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Unformatted text preview: M y = ∂M ( x ) ∂y = 0 and N x = 0. Since they are the same, any separable equation is also exact (since it can be written in the above form). 2.5.17 (a) Solve the Gompertz equation dy dt = ry ln ± K y ² , subject to the initial condition y (0) = y . Separating gives us Z dy y ln ³ K y ´ = Z r dt. For the integral on the left, if we let u = ln( K/y ) as suggested, then du =dy/y , so the integral becomesR u1 du =ln  ln( K/y )  . So ln  ln( K/y )  =rt + C or ln( K/y ) = Dert . 1 Finally, solving for the constant D , and then solving for y we get y ( t ) = Keln( K/y ) ert = Ke ln( y /K ) ert . (b) If r = 0 . 71 per year, K = 80 . 5 × 10 6 kg, and y /K = 0 . 25, then y (2) = 57 . 580 × 10 6 . (c) y ( τ ) = 0 . 75 K when e ln(0 . 25) e. 71 τ = 0 . 75 , i.e. when τ =1 . 71 ln ln 0 . 75 ln 0 . 25 ≈ 2 . 2148 . 2...
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 Spring '07
 COSTIN
 Differential Equations, Equations, Fractions, Derivative, Partial Differential Equations, 2k, 0.75k, 106 kg

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