Solutions to Math 2250-3 Midterm Exam 1, Fall 2007, Exams of Mathematics

The solutions to the math 2250-3 midterm exam held in fall 2007. It includes step-by-step solutions to eight problems covering topics such as first-order odes, separable differential equations, linear differential equations, and equilibrium solutions.

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Pre 2010

Uploaded on 08/30/2009

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Solutions to Midterm Exam 1
Math 2250-3, Fall 2007
Instructions: No calculators of any kind are permitted for use on the test, nor is scratch paper allowed.
You have 50 minutes in which to answer all 8 questions. The value of each question as a percentage of the
value of the entire test is indicated next to each question number (this is also approximately the percentage
of 50 minutes each question should require: 5% = 2.5 minutes). Please show all your work in the space
provided below each question.
1. (5%) If the first-order ODE
dx
dt =f(x, t)
is autonomous, what does this imply about the function f?
fis autonomous if and only if f(x, t) = f(x), i.e., fis a function of xonly.
2. (15%) Solve the initial value problem
dy
dx =y1/2
x+ 1 , y (0) = 0.
This is separable!
Separate variables: y1/2dy =dx
x+ 1
Integrate: 2y1/2= ln(x+ 1) + C
Solve for y:y(x) = (ln(x+ 1) + C)2
4
Solve for C: 0 = y(0) = (ln(1) + C)2
4=C2
4=> C = 0
Particular Solution: y(x) = (ln(x+ 1))2
4
3. (15%) Solve the initial value problem
dy
dx =2y
x+cos(x)
x2, y(π) = 1.
This is linear!
Find Pand Q:P(x) = 2
x, Q(x) = cos(x)
x2
Integrating factor: eRP(x)dx = e2Rdx/x = e2 ln(x)=x2
Integral with Q:ZeRP(x)dxQ(x)dx =Zx2cos(x)
x2dx =Zcos(x)dx = sin(x) + C
General solution: y(x) = eRP(x)dx ZeRP(x)dx Q(x)dx +C=sin(x) + C
x2
Solve for C: 1 = y(π) = sin(π) + C
π2=C
π2C=π2
Particular Solution: y(x) = sin(x) + π2
x2
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Solutions to Midterm Exam 1

Math 2250-3, Fall 2007

Instructions: No calculators of any kind are permitted for use on the test, nor is scratch paper allowed.

You have 50 minutes in which to answer all 8 questions. The value of each question as a percentage of the

value of the entire test is indicated next to each question number (this is also approximately the percentage

of 50 minutes each question should require: 5% = 2.5 minutes). Please show all your work in the space

provided below each question.

  1. (5%) If the first-order ODE

dx

dt

= f (x, t)

is autonomous, what does this imply about the function f?

f is autonomous if and only if f (x, t) = f (x), i.e., f is a function of x only.

  1. (15%) Solve the initial value problem

dy

dx

y 1 / 2

x + 1

, y(0) = 0.

This is separable!

Separate variables: y

− 1 / 2 dy =

dx

x + 1

Integrate: 2 y

1 / 2 = ln(x + 1) + C

Solve for y: y(x) =

(ln(x + 1) + C)

2

Solve for C: 0 = y(0) =

(ln(1) + C)

2

C

2

=> C = 0

Particular Solution: y(x) =

(ln(x + 1)) 2

  1. (15%) Solve the initial value problem

dy

dx

2 y

x

cos(x)

x 2

, y(π) = 1.

This is linear!

Find P and Q: P (x) =

x

, Q(x) =

cos(x)

x 2

Integrating factor: e

R P (x)dx = e

2

R dx/x = e

2 ln(x) = x

2

Integral with Q:

e

R P (x)dx Q(x)dx =

x

2 cos(x)

x 2

dx =

cos(x)dx = sin(x) + C

General solution: y(x) = e

R P (x)dx

[∫

e

R P (x)dx Q(x)dx + C

]

sin(x) + C

x 2

Solve for C: 1 = y(π) =

sin(π) + C

π^2

C

π^2

⇒ C = π

2

Particular Solution: y(x) =

sin(x) + π 2

x 2

  1. (5%) Find the equilibrium solutions of

dx

dt

= x

2 (1 + x).

Equilibrium solutions are the zeros of f (x) = x

2 (1 + x), which are x = 0, −1.

  1. (10%) x = e is an equilibrium solution of

dx

dt

= ln(x) −

x

e

What is the stability of x = e?

First, take derivative of f (x) = ln(x) − x/e with respect to x:

f

′ (x) =

x

e

Then determine the sign of f ′ at the equilibrium solution x = e:

f

′ (e) =

e

e

Therefore, x = e is neutrally stable.

  1. (10%) Find the determinant of the matrix A, where

A =

Expand along 1st row:

|A| =

1+

1+

2+

2+