# Quiz1 - φ t = t Solution F = t ln y t = c Method 2 Separable dy y(1 ln y = dt t ln(1 ln y = ln t c 2 Solve the following linear equation dy dx =-3

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Quiz 1: Name: MAP 2302 (Sec 0693) This Quiz contains 2 problems 11 points each. 3 points for writing your name. Calculators are allowed. Total Points: 25 1. Solve the following equation: dy dt = - 1+ln( y ) t/y Method 1: t/ydy + (1 + ln( y )) dt = 0 Ndy + Mdt = 0 We observe ∂M ∂y = 1 y = ∂N ∂t . Hence this is an exact equation. F = Z t/ydy + φ ( t ) = t ln( y ) + φ ( t ) ∂F ∂t = ln( y ) + φ 0 ( t ) = 1 + ln( y ) φ 0 ( t ) = 1

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Unformatted text preview: φ ( t ) = t Solution F = t ln( y ) + t = c Method 2: Separable : dy y (1 + ln( y )) = dt t ln(1 + ln( y )) = ln( t ) + c 2. Solve the following linear equation. dy dx =-3 y x-2-3 x dy dx + 3 y x =-2-3 x I.F = μ = e 3 ln( x ) = x 3 d ( yx 3 ) = (-2-3 x ) x 3 dx yx 3 =-x 4 2-3 x 5 5 + c...
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## This note was uploaded on 12/15/2011 for the course MAP 2302 taught by Professor Tuncer during the Spring '08 term at University of Florida.

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Quiz1 - φ t = t Solution F = t ln y t = c Method 2 Separable dy y(1 ln y = dt t ln(1 ln y = ln t c 2 Solve the following linear equation dy dx =-3

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