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FirstEqn3

# FirstEqn3 - Math 246 Fall 2009 Professor David Levermore 7...

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Math 246, Fall 2009, Professor David Levermore 7. Exact Differential Forms and Integrating Factors Let us ask the following question. Given a first-order ordinary equation in the form (7.1) d y d x = f ( x, y ) , when do its solutions satisfy a relation of the form H ( x, y ) = c where c is an arbitrary constant ? Such an H ( x, y ) is called an integral of (7.1). This question is easily answered if we assume that all functions involved are as differntiable as we need. Suppose that such an H ( x, y ) exists, and that y = Y ( x ) is a solution of differential equation (7.1). Then H ( x, Y ( x )) = H ( x I , Y ( x I )) , where x I is any point in the interval of definition of Y . By differentiating this equation with respect to x we find that x H ( x, Y ( x )) + Y ( x ) y H ( x, Y ( x )) = 0 . Therefore, wherever y H ( x, Y ( x )) negationslash = 0 we see that Y ( x ) = x H ( x, Y ( x )) y H ( x, Y ( x )) . For this to hold for every solution of (7.1), we must have d y d x = x H ( x, y ) y H ( x, y ) , or equivalently (7.2) f ( x, y ) = x H ( x, y ) y H ( x, y ) , wherever y H ( x, y ) negationslash = 0. The question then arises as to whether we can find an H ( x, y ) such that (7.2) holds for any given f ( x, y )? It turns out that this cannot always be done. In this section we explore how to seek such an H ( x, y ). 7.1. Exact Differential Forms. The starting point is to write equation (7.1) in a so-called differential form (7.3) M ( x, y ) d x + N ( x, y ) d y = 0 , where f ( x, y ) = M ( x, y ) N ( x, y ) . There is not a unique way to do this. Just pick one that looks natural. If you are lucky then there will exist a function H ( x, y ) such that (7.4) x H ( x, y ) = M ( x, y ) , y H ( x, y ) = N ( x, y ) . When this is the case the differential form (7.3) is said to be exact . 1

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2 It turns out that there is test you can easily apply to find out if you are lucky. It de- rives from the fact that “mixed partials commute” — namely, the fact that for any twice differentiable H ( x, y ) one has y ( x H ( x, y ) ) = x ( y H ( x, y ) ) . This fact implies that if (7.4) holds for some twice differentiable H ( x, y ) then M ( x, y ) and N ( x, y ) satisfy y M ( x, y ) = y ( x H ( x, y ) ) = x ( y H ( x, y ) ) = x N ( x, y ) . In other words, if the differential form (7.3) is exact then M ( x, y ) and N ( x, y ) satisfy (7.5) y M ( x, y ) = x N ( x, y ) . The remarkable fact is that the converse holds too. Namely, if the differential form (7.3) satisfies (7.5) for every ( x, y ) then it is exact — i.e. there exists an H ( x, y ) such that (7.4) holds. Moreover, the problem of finding H ( x, y ) is reduced to evaluating two integrals. We illustrate this fact with examples. Example: Solve the initial-value problem d y d x + e x y + 2 x 2 y + e x = 0 , y (0) = 0 . Solution: Express this equation in the differential form ( e x y + 2 x ) d x + (2 y + e x ) d y = 0 . Because y ( e x y + 2 x ) = e x = x (2 y + e x ) = e x , this differential form satisfies (7.5) and is thereby exact . We can therefore find H ( x, y ) such that (7.6) x H ( x, y ) = e x y + 2 x , y H ( x, y ) = 2 y + e x .
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FirstEqn3 - Math 246 Fall 2009 Professor David Levermore 7...

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