Mathematics 255
15 October 2008
Quiz # 4
Name
Score: out of 30 points This quiz is due at the beginning of class on Friday, 17 October. The work you hand in must be your own. (7) 1. For t > 0, nd the general solution of t d2 y dy + (1 2t) + (t 1)y = 0, 2
Improved Eulers method
Again consider the initial-value problem dy = f (t, y ), dt
Mathematics 255: Lecture 10
Improved Euler Method
y (t0 ) = y0 .
As before, we want to approximate the solution on the interval a [t0 , t0 + a] using N steps of size h = N
Mathematics 255
17 September 2008
Quiz # 2
Name
Score: out of 30 points This quiz is due at the beginning of class on Friday, 19 September. The work you hand in must be your own. Show all your work! (5) 1. Solve the initial-value problem y + t + 2 y dy =
Integral equations
Mathematics 255: Lecture 8
Picard Iteration Dan Sloughter
Furman University
Note: the dierential equation dy = f (t, y ), dt is equivalent to the integral equation
t
y (t0 ) = y0 ,
y (t) = y0 + September 16, 2008
t0
f (s, y (s)ds.
Dan S
Exponential growth and decay
Mathematics 255: Lecture 3
Exponential Growth and Decay
The initial-value problem dy = ky , dt arises in many applications. From our work above, we nd the solution
Rt
0
y (0) = y0
Dan Sloughter
Furman University
y = y0 e Septe
Definition
Mathematics 255: Lecture 4
Separation of Variables We say a differential equation of the form g (t) dy = dt f (y ) is separable.
Dan Sloughter
Furman University
September 5, 2008
Dan Sloughter (Furman University)
Mathematics 255: Lecture 4
Sept