Math 308, Sections 301, 302,
Summer 2008
Lecture 3.
06/2/2008
Chapter 2. First-Order Differential Equations
In this chapter we learn how to obtain solutions for some specyfic
types of first-order equations. We begin by studying separable
equations, then the linear equations.
Section 2.1 Introduction: Motion Of a Falling Body (Home
reading)
Sections 2.4–2.6 are skipped.
Section 2.2 Separable Equations
dy
= f (x, y )
(1)
dx
Sometimes function f (x, y ) can be represented as a product of two
functions, one of which depens ONLY on x, another depens ONLY
on y , or f (x, y ) = g (x)h(y ). Then
dy
= g (x)h(y ).
dx
Lets multiply equation (5) by
(2)
dx
h(y )
dy
= g (x)dx,
h(y )
Z
Z
dy
= g (x)dx,
h(y )
The solution to an equation (5) is H(y ) = G (x) + C , here H(y )
is an antiderivative of 1/h(y ), G (x) is an antiderivative of g (x), C
ia a constant.
Example 1. Determine whether the given equation is separable.
(a) (x − 2y )2 y ′ = 2
(b) y 4 ey + (x 3 + 1)y ′ = y ′ (x 3 + 1)e2y
√
(c) yx ln xdx − y dy + x ln xdx = 0
(d) y ′ = cot2 x2 + y − 1 + 12
Example 2. Solve the equations/initial value problems:
(a) xydx + (x + 1)dy = 0
(b) (x 2 − 1)y ′ + 2xy 2 = 0, y (0) = 1
√
(c) xydx − x 2 + 1 ln2 ydy = 0
(d) x cos2 ydx − ex sin 2y dy = 0,
y (0) = 0
Section 2.3 Linear Equations
A linear first-order equation is an equation that can be expressed
in the form
dy
+ a0 (x)y = b(x),
(3)
dx
where a0 (x), a1 (x), b(x) depend only on x.
We will assume that a0 (x), a1 (x), b(x) are continuous functions of
x on an interval I .
For now, we are interested in those linear equations for which
a1 (x) is never zero on I . In that case we can rewrite (4) in the
standard form
a1 (x)
dy
+ P(x)y = Q(x),
(4)
dx
where P(x) = a0 (x)/a1 (x) and Q(x) = b(x)/a1 (x) are continuous
on I .
Equation (5) can be solved by finding an integrating factor µ(x)
such that
d
dy
dµ
[µy ] =
µ+
y = µQ(x).
dx
dx
dx
To solve the first order linear equation
(a) Write the equation in the standart form (5).
(b) Find the integrating factor µ(x) solving differential equation
dµ
− P(x)µ = 0.
dx
(c) Integrate the equation
d
[µy ] = µQ(x)
dx
and solve for y by dividing by µ(x).
The solution
Z
1
µ(x)Q(x)dx + C
µ(x)
where C is a constant, to equation (5) is called the general
solution.
y (x) =
(5)
Example 3. Find the general solution to the equation
xy ′ − y = − ln x.
Example 4. Solve the initial value problem
dy − ydx − 2xex dx = 0, y (0) = e − 2
Existence and uniqueness of solution
Theorem Suppose P(x) and Q(x) are continuous on some
interval I that contains the point x0 . Then for any choice of initial
value y0 , there exists a unique solution y (x) on I to the initial
value problem
y ′ + P(x)y = Q(x),
y (x0 ) = y0 .
In fact, the solution is given by (6) for a suitable value of C .
(6)