73_pdfsam_math 54 differential equation solutions odd

73_pdfsam_math 54 differential equation solutions odd -...

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Exercises 2.4 Thus, N ( x, t )=2 x 2 t and M ( s, y 0 )=2 s · 0 2 + 1 = 1, and (18) yields F ( x, y )= y Z 0 ( 2 x 2 t ) dt + x Z 0 1 · ds = x 2 y Z 0 2 tdt + x Z 0 ds = x 2 t 2 ± ± ± t = y t =0 + s ± ± ± s = x s =0 = x 2 y 2 + x. (b) Since M ( x, y )=2 xy sec 2 x and N ( x, y )= x 2 +2 y ,wehave N ( x, t )= x 2 +2 t and M ( s, y 0 )=2 s · 0 sec 2 s = sec 2 s, F ( x, y )= y Z 0 ( x 2 +2 t ) dt + x Z 0 ( sec 2 s ) ds = ( x 2 t + t 2 ) ± ± ± t = y t =0 tan s ± ± ± s = x s =0 = x 2 y + y 2 tan x. (c) Here, M ( x, y )=1+ e x y + xe x y and N ( x, y )= xe x + 2. Therefore, N ( x, t )= xe x +2 and M ( s, y 0 )=1+ e s · 0+ se s · 0=1 , F ( x, y )= y Z 0 ( xe x +2) dt + x Z 0 1 · ds =( xe x +2) t ± ± ± t = y t =0 + s ± ± ± s = x s =0 =( xe x +2) y + x, which is identical to F ( x, y ) obtained in Example 3. 32. (a) The slope of the orthogonal curves, say m ,mustbe 1 /m ,where m is the slope of the original curves. Therefore, we have m
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Unformatted text preview: original curves. Therefore, we have m ⊥ = F y ( x, y ) F x ( x, y ) ⇒ dy dx = F y ( x, y ) F x ( x, y ) ⇒ F y ( x, y ) dx − F x ( x, y ) dy = 0 . (b) Let F ( x, y ) = x 2 + y 2 . Then we have F x ( x, y ) = 2 x and F y ( x, y ) = 2 y . Plugging these expressions into the Fnal result of part (a) gives 2 y dx − 2 x dy = 0 ⇒ y dx − x dy = 0 . 69...
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at Berkeley.

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