MATH 150 Separable Equations
Cheung Man Wai Tutorial 2
An equation is said to be separable if it can be written in the form M (x)dx + N (y )dy = 0. Moreover, if we require the initial condition, namely y (x0 ) = y0 , then we can solve it as x y M (x)dx +
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Final Exam
MATH 150: Introduction to Ordinary Differential Equations
Mathematical Modeling Using Differential Equations: Models from Physical Laws and Conservation Laws
September 25, 2008
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Mathematical modeling is a basic tool for scientists and engineers in doing their re
Separable and linear equations: some problems
September 18, 2008
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Separable equations are among the very simple types of ODEs for which we can easily recognize and nd the solutions. Knowing how to compute
Review of Calculus: some examples and problems
September 11, 2008
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Name: WANG Wenjie (
#)
Ofce: Room 3474 , Desk 4 E-mail: wjwang@ust.hk Ofce Hours: Tuesday , 4:00pm-6:00pm Tel: 23350839 or 63718858
Fir
Midterm Exam
MATH 150: Introduction to Ordinary Dierential Equations J. R. Chasnov
24 October 2007 Answer ALL questions Full mark: 50; each question carries 10 marks. Time allowed one hour
Directions This is a closed book exam. You may write on the front
MATH150 Introduction to Ordinary Differential Equations Midterm Exam Suggested Solution Prepared by CHAU Suk Ling, Pat
1.
Find the solution y = y ( x) of the following initial value problem:
yy ' = (e x e x ) , y (0) = 2 .
Solution
y dy = e x e x dx ydy =
Midterm Exam
MATH 150: Introduction to Ordinary Dierential Equations J. R. Chasnov
7 November 2008 Answer ALL questions Full mark: 40; each question carries 10 marks. Time allowed 1 hour
Directions This is a closed book exam. You may write on the front an
MATH 150 Systems of rst-order linear equations
Cheung Man Wai Tutorial 10
Example 1. Find the general solution of x= 3 4 1 1 x.
First we use the ansatz x = vet as what we did before, then we get the characteristic equation det(A I) = 3 1 4 1 = ( 3)( + 1)
MATH 150 Series
Cheung Man Wai Tutorial 9
The purpose of this note is to nd two independent solutions around x0 of the following P (x)y + Q(x)y + R(x)y = 0, (1) where P (x), Q(x) and R(x) are polynomials with no common factors. We shall consider two cases
MATH150 Introduction to Ordinary Differential Equations Homework 1 Suggested Solution Prepared by CHAU Suk Ling, Pat
1.
y' =
2x y + x2 y
dy 2x = dx y (1 + x 2 ) 2x ydy = dx 1 + x2 y x 2x 2 ydy = 0 1 + x 2 dx
x y2 1 d (1 + x 2 ) = 0 2 2 1 + x2 x y2 2 = ln(
MATH150 Introduction to Ordinary Differential Equations Homework 3 Suggested Solution
1.
Let y ( x) = an x n . Then y '( x) = nan x n 1 and y '( x) = n(n 1)an x n 2 .
n=0 n =1 n=2
Putting these into y ' xy ' y = 0 ,
n(n 1)an x n2 x nan x n1 an x n = 0
n=
HW3 solution for Q4-Q7 4. Apply the Laplace transform and the initial conditions, we obtain s2 X + 4 X = thus X = = 1 e2s (s2 + 1)(s2 + 4) 1 e2s 1 1 3 s2 + 1 s2 + 4 1 e2s 2 , s2 + 1 s + 1
.
Thus by Table 5.1 in the lecture note, x(t) = 1 [1 u2 (t)](2 sin
Math 150, Final Exam, Spring 2001
1. Solve the following Euler's equation (8pts)
t y
2
00
+ 5 + 13 = 0
ty
0
y
t >
0
:
2. Find the solution of the following initial value problem (8pts)
y
00
+ 2 + 5 = ( ; 2) + 5
y
0
y
t
u2
()
t
y
(0) = 1
y
0
(0) = ;1
:
3.
Section I - Introduction
1. Preliminaries 1.1 Function of single variable, y = f (x) . y is called dependent variable. x is called independent variable. Derivative of y dened as:
dy dx dy dx
= lim
f (x+
x0
x)f (x) x
is also called ordinary derivative of
and if follows that (as + b)(s2 + 1) + (cs + d)(s2 1) = s2 for all s. By setting s = 1 and s = 1, respectively in the last equation, we obtain the pair of equations 2(a + b) = 1, 2(a + b) = 1,
and therefore a = 0 and b = 1/2. If we set s = 0, then b d = 0
Thus by (3) the general solution of the ode is x(t) = e2t (A cos t + B sin t). The derivative is x(t) = 2e2t (A cos t + B sin t) +e2t (A sin t + B cos t). Applying the initial conditions x(0) = 1 = A, x(0) = 0 = 2A + B, which give us A = 1, Therefore, x(t
The unknown constants c1 and c2 can be determined by the initial conditions x(t0 ) = x0 and x(t0 ) = u0 . Let us give some examples of this method. Example 1. Find the solution of the given initial value problem x + x 2x = 0, x(0) = 2, x(0) = 1
and descr
Applications: Modeling with First Order Equations
In this note we present examples of a few simple mathematical models with rst order dierential equations, like compound interest with deposits or withdrawals, velocity of falling mass, escape velocity and
2. Solve the initial value problem y = (1 + 3x2 )/(3y 2 6y ), y (0) = 1
and determine the interval in which the solution is valid. As before, we rst manipulate the equation to the form (3y 2 6y )dy = (1 + 3x2 )dx. By the initial condition, one has
y 1
(3y
homework 2
October 29, 2007
1. The general solution is x = c1 et + c2 et , and the initial conditions are satised if c1 = 1 , c2 = 1. Thus 4 x= 1t e + et 2 4 1t e et. 4
Then the minimum value of the solution is xmin = x(ln2) = 1. 2. The characteristic equ