Page 180 - Numerical Analysis Using MATLAB and Excel

Using the Method of Undetermined Coefficients for the Forced Response

– 2t
reduces to kte . Therefore the forced response will have the form

y = ( k t + k ) 4 e – 2t (5.63)
F
3
– 2t – 2t 2
But the terms e and te are also present in (5.61); therefore, we multiply (5.62) by t to
obtain a suitable form for the forced response which now is
3
y = ( k t + k t ) 4 2 e – 2t (5.64)
F
3
Now, we need to evaluate the constants k 3 and k 4 . This is done by substituting (5.64) into the
given ODE of (5.61) and equating with the right side. We use MATLAB do the computations as
shown below.

syms t k3 k4 % Define symbolic variables
f0=(k3*t^3+k4*t^2)*exp(−2*t); % Forced response (5.64)
f1=diff(f0); f1=simple(f1) % Compute and simplify first derivative
f1 =
-t*exp(-2*t)*(-3*k3*t-2*k4+2*k3*t^2+2*k4*t)
f2=diff(f0,2); f2=simple(f2) % Compute and simplify second derivative

f2 =
2*exp(-2*t)*(3*k3*t+k4-6*k3*t^2-4*k4*t+2*k3*t^3+2*k4*t^2)
f=f2+4*f1+4*f0; f=simple(f) % Form and simplify the left side of the given ODE

f = 2*(3*k3*t+k4)*exp(-2*t)
Finally, we equate f above with the right side of the given ODE, that is

(
23k t + k ) 4 e – 2t = te – 2t – e – 2t (5.65)
3
⁄
⁄
and we find k = 16 and k = – 12 . By substitution of these values into (5.64) and combining
3
4
the forced response with the natural response, we obtain the total solution
--t e
-
--t e
-
yt() = k e – 2t + k te – 2t + 1 3 – 2t – 1 2 – 2t (5.66)
2
1
6
2
We verify this solution with MATLAB as follows:
z=dsolve('D2y+4*Dy+4*y=t*exp(−2*t)−exp(−2*t)')
z =
1/6*exp(-2*t)*t^3-1/2*exp(-2*t)*t^2
+C1*exp(-2*t)+C2*t*exp(-2*t)

Numerical Analysis Using MATLAB® and Excel®, Third Edition 5−19

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