Page 212 - Numerical Analysis Using MATLAB and Excel

Solutions to End−of−Chapter Exercises

We will use MATLAB to find the first and second derivatives of this expression.

syms t k1 k2 k3 k4 k5 % Define symbolic variables
y0=k3*cos(2*t)+k4*sin(2*t)+k5; % Assumed form of total solution
y1=diff(y0); f1=simple(y1) % Compute and simplify first derivative
f1 =
-2*k3*sin(2*t)+2*k4*cos(2*t)
Thus, the first derivative of y F is

dy ⁄ dt = – 2k sin 2t + 2k cos 2t
3
4
F
y2=diff(y0,2); f2=simple(y2) % Compute and simplify second derivative
f2 =
-4*k3*cos(2*t)-4*k4*sin(2*t)
and the second derivative of is
y

2
2
d y ⁄ dt = – 4k cos 2t 4k sin 2t
–
4
F
3
f=y2+2*y1+y0; f=simple(f) % Form and simplify the left side of the given ODE
f =
-3*k3*cos(2*t)-3*k4*sin(2*t)-4*k3*sin(2*t)+4*k4*cos(2*t)+k5
Simplifying this expression and equating with the right side of the given ODE we obtain:

– ( 3k + 4k ) 4 cos 2t – ( 4k + 3k ) 4 sin 2t + k = cos 2t 1 -
--
------------- +
3
3
5
2
2
Equating like terms and solving for the terms we obtain
k
⁄
– 3k + 4k = 12
3
4
– 4k – 3k = 0
4
3
⁄
k = 12
5
⁄
⁄
Simultaneous solution of the first two equations above yields k = – 3 50 and k = 450 .
3
4
Therefore, the forced response is
)
⁄
⁄
⁄
)
y = – ( 3 50 cos 2t + ( 4 50 sin 2t + 1 2
F
and the total response is
t –
t –
-- –
y = k e + k te + 1 - 3cos 2t – 4sin 2t
---------------------------------------
2
1
50
2
Check with MATLAB:
Numerical Analysis Using MATLAB® and Excel®, Third Edition 5−51
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