You are given the following differential equation with the initial condition, vt = 0 = 0, 2 relativ e error a The force balance is

1 CHAPTER 1

1.1 You are given the following differential equation with the initial condition, vt = 0 = 0,

2 v m c g dt dv d − = Multiply both sides by mc d 2 v g c m dt dv c m d d − = Define d c mg a = 2 2 v a dt dv c m d − = Integrate by separation of variables, dt m c v a dv d ∫ ∫ = − 2 2 A table of integrals can be consulted to find that a x a x a dx 1 2 2 tanh 1 − = − ∫ Therefore, the integration yields C t m c a v a d + = −1 tanh 1 If v = 0 at t = 0, then because tanh –1 0 = 0, the constant of integration C = 0 and the solution is t m c a v a d = −1 tanh 1 This result can then be rearranged to yield ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = t m gc c gm v d d tanh 1.2 This is a transient computation. For the period from ending June 1: 2 Balance = Previous Balance + Deposits – Withdrawals Balance = 1512.33 + 220.13 – 327.26 = 1405.20 The balances for the remainder of the periods can be computed in a similar fashion as tabulated below: Date Deposit Withdrawal Balance 1-May 1512.33 220.13 327.26 1-Jun 1405.20 216.80 378.61 1-Jul 1243.39 350.25 106.80 1-Aug 1586.84 127.31 450.61 1-Sep 1363.54 1.3 At t = 12 s, the analytical solution is 50.6175 Example 1.1. The numerical results are: step v12 absolute relative error 2 51.6008 1.94 1 51.2008 1.15 0.5 50.9259 0.61 where the relative error is calculated with 100 analytical numerical analytical error relative absolute × − = The error versus step size can be plotted as

0.0 1.0

2.0

0.5 1

1.5 2

2.5 relativ e error

Thus, halving the step size approximately halves the error.

1.4 a The force balance is

3 v m c g dt dv − = Applying Laplace transforms, V m c s g v sV − = − Solve for m c s v m c s s g V + + + = 1 The first term to the right of the equal sign can be evaluated by a partial fraction expansion, m c s B s A m c s s g + + = + 2 m c s s Bs m c s A m c s s g + + + = + Equating like terms in the numerators yields A m c g B A = = + Therefore, c mg B c mg A − = = These results can be substituted into Eq. 2, and the result can be substituted back into Eq. 1 to give m c s v m c s c mg s c mg V + + + − = Applying inverse Laplace transforms yields t m c t m c e v e c mg c mg v − − + − = or 4 t m c t m c e c mg e v v 1 − − − + = where the first term to the right of the equal sign is the general solution and the second is the particular solution. For our case, v0 = 0, so the final solution is t m c e c mg v 1 − − = b The numerical solution can be implemented as 62 . 19 2 1 . 68 5 . 12 81 . 9 2 = ⎥⎦ ⎤ ⎢⎣ ⎡ − + = v 2087 . 6 2 62 . 19 1 . 68 5 . 12 81 . 9 62 . 19 4 = ⎥⎦ ⎤ ⎢⎣ ⎡ − + = v The computation can be continued and the results summarized and plotted as: t v dvdt 0 0 9.81 2 19.6200 6.2087 4 32.0374 3.9294 6 39.8962 2.4869 8 44.8700 1.5739 10 48.0179 0.9961 12 50.0102 0.6304 20 40 60 4 8 12 Note that the analytical solution is included on the plot for comparison. 5

1.5 a The first two steps are