Page 184 - Numerical Analysis Using MATLAB and Excel

Using the Method of Variation of Parameters for the Forced Response

Also, from (5.69),

du du
2
1
--------y + --------y = 0
dt 1 dt 2
or
du du
1
2
-------- cos 2t + -------- sin 2t = 0 (5.86)
dt dt
and from (5.70),
du dy du dy du du
1
1
2
2
1
(
⋅
)
⋅
)
-------- -------- + -------- -------- = ft() = -------- – 2sin 2t + -------- 2cos( 2 2t = tan 2t (5.87)
dt dt dt dt dt dt
Next, we find du dt⁄ and du ⁄ dt by Cramer’s rule as follows:
2
1
0 sin 2t sin 2 2t
du tan 2t 2cos 2t – --------------- – sin 2 2t
cos
2t
1
-------- = ------------------------------------------------------ = -------------------------------------------- = ------------------ (5.88)
dt cos 2t sin 2t 2cos 2 2t + 2sin 2 2t 2cos 2t
– 2sin 2t 2cos 2t
and
cos 2t 0
du – 2sin 2t tan 2t sin 2t
2
-------- = -------------------------------------------------- = ------------ (5.89)
dt 2 2
Now, integration of (5.88) and (5.89) and substitution into (5.85) yields
2
1 sin 2t sin 2t 1
∫
-
)
-
u = – -- --------------- td = ------------ – -- ln ( sec 2t + tan 2t + k 1 (5.90)
1
2t
4
2 cos
4
1 cos 2t
∫
-
d
u = -- sin 2t t = – ------------- + k 2 (5.91)
2
2
4
sin
2tcos
2t
)
--------------------------- +
--- cos
y = u y + u y = --------------------------- – 1 2t ( ln sec 2t + tan 2t + k cos 2t – sin 2tcos 2t k sin 2t
1
1 1
2
2 2
4
4
4
(5.92)
1
-
)
– -- cos 2t (= ln sec 2t + tan 2t + k cos 2t + k sin 2t
4 1 2
Check with MATLAB:
y=dsolve('D2y+4*y=tan(2*t)')
y =
-1/4*cos(2*t)*log((1+sin(2*t))/cos(2*t))+C1*cos(2*t)+C2*sin(2*t)
Numerical Analysis Using MATLAB® and Excel®, Third Edition 5−23
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