
3.9 Number 4
The complementary solution is
u_{h} = c_{1} e^{t} + c_{2} e^{t}.
Since e^{t} is already part of the complementary solution, multiply by
t, and look for a particular solution of the form u=a t e^{t}.
Substituting into the differential equation gives
a (2+t)e^{t}  a te^{t} = e^{t}
and simplifying gives 2a=1, so a=1/2. Thus

3.9 Number 10
The complementary solution is
To find particular solution, look at the complexified equation
u'' + u = 4te^{it}.
Since one of the terms of
a t e^{it} + b e^{it} is already part of the
complementary solution, multiply through by t, and look for a particular
solution of the form
u = a t^{2} e^{it} + b t e^{it}.
Substituting into the differential equation gives
(at^{2} + (4iab)t + (2a+2ib)) e^{it} + (at^{2} +bt)) e^{it} = 4t e^{it}.
and simplifying gives
(4ai) t + (2a+2ib) = 4t.
Equating coefficients on the left and right gives the system
Solving for a and b gives
a=i and b=1, so the complexified solution is
The required particular solution is the imaginary part:

3.10.2 Number 4
The mass is
where g is the acceleration due to gravity, so the differential equation
for the velocity is
with the initial condition v(0)=0. Putting
and
simplifying gives
v' + .049 v = 9.8
The complementary solution is
v_{h} = c e^{0.049 t}, and a particular
solution is v_{p} = 200, so the general solution is
v = 200 + c e^{0.049t}.
The initial condition gives c=200 so
v = 200 (1e^{0.049t})
Clearly the terminal velocity is
.
To find when the object
reaches 99% of terminal velocity, set
v = (0.99)(200) = 198 and solve:

3.10.2 Number 6
The equation of motion is
with initial condition
y(0)=y'(0)=0. The complexified equation is
2z'' + 32 z = 0.1 e^{4it}
Look for a solution of the form
z=ate^{4it}. Differentiating twice gives
Inserting this into the equation and cancelling the exponentials gives
16 i a= 0.1
so
and
Taking the imaginary part gives the particular solution
You can easily verify that this solution already satisfies the initial
conditions, so you don't need to bother with a complementary solution.
The solution is thus a cosine wave modulated by the amplitude 0.00625 t.
The system will fail when the amplitude reaches 0.5, which happens when
t=0.5/0.00625 = 80 seconds.

3.10.3 Number 2
The equation of motion is
which complexifies to
3 z'' + 5 z' + 12 z = 20 e^{2it}
Look for a solution of the form
z = a e^{2it}
Substituting gives
10 i a = 20
so
,
and
Taking the imaginary part gives
Thus the amplitude is 2 and the phase lag is .

3.10.3 Number 5
The equation of motion is
which complexifies to
Look for a solution of the form
.
Plug it to get
so
The number a is the complex amplitude, which encodes both the real
amplitude and a phase lag. The real amplitude is
Plugging in the four given frequencies gives amplitudes of 1.3558,
2.3783, 5, and 1.0963 respectively.