Differential Equations, Probability, and Statistics
MATH 2A

Fall 2013
MATH 2A 2013 HOMEWORK 5 SOLUTIONS
Problem 1 10 points
1
Find the general solution of the following equation:
d2 x
d4 x
5 2 + 4 x = et
dt4
dt
4
2
x
x
Solution. First consider the associated homogenous
Differential Equations, Probability, and Statistics
MATH 2A

Fall 2013
85
PART IV: LINEAR EQUATIONS (ADDITIONAL TOPICS)
32. Wronskian
32.1. Denition. The Wronskian of two functions y1 , y2 C 1(I ) is the function
W=
y1
y1
y2
C (I ).
y2
The Wronskian of n functions yj C
Differential Equations, Probability, and Statistics
MATH 2A

Fall 2013
Ma2a (analytical)
Fall 2011
PART I: FIRST ORDER ODEs
1. Introduction
1.1. Fundamental theorem of calculus. The simplest (trivial) d.e.:
y (x) = f (x).
Examples:
(i) Find all (maximal) solutions of the
Differential Equations, Probability, and Statistics
MATH 2A

Fall 2013
MATH 2A 2013 HOMEWORK 6 SOLUTIONS
Problem 1 10 pts
1
Find the eigenvalues and eigenfunctions of the BVPs
x2 u xu + u = 0, u(1) = u(2) = 0
Solution.
dx
Since x [1, 2], we can put x = ez i.e. consider
Differential Equations, Probability, and Statistics
MATH 2A

Fall 2013
MATH 2A 2013 HOMEWORK 1 SOLUTIONS
Problem 110 pts (Problem 8.7) Show that for k = 0 the solution of the IVP
x = kx x2 ,
x(0) = x0
is
kekt x0
x0 (ekt 1) + k
Using this explicit solution describe the b
Differential Equations, Probability, and Statistics
MATH 2A

Fall 2013
MATH 2A 2013 HOMEWORK 2 SOLUTIONS
blems 1 and 220 pts Plot a slope eld for the following equations:1
y = y 2 x Solution. We can plot it using the command (in Mathematica 8.0)
VectorPlot[cfw_1, y^2 
Differential Equations, Probability, and Statistics
MATH 2A

Fall 2013
MATH 2A 2013 HOMEWORK 3 SOLUTIONS
Problem 110 pts (Problem 10.2) Find an integrating factor depending only on x that makes the equation
ey sec x + 2 cot x ey
dy
=0
dx
exact, and hence nd its solution
Differential Equations, Probability, and Statistics
MATH 2A

Fall 2013
MATH 2A 2013 HW4 SOLUTIONS
Problem X110 pts Solve the equation
1
x=
t
t 2 x + x3
Solution. Rewriting the equation as (t2 x + x3 )dx tdt = 0, we can check that t (t2 x + x3 ) = 2tx =
1
0 = x t, so thi