theories are as shown. This will allow you to add the f=f(y) case to the notebook later on.
The main equations are in the M and N Props. By setting the notebook up this way, you can test other problems by changing M and N with Auto-Remanipulation turned on. Make sure all variables are declared as User Defined Variables.
The equation below shows the new coefficients applied. Using the methods discussed earlier, you can now solve the equation.
NOTE: It is important to again note that the function u is in terms of x alone. This is a requirement for this factor to work. If it is not, try one of the other methods shown below.
The previous example showed that, if a factor in x is obtainable, you can make a non-exact equation exact by applying the integrating factor just derived. You can derive another case in which a factor in y can turn a non-exact equation into one that is. It is derived in the same manner as above and is shown below.
Three more cases can be derived and they are:
This analysis of integrating factors leads you into the next solution method. A very important differential equation is one that has both the dependent variable and its derivative of the first degree.