# Math 0250 Homework 11

• 3.9 Number 4
The complementary solution is

uh = c1 e-t + c2 et.

Since et is already part of the complementary solution, multiply by t, and look for a particular solution of the form u=a t et. Substituting into the differential equation gives

a (2+t)et - a tet = et

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 = 4teit.

Since one of the terms of a t eit + b eit is already part of the complementary solution, multiply through by t, and look for a particular solution of the form

u = a t2 eit + b t eit.

Substituting into the differential equation gives

(-at2 + (4ia-b)t + (2a+2ib)) eit + (at2 +bt)) eit = 4t eit.

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 vh = c e-0.049 t, and a particular solution is vp = 200, so the general solution is

v = 200 + c e-0.049t.

The initial condition gives c=-200 so

v = 200 (1-e-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 e4it

Look for a solution of the form z=ate4it. 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 e2it

Look for a solution of the form

z = a e2it

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.