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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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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.
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.
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
2
Particular Solution: y(x) =
(ln(x + 1)) 2
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
π^2
⇒ C = π
2
Particular Solution: y(x) =
sin(x) + π 2
x 2
dx
dt
= x
2 (1 + x).
Equilibrium solutions are the zeros of f (x) = x
2 (1 + x), which are x = 0, −1.
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.
Expand along 1st row:
1+
1+
2+
2+