notes 2-6

# notes 2-6 - (3 x 2 y 2 4 e y 2(2 x 3 y 4 xe y-9 y 2 y = 0...

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SECTION 2.6 Exact Equations and Integrating Factors Remember (maybe, I hope) that if y is a function of x and F is a function of both x and y , then by the Chain Rule for such things, dF dx = ∂F ∂x + ∂F ∂y dy dx so that if F ( x,y ( x )) = C , where C is a constant, then we’d get dF dx = ∂F ∂x + ∂F ∂y dy dx = 0 . So, if we start with ∂F ∂x + ∂F ∂y dy dx = 0 , then by integrating both sides we see that the solution of this diﬀerential equation is F ( x,y ( x )) = C. Diﬀerential equations of this type are called exact . But of course we don’t start with the F , either! We start with M ( x,y ) + N ( x,y ) y 0 = 0 . How do we know when there is an F , that is, how do we know whether the diﬀerential equation is exact? Theorem 2.6.1 on page 95 gives the answer, which is essentially that the equation is exact when M y ( x,y ) = N x ( x,y ). EXAMPLE. Show that the diﬀerential equation

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Unformatted text preview: (3 x 2 y 2 + 4 e y + 2) + (2 x 3 y + 4 xe y-9 y 2 ) y = 0 is exact and ﬁnd the solution. Sometimes a diﬀerential equation is not exact but becomes exact when you multiply by some function μ ( x,y ). Then μ ( x,y ) is called an integrating factor . There are various tricks that sometimes work to produce integrating factors, e.g., Problems 23 and 24 on page 100, but we will be satisﬁed with knowing what integrating factors are and what they are good for. EXAMPLE. Show that the given equation is not exact but becomes exact when multiplied by the given integrating factor. Then solve the equation. " 3 x + 6 y # dx + " x 2 y + 3 y x # dy = 0 , μ ( x,y ) = xy HOMEWORK: SECTION 2.6...
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notes 2-6 - (3 x 2 y 2 4 e y 2(2 x 3 y 4 xe y-9 y 2 y = 0...

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