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 0 ( 2 x 2 t ) dt + x 0 1 · ds = x 2 y 0 2 t dt + x 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 , we have 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 0 ( x 2 + 2 t ) dt + x 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 0 ( xe x + 2) dt + x 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 , must be 1 /m , where m is the slope of the original curves. Therefore, we have
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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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