Métodos numéricos

Solutions Manual
to accompany

Applied Numerical Methods
With MATLAB for Engineers and Scientists

Steven C. Chapra Tufts University

CHAPTER 1
1.1 You are given the following differential equation with the initial condition, v(t = 0) = 0,

c dv = g − d v2 dt m
Multiply both sides by m/cd

m dv m = g − v2 c d dt c d
Define a = mg / c d

m dv = a2 − v2 c d dt Integrate byseparation of variables,

∫ a 2 − v 2 = ∫ m dt
A table of integrals can be consulted to find that

dv

cd

∫a

2

dx x 1 = tanh −1 2 a a −x

Therefore, the integration yields
1 v c tanh −1 = d t + C a a m

If v = 0 at t = 0, then because tanh–1(0) = 0, the constant of integration C = 0 and the solution is
1 v c tanh −1 = d t a a m This result can then be rearranged to yield

v=

⎛gc d ⎞ gm tanh ⎜ t⎟ ⎜ m ⎟ cd ⎠ ⎝

1.2 This is a transient computation. For the period from ending June 1:

1

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 1-May $ 220.13 1-Jun $ 216.80 1-Jul $ 350.25 1-Aug $ 127.31 1-Sep $ 450.61 $1363.54 $ 106.80 $ 1586.84 $ 378.61 $ 1243.39 $ 327.26 $ 1405.20 Deposit Withdrawal Balance $ 1512.33

1.3 At t = 12 s, the analytical solution is 50.6175 (Example 1.1). The numerical results are:
step 2 1 0.5 v(12) 51.6008 51.2008 50.9259 absolute relative error 1.94% 1.15% 0.61%

where the relative error is calculated with

absolute relative error =

analytical − numerical × 100%analytical

The error versus step size can be plotted as
2.0%

1.0% relative error 0.0% 0 0.5 1 1.5 2 2.5

Thus, halving the step size approximately halves the error.
1.4 (a) The force balance is

2

dv c' =g− v dt m Applying Laplace transforms, sV − v(0) = Solve for V= g v ( 0) + s ( s + c ' / m) s + c ' / m
(1)

g c' − V s m

The first term to the right of the equal sign can beevaluated by a partial fraction expansion,

g A B = + s ( s + c ' / m) s s + c ' / m
g A( s + c' / m) + Bs = s ( s + c ' / m) s ( s + c ' / m)

(2)

Equating like terms in the numerators yields
A+ B=0 g= c' A m

Therefore, A= mg c' B=− mg c'

These results can be substituted into Eq. (2), and the result can be substituted back into Eq. (1) to give V= mg / c' mg / c' v ( 0) − + s s + c' / ms + c' / m

Applying inverse Laplace transforms yields v= or mg mg −( c '/ m )t − e + v ( 0) e − ( c ' / m ) t c' c'

3

v = v(0)e −( c '/ m )t +

mg 1 − e −( c '/ m )t c'

(

)

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 v= mg 1 − e −( c '/ m )t c'

(

)(b) The numerical solution can be implemented as
12.5 ⎤ ⎡ v(2) = 0 + ⎢9.81 − (0) 2 = 19.62 68.1 ⎥ ⎣ ⎦

12.5 ⎡ ⎤ v(4) = 19.62 + ⎢9.81 − (19.62)⎥ 2 = 6.2087 68.1 ⎣ ⎦ The computation can be continued and the results summarized and plotted as:
t 0 2 4 6 8 10 12 v 0 19.6200 32.0374 39.8962 44.8700 48.0179 50.0102 dv/dt 9.81 6.2087 3.9294 2.4869 1.5739 0.9961 0.6304

60

40

20

0 0 4 812

Note that the analytical solution is included on the plot for comparison.

4

1.5 (a) The first two steps are

c(0.1) = 10 − 0.2(10)0.1 = 9.8 Bq/L c(0.2) = 9.8 − 0.2(9.8)0.1 = 9.604 Bq/L The process can be continued to yield
t 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 c 10.0000 9.8000 9.6040 9.4119 9.2237 9.0392 8.8584 8.6813 8.5076 8.3375 8.1707 dc/dt -2.0000 -1.9600 -1.9208 -1.8824-1.8447 -1.8078 -1.7717 -1.7363 -1.7015 -1.6675 -1.6341

(b) The results when plotted on a semi-log plot yields a straight line
2.4 2.3 2.2 2.1 2 0 0.2 0.4 0.6 0.8 1

The slope of this line can be estimated as
ln(8.1707) − ln(10) = −0.20203 1

Thus, the slope is approximately equal to the negative of the decay rate.
1.6 The first two steps yield

400 ⎤ ⎡ 400 y (0.5) = 0 + ⎢3 sin 2 (0)...