Ordinary and Partial Differential Equations
Lecture 2. First-order differential equations - Part 1.
Separable equations. Linear equations.
Bibliography: Nagle, Snaff, Sieder - ”Fundamentals of DEs and BVPs” (6th ed.) - sections 2.1, 2.2, 2.3
INTRODUCTION: Motion of a falling body
Problem: An object falls through the air toward Earth. Assuming that the only forces acting on the
object are gravity and air resistance, determine the velocity of the object as a function of time.
Newton’s second law can be applied to the falling object, leading to the differential equation:
dv
=F
dt
where v = v(t) is the velocity of the body at the moment of time t, m is the mass of the body, and F is the
dv
total force acting on the body. The first-order derivative
of the velocity represents the acceleration.
dt
Gravity can be expressed by mg, where g is the gravitational acceleration. Air resistance can be reasonably
represented by −bv, where b is a positive constant depending on the density of the air and the shape of
the object, and v is the velocity of the object. We use the negative sign because air resistance is a force
that opposes the motion.
Hence, our differential equation becomes:
m
dv
= mg − bv
(1)
dt
This is a first-order linear differential equation, with the dependent variable v = v(t).
We solve this equation using a technique called separation of variables.
Isolating the variables v and t on opposite sides of the equation, we obtain:
∫
∫
dv
dt
dv
dt
1
t
mg
=
⇒
=
⇒ − ln |mg − bv| =
+c ⇒ v =
− C · e−bt/m
mg − bv
m
mg − bv
m
b
m
b
mg
Associating the initial condition v(0) = v0 to the differential equation (1), we find C = v0 −
.
b
(
)
mg
mg
The solution of this IVP is: v(t) =
− v0 −
· e−bt/m .
b
b
m
SEPARABLE EQUATIONS
♣ A separable differential equation is and equation of the form:
dy
= g(x)p(y)
(2)
dx
where the function g depends only on the independent variable x and the function p depends only on the
dependent variable y.
Method for solving separable equations:
• separate the variables on opposite sides of the equation:
1
, and so: h(y)dy = g(x)dx
p(y)
∫
∫
• integrate on both sides:
h(y)dy = g(x)dx
1
dy = g(x)dx
p(y)
• denote h(y) =
⇒
H(y) = G(x) + c
where H and G are primitives of the functions h and g, respectively.
• the equation H(y) = G(x) + c represents an implicit solution for the separable equation (2).
Caution: Constant functions y ≡ c such that p(c) = 0 are also solutions to (2), which may or may not
be found by this method.
c 2011/2012 Eva Kaslik - West University of Timişoara, Romania
⃝
LINEAR EQUATIONS
A first-order linear differential equation is of the form
a1 (x)
dy
+ a0 (x)y = b(x)
dx
(3)
where the function a0 (x), a1 (x) and b(x) do not depend on the dependent variable y.
Method for solving linear equations:
• write eq. (3) in the standard form:
dy
a0 (x)
b(x)
+ P (x)y = Q(x), where P (x) =
and Q(x) =
dx
a1 (x)
a1 (x)
• calculate the integrating factor µ(x) = e
• notice that
∫
P (x)dx
∫
dµ
= P (x)e P (x)dx = P (x)µ(x)
dx
• multiply the equation by µ(x) to obtain
µ(x)
dy
+ P (x)µ(x)y = Q(x)µ(x)
dx
⇔
µ(x)
dy dµ
+
y = Q(x)µ(x)
dx dx
⇔
d
[µ(x)y] = Q(x)µ(x)
dx
∫
• integrate the last equation: µ(x)y =
Q(x)µ(x)dx + C
• obtain the explicit solution:
(∫
) (∫
)
[ ∫
]
1
y(x) =
Q(x)µ(x)dx + C =
Q(x)µ(x)dx + C exp − P (x)dx
µ(x)
(4)
Existence and Uniqueness of the Solution
Suppose P and Q and are continuous on an interval (a, b) that contains the point x0 . Then for any choice
of initial value y0 , there exists a unique solution on (a, b) to the initial value problem
dy
+ P (x)y = Q(x)
dx
(∫
The solution is:
)
x
y(x) =
;
Q(s)µ(s)ds + y0
x0
y(x0 ) = y0
[ ∫
exp −
P (s)ds
x0
2
]
x