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lecture_10 (dragged) 1 - 2 MA 36600 LECTURE NOTES WEDNESDAY...

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Unformatted text preview: 2 MA 36600 LECTURE NOTES: WEDNESDAY, FEBRUARY 4 Assume first that statement p is true. Then we have ￿￿ ￿￿ ∂M ∂ ∂f ∂2f ∂2f ∂ ∂f ∂N = = = = = . ∂y ∂y ∂x ∂y ∂x ∂x ∂y ∂x ∂y ∂x (We can interchange the partial derivatives when f (x, y ) has partial derivatives which are continuous functions. This is known as Clairaut’s Theorem.) This shows that statement q is true. Now assume that statement q is true. We explain how to construct the function f (x, y ): #1: Choose a function g (x, y ) according to the partial differential equation ￿x ∂g = M (x, y ) =⇒ g (x, y ) = M (σ, y ) dσ. ∂x #2: Choose a function h(y ) according to the ordinary differential equation ￿ ￿ y￿ dh ∂g ∂g = N (x, y ) − =⇒ h(y ) = N (x, τ ) − (x, τ ) dτ . dy ∂y ∂y #3: Choose the function f (x, y ) = g (x, y ) + h(y ). The function h(y ) is not a function of x because ￿ ￿ ￿ ￿y ￿ y￿ ￿ ￿ ∂￿ ∂ ∂g ∂N ∂2g h(y ) = N− dτ = − dτ = ∂x ∂x ∂y ∂y ∂x ∂y Then f (x, y ) is the desired function because ∂f ∂g ∂h = + = M (x, y ), ∂x ∂x ∂x y ￿ ￿ ∂N ∂M − dτ = 0. ∂x ∂y ∂f ∂g ∂h = + = N (x, y ). ∂y ∂y ∂y This shows that statement p is true. Example 1. Consider the differential equation dy 2 x + y2 =− dx 2xy ￿ =⇒ ￿ 2 x + y 2 dx + (2 x y ) dy = 0. We have the two functions M (x, y ) = 2 x + y 2 and N (x, y ) = 2 x y . This is an exact equation because we have ∂M ∂N ∂M ∂N = 2y and = 2y =⇒ = . ∂y ∂x ∂y ∂x We use the algorithm above to construct the function f (x, y ). First, construct g (x, y ) by the equation ∂g = M (x, y ) = 2 x + y 2 =⇒ g (x, y ) = x2 + x y 2 . ∂x (We don’t care about a constant of integration.) Second, construct h(y ) by the equation dh ∂g = N (x, y ) − = (2 x y ) − (2 x y ) = 0 dy ∂y =⇒ h(y ) = 0. Hence the desired function is f (x, y ) = g (x, y ) + h(y ) = x2 + x y 2 . Example 2. Now consider the differential equation dy 3 x y + y2 =− 2 dx x + xy ￿ =⇒ We have the two functions M (x, y ) = 3 x y + y 2 ￿ ￿ ￿ 3 x y + y 2 dx + x2 + x y dy = 0. and N (x, y ) = x2 + x y. This is not an exact equation because we have ∂M = 3x + 2y ∂y and ∂N = 2x + y ∂x =⇒ ∂M ∂N ￿= . ∂y ∂x ...
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