HW 3 Mathematics 503, Mathematical Modeling, CSUF
June 2 2007
Nasser M. Abbasi
June 15, 2014
Contents
1
Problem 1 (section 1.3,#1, page 40)
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2
Problem 2 (section 1.3,#1, page 40)
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1
Problem 1 (section 1.3,#1, page 40)
problem: Find the general solution of the following differential equations
(m) u00 + ω 2 u = sin βt, ω 6= β
solution:
We start by assuming ω is real, hence ω 2 must be positive.
Now, the general solution is
u (t) = uh (t) + u p (t)
= c1 u1 (t) + c2 u2 (t) + u p (t)
Where u1 (t) , u2 (t) are the 2 independent solutions of the homogeneous differential equation
u00 + ω 2 u = 0
and u p (t) is the particular solution.
To find u1 (t) and u2 (t), we assume the homogeneous solution is uh (t) = Aemt for some constants
A, m and substitute this assumed solution in the ODE. We obtain the characteristic equation Am2 emt +
ω 2 Aemt = 0 → m2 + ω 2 = 0 → m2 = −ω 2 or m = ±iω, hence u1 (t) = A1 eiωt and u2 (t) = A2 e−iωt .
Since the homogeneous solution is a linear combination of all the independent solutions, we take
the sum and the difference of these solutions, and using Euler relation which converts the complex
exponential to the trigonometric sin and cos functions we obtain
u1 (t) = cos ωt
u2 (t) = sin ωt
1
and we now write
uh (t) = c1 cos ωt + c2 sin ωt
Now to obtain u p (t), we use the method of undetermined coefficients. Assume
u p (t) = a cos βt + b sin βt
and plug into the original ODE, we obtain
(a cos βt + b sin βt)00 + ω 2 (a cos βt + b sin βt) = sin βt
(−β a sin βt + β b cos βt)0 + ω 2 (a cos βt + b sin βt) = sin βt
−β 2 a cos βt − β 2 b sin βt + ω 2 (a cos βt + b sin βt) = sin βt
−β 2 a + ω 2 a cos βt + −β 2 b + ω 2 b sin βt = sin βt
Hence by comparing coefficients we obtain
a ω2 − β 2 = 0
b ω2 − β 2 = 1
Since ω 6= β (given), then ω 2 − β 2 6= 0 , hence this must mean the following
a=0
b=
1
ω2 − β 2
Therefor, the particular solution now can be written as
u p = b sin βt
sin βt
= 2
ω −β2
Hence the general solution, which is y (t) = yh (t) + y p (t) is given by
βt
y (t) = c1 cos ωt + c2 sin ωt + ωsin
2 −β 2
Where c1 and c2 are constants that can be found from the initial conditions.
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3
2
Problem 2 (section 1.3,#1, page 40)
problem: Find the general solution of the following differential equations
(n) u00 + ω 2 u = cos ωt
solution:
First, let the forcing function be called f (t), hence f (t) = cos ωt in this example.
From (m) we found the homogeneous solution to be uh (t) = c1 u1 (t) + c2 u2 (t) where
u1 (t) = cos ωt
and
u2 (t) = sin ωt
Now to find the particular solution we can not use the method of undetermined coefficients since the
forcing frequency is the same as the undamped natural frequency of the system ω and this will lead
to the denominator going to zero. Hence use the method of variation of parameters which is a general
method to find particular solution which will work with this case.
u2 (t) f (t)
u1 (t) f (t)
u p (t) = −u1 (t)
dt + u2 (t)
dt
W
W
Where W is the Wronskian of u1 , u2 given by W = u1 u02 − u2 u01
Hence W (t) = cos ωt (ω cos ωt) − sin ωt (−ω sin ωt) = ω cos2 ωt + sin2 ωt = ω
Hence
Z
Z
sin ωt
cos ωt
sin (ωt) f (t) dt +
cos (ωt) f (t) dt
u p (t) = −
ω
ω
Z
Z
cos ωt
sin ωt
=−
sin (ωt) cos (ωt) dt +
cos (ωt) cos (ωt) dt
ω
ω
= I1 + I2
Z
Z
We now evaluate I1 and I2 . Start with the easy one, I2
sin ωt
cos (ωt) cos (ωt) dt
ω
Z
sin ωt
=
cos2 (ωt) dt
ω sin ωt 1
=
(sin 2tω + 2tω)
ω
4ω
sin ωt
=
(sin 2tω + 2tω)
4ω 2
Z
I2 =
and now I1
cos ωt
sin (ωt) cos (ωt) dt
ω
We can use integration by parts (do it twice) or use an trigonometric identity. From tables, Using the
formula of
λ +ζ
λ −ζ
2 sin
cos
= sin (λ ) + sin (ζ )
2
2
I1 = −
Z
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so if we let λ = 2ωt and ζ = 0 we obtain the integrand above, hence
2 sin (ωt) cos (ωt) = sin (2ωt)
Substitute into I1
cos ωt 1
I1 = −
sin (2ωt) dt
ω 2
cos ωt −1
=−
cos (2ωt)
2ω
2ω
cos ωt
=
cos (2ωt)
4ω 2
Z
Therefor
u p (t) = I1 + I2
cos ωt
sin ωt
=
cos (2ωt) +
(sin 2ωt + 2ωt)
2
4ω
4ω 2
cos (ωt) cos (2ωt) + sin ωt (sin 2ωt + 2ωt)
=
4ω 2
Hence the general solution is
u (t) = c1 u1 (t) + c2 u2 (t) + u p (t)
cos (ωt) cos (2ωt) + sin ωt (sin 2ωt + 2ωt)
= c1 cos ωt + c2 sin ωt +
4ω 2
Verify using Mathematica
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