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4: Exploring Exact First Order Differential Equations and Euler s Method
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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
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Finding an integrating factor
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How can you find an integrating factor that makes differential equations exact Just follow the steps in the following sections
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Multiplying by the factor you want to find
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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
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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
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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
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Part I: Focusing on First Order Differential Equations
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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
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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
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Completing the process
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Remember the differential equation that you re trying to solve If not, here it is: y-x dy =0 dx
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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
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4: Exploring Exact First Order Differential Equations and Euler s Method
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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!
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