Solutions Manual to accompany Applied Numerical Methods With MATLAB for Engineers and Scientists

Solutions Manual to accompany Applied Numerical Methods With MATLAB for Engineers and Scientists  CHAPTER 1 1.1 You are given the following differential equation with the initial condition, v(t = 0) = 0, 2 v m c g dt dv d = − Multiply both sides by m/cd 2 g v c m dt dv c m d d = − Define d a = mg / c 2 2 a v dt dv c m d = − Integrate by separation of variables, dt m c a v 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 t C 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 v(12) 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 analytical numerical absolute relative error × − = The error versus step size can be plotted as 0.0% 1.0% 2.0% 0 0.5 1 1.5 2 2.5 relative 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 sV v ' − (0) = − Solve for s c m v s s c m g V '/ (0) ( '/ ) + + + = (1) The first term to the right of the equal sign can be evaluated by a partial fraction expansion, s c m B s A s s c m g ( '/ ) + '/ = + + (2) ( '/ ) ( '/ ) ( '/ ) s s c m A s c m Bs s s c m g + + + = + Equating like terms in the numerators yields A m c g A B ' 0 = + = 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 s c m v s c m mg c s mg c V '/ (0) '/ / ' / ' + + + = − Applying inverse Laplace transforms yields c m t c m t e v e c mg c mg v ( '/ ) ( '/ ) (0) ' ' − − = − + or 4 ( ) c m t c m t e c mg v v e ( '/ ) ( '/ ) 1 ' (0) − − = + − 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, v(0) = 0, so the final solution is ( ) c m t e c mg v ( '/ ) 1 ' − = − (b) The numerical solution can be implemented as (0) 2 19.62 68.1 12.5 (2) 0 9.81 = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ v = + − (19.62) 2 6.2087 68.1 12.5 (4) 19.62 9.81 = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ v = + − The computation can be continued and the results summarized and plotted as: t v dv/dt 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 0 20 40 60 0 4 8 12 Note that the analytical solution is included on the plot for comparison.