Study your flashcards anywhere!

Download the official Cram app for free >

  • Shuffle
    Toggle On
    Toggle Off
  • Alphabetize
    Toggle On
    Toggle Off
  • Front First
    Toggle On
    Toggle Off
  • Both Sides
    Toggle On
    Toggle Off
  • Read
    Toggle On
    Toggle Off

How to study your flashcards.

Right/Left arrow keys: Navigate between flashcards.right arrow keyleft arrow key

Up/Down arrow keys: Flip the card between the front and back.down keyup key

H key: Show hint (3rd side).h key

A key: Read text to speech.a key


Play button


Play button




Click to flip

5 Cards in this Set

  • Front
  • Back
Given a general differential equation in standard form, dy/dx = ay - b, and given the initial condition y(0) = y_0, find the general solution and solve the initial value problem. Also find the parameter value(s) corresponding to the equilibrium solution.
general solution: y = b/a + ce^(at), a != 0, c = +or- e^C
particular solution: y = b/a + [y_0 - b/a]e^(at)
equilibrium solution: y = b/a, c= 0
The geometric representation of the general solution is an infinite family of curves called ________. Each one is associated with a particular ________.
Satisfying an initial condition amounts to ________.
integral curves


finding the integral curve that corresponds to the given initial point
Solve the general predator-prey (owl-mouse) problem given p(0) = p_0.
p = k/r + [p_0 - k/r]e^(rt)
Explain the falling object and owl-mouse models in terms of the general equation.

a = -gamma/m
b= -g

v = gm/gamma + [v_0 - gm/gamma]e^-(tgamma/m)

As t increases, the second term dies off and we converge to equilibrium. The higher the drag coefficient relative to the mass of the object, the faster this occurs.

a = k
b = r

p = k/r + [p_0 - k/r]e^(rt)

If the population begins at the equilibrium value, it stays. Otherwise, the second term grows exponentially and the mouse population either dies off or grows infinitely.
Consider an electric circuit containing a capacitor, resistor, and battery. The Charge Q(t) on the capacitor satisfies the equation

RdQ/dt + Q/C = V

where R is the resistance, C is the capacitance, and V is the constant voltage supplied by the battery.

a)If Q(0) = 0, find Q(t) at any time t.

b)Find the limiting value Q_L that Q(t) approaches after a long time.

c) Suppose that Q(t_1) = Q_L and that the battery is removed from the circuit at t = t_1. Find Q(t) for t > t_1.
a) Q(t) = CV(1 - e^(-t/CR))
b)Q_L = lim t->inf Q(t) = CV
c)Q(t) = CVe^-(t-t_1)/CR