# Math 0250 Homework 10

• 3.6 Number 11
The characteristic equation is , which has as a double root. This gives the two solutions to the differential equation u1=e0 = 1 and u2=t, so the general solution is u = c1+c2t.

• 3.6 Number 13
The characteristic equation is which factors as . The two independent solutions to the differential equation are u1 = e2t and u2=te2t. The general solution is u = c1 e2t + c2 t e2t.

• 3.6 Number 16
From number 13, the general solution is

u = c1 e2t + c2 t e2t.

Differentiating gives

u' = 2 c1 e2t + c2(2te2t + e2t)

And plugging in the initial values gives

Solving gives c1=0 and c2=-2 so the desired solution is

u = -2 t e2t.

• 3.6 Number 18
The characteristic equation is , which has roots 10i -10i. The root gives the complex valued solution

The real and imaginary parts are real valued solutions, so the general solution is

• 3.6 Number 22
The characteristic polynomial is , which has roots . These give the complex solutions

Since the real and imaginary parts are solutions, this gives the general solution

• 3.6 Number 23
The characteristic equation is , which has roots . These give complex solutions

Since the real and imaginary parts are independent solutions, this gives the general solution

Differentiating gives

Plugging in the initial conditions gives

Solving gives c1=1 and c2=1/3, so the desired solution is

• 3.7.1 Number 2
The general solution in every case will be

with

The frequency of oscillation is .

• 3.7.2 Number 1
In each case, you want to solve the equation

4 y'' + c y' + 48y = 0

with the initial conditions y(0) = 0.5 and y'(0)=0. The characteristic equation has roots

The system is critically damped when , overdamped when c>27.7, and underdamped (leading to oscillation) when c<27.7. Thus, in all the given cases, the system is overdamped, and will not oscillate.

I'll omit detailed calculations, but for each case, the above formula gives two distinct real roots, say and , and the general solution has the form

Just plug in the initial conditions and solve for c1 and c2.

• 3.7.2 Number 5
a
The equation of motion is

4 y'' + 40 y' + 64 y = 0

with initial conditions y(0)=0 and y'(0)=40. The characteristic roots are -8 and -2, and the general solution is

y = c1 e-8t + c2 e-2t

Plugging in the initial conditions gives

c2 = -c1 = 20/3

so

To find the maximum displacement, put y'=0 and solve for t to get that the maximum displacement occurs when . The maximum displacement is therefore

b
The equation of motion is

2 y'' + 28 y' + 24 y = 0

with the same initial conditions as the previous problem. The characteristic roots are , and the general solution is

The initial conditions give c1=0 and c2=8, so

The maximum displacement occurs the first time that the derivative vanishes, which happens at . Substituting this value of t in the formula for y gives a maximum displacement of 0.0042 meters.

• 3.9 Number 3
Look for a solution of the form u = a e-t. Plugging in and cancelling the exponentials gives 2a=1, so a=1/2, and the required solution is

• 3.9 Number 5
Complexify to get

z'' + 10 z = 5 eit.

Look for a solution of the form z=aeit. Plug in and cancel the exponentials to get 9a=5, so a = 5/9 and the complex solution is

Finally, taking the imaginary part gives the real solution

• 3.9 Number 9
Comlexify to get

z''+z = 4e(1+i)t

and look for a solution of the form z=ae(1+i)t. Plug in and cancel exponentials to get (1+2i)a=4 or

so

Taking teh imaginary part gives the real solution

• 3.9 Number 13
Look for a solution of the form u = (a + bt)et. Differentiating twice gives

Plug in, and cancel the exponentials to get

(9a+6b) + (9bt) = t

Equating coefficients gives the linear system

Solving gives b=1/9 and a=2/27, so

• 3.9 Number 15
The complementary solution is

Look for a particular solution of the form u = ae2t. Plug in and cancel to get 5a=1, so a=1/5, and up=(1/5)e2t, so the general solution is

• 3.9 Number 17
The complementary solution is

Look for a particular solution of the form u=at2 + b t + c. Plug in to get

9at2 + 9bt + (2a+9c) = t2.

Equating coefficients gives the linear system

Solving gives a=1/9, b=0, and c=2/81, so

and the general solution is

• 3.9 Number 25
The complementary equation has a double characteristic root of -2, so the complemenary solution is

uh = (c1 + c2 t) e-2t

Look for a particular solution of the form

u = at2 + b t + c

Plugging in gives

4at2 + (8a+4b) t + (2a + 4b + 4c) = t2

and equating coefficients gives the linear system

Solving gives a=1/4, b=-1/2, and c = 3/8, so

and

The derivative is

Inserting the given initial conditions gives

and solving gives c1=-3/8 and c2=1/4, so

Frank Beatrous
1998-11-23