4: Exploring Exact First Order Differential Equations and Euler s Method in Java

Drawer Denso QR Bar Code in Java 4: Exploring Exact First Order Differential Equations and Euler s Method

and 2N _ x, y i = -1 2x As you can see, the original equation isn t exact However, the two partial derivatives, M/ x and N/ y, differ only by a minus sign Isn t there some way of making this differential equation exact Yup, you re right There is a way! You just have to use an integrating factor As I discuss in 2, integrating factors are used to multiply differential equations to make them easier to solve In the following sections, I explain how to use an integrating factor to magically turn a nonexact equation into an exact one

How can you find an integrating factor that makes differential equations exact Just follow the steps in the following sections

Using the example from earlier, here I show you how to find an integrating factor that makes differential equations exact Say you have this differential equation: M _ x, y i + N _ x, y i dy =0 dx

Multiplying this equation by the integrating factor (x, y) (which you want to find) gives you this: _ x, y i M _ x, y i + _ x, y i N _ x, y i This equation is exact if: 2 a _ x, y i M _ x, y i k 2y = 2 a _ x, y i N _ x, y i k 2x dy =0 dx

which means that (x, y) must satisfy this differential equation: V R 2 _ x, y i 2 _ x, y i S a 2M _ x, y i k 2N _ x, y i W W _ x, y i = 0 M _ x, y i - N _ x, y i +S 2y 2x 2y 2x W S W S X T

Well, yipes That doesn t seem to have bought you much simplicity In fact, this equation looks more complex than before What you have to do now is to assume that (x, y) is a function of x only that is, (x, y) = (x) which also means that: 2 ^ x h =0 2y This turns the lengthy and complex equation into the following: N J 2 ^ x h K 2M _ x, y i 2N _ x, y i O + x =0 O ^ h K 2x 2y 2x P L Or in other words: - N _ x, y i N _ x, y i N J 2 ^ x h K 2M _ x, y i 2N _ x, y i O = x O ^ h K 2x 2y 2x P L

Dividing both sides by N(x, y) to simplify means that: J 2M _ x, y i 2N _ x, y i N 2 ^ x h 1 O ^ x h K = O 2x 2y 2x N _ x, y i K P L Well, that looks somewhat better You can see how to finish the example in the next section

In this case, M(x, y) = y and N(x, y) = x Plug these into your simplified equation from the previous section, which becomes: 2 ^ x h -1 = x `1 - ^-1hj ^ x h 2x

or: 2 ^ x h - 2 = x ^ x h 2x Now this equation can be rearranged to get this one: 2 ^ x h = -2 2x x ^ x h Integrating both sides gives you: ln|>(x)| = 2 ln|x| And finally, exponentiating both sides results in this equation: ! ^ x h = 12 x For this differential equation: 2M _ x, y i =1 2y and 2N _ x, y i = -1 2x Because these equations differ by a sign, you may suspect that you need the negative version of (x), meaning that: ^ x h = - 12 x So that, at last, is your integrating factor Whew!