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Extra info for Global existence for the heat equation with nonlinear dynamical boundary conditions

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E. e. A = 2 Equating the coefficients of x 2 terms gives: From equation (1), since A = 2, C = −4 7 = A + C, and since C = 5, A = 2 Equating the coefficients of x terms gives: Hence 5 = A + B, and since A = 2, B = 3 2 3 + 6x + 4x2 − 2x3 1 −4x + 3 ≡ + 2+ 2 2 2 x x x +3 x (x + 3) ≡ [Check: equating the constant terms gives: 13 = B + 2C When B = 3 and C = 5, partial fractions. e. they are repeated linear factors. Let B Cx + D A 3 + 6x + 4x 2 − 2x 3 ≡ + 2+ 2 x 2 (x 2 + 3) x x (x + 3) Ax(x 2 + 3) + B(x 2 + 3) + (Cx + D)x 2 ≡ x 2 (x 2 + 3) 1.

3! 4! 5! Grouping like terms gives: ex (x 2 − 1) = −1 − x + x 2 − + x4 x4 − 2! 4! + x2 2! + x3 − x5 x5 − 3! 5! x3 3! + ··· 5 11 19 5 1 x = − 1 − x + x2 + x3 + x4 + 2 6 24 120 when expanded as far as the term in x 5 . LOGARITHMS AND EXPONENTIAL FUNCTIONS 31 Now try the following exercise. A Exercise 19 Further problems on the power series for ex 1. 0601] using the power series for ex . 2. 3 and check your result by using a calculator. 7408] 3. Expand (1 − 2x) e2x as far as the term in x 4 . 1 − 2x 2 − 2 8x 3 − 2x 4 3 1 4.

The equation y = a ekx is reduced to linear form. e. 55 1 0 10 20 30 40 50 Time, t ms 60 70 80 Problem 29. 11 the law v = V e T , where V and T are constants. 0, correct to 3 significant figures Since the straight line does not cross the vertical axis at t = 0 in Fig. 0 Show that the law relating voltage and time is as stated and determine the approximate values of V and T . 0 V. = t into v = V e T . t Since v = V e T then ln v = T1 t + ln V which is of the form Y = mX + c. Using ‘log 3 cycle × linear’graph paper, the points are plotted as shown in Fig.

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Global existence for the heat equation with nonlinear dynamical boundary conditions by Vitillaro E.


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