hw_2_soln_corrected

f x y 2 y 4 4x y g x fx x y 5 x 4 y and

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Unformatted text preview: 4 y 8 y 3 − 4x = 5x + 4 y = 0 = 0 This differential is exact: ￿ ∂￿ −5x − 4 y = −4; ∂y ￿ ∂￿ 3 8 y − 4x = −4. ∂x ￿ ￿ ￿ ￿ We want f such that d f x , y = (2x + 3) d x + 2 y − 2 d y , that is ￿ ￿ ￿ ￿ fx x, y = −5x − 4 y and f y x , y = 8 y 3 − 4x . ￿ ￿ . f ￿x , y ￿ = 2 y 4 − 4x y + g (x ) ￿ ￿. fx x, y = −5 x − 4 y and f x x , y = −4 y + g ￿ (x ) (3) Combine the last pair of equations: −4 y + g ￿ (x ) ￿ g (x ) ￿￿ gy Substitute (4) into (3): The general solution is = = = −5 x − 4 y −5 x 5 − x2 2 (4) ￿ ￿ 5 f x , y = 2 y 4 − 4x y − x 2 2 5 2 y 4 − 4x y − x 2 = C 2 ...which will be left implicit. 3. Rewrite in differential form: ￿ ￿ dy e x sin y − 2 y sin x + e x cos y + 2 cos x ￿x ￿ ￿ ￿d x e sin y − 2 y sin x d x + e x cos y + 2 cos x d y This differential is exact: ￿ ∂￿x e sin y + 2 y sin x ∂y ￿ ∂￿x e cos y − 2 cos x ∂x 2 = 0 = 0 = e x cos y + 2 sin x ; = e x cos +2 sin x . We want f that is ￿ ￿ fx x, y ￿ ￿ f ￿x , y ￿ f y x, y ￿ ￿￿ ￿ ￿ ￿ such that d f x , y = e x sin y − 2...
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This note was uploaded on 01/26/2014 for the course MATH 527 taught by Professor Boucher during the Spring '07 term at New Hampshire.

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