The University of Sydney
School of Mathematics and Statistics
Lectures Week 2 – Separable Differential Equations & Newtonian
Dynamics
MATH1023: Multivariable Calculus and Modelling Semester 1, 2023
1. Existence and uniqueness of solutions
2. Simple 1st Order DEs
3. Separable Equations
4. Describe the model of Newtonian Dynamics
5. Newton’s law of gravitation
6. Newton’s law contains Galileo’s law as a special case
Existence and uniqueness of solutions
Given an initial value problem (IVP) for a 1st-order DE
dy
dx
= f(x, y) y(x0) = y0
can we always find a solution? If yes, is the solution unique?
There are theorems that show that a solution indeed exists over a certain domain of x and y
if f(x, y) is continuous in that domain. The solution is unqiue if the values of f do not fluctuate
too widly.
Uniqueness implies that solution curves cannot intersect or even touch at a point.
First-Order Differential Equations
Four types of 1st order DE’s in standard form are:
The right-hand-side is a function of x only. By the Fundamental Theorem of
Calculus this equation has a solution if f(x) is continuous.
(Type 2) Example dy
dx
= 3y2.
We rewrite it by taking the reciprocal, where it transforms into a Type 1 equation.
(Type 3) - Separable Differential Equations
Have the form dy
dx
= g(x)h(y), where can separate the variables
dy
h(y)
= g(x) dx and
integrate both sides to yield
∫
dy
h(y)
=
∫
g(x) dx + C.
In general, this yields an implicit function of x involving one arbitrary constant.
Sometimes it may be possible to solve for y explicitly as a function of x.
Example
Find the general solution of cos xdy
dx
y2 tanx = 0.
Example – Exponential growth
Use separation of variables to find a solution to the equation of exponential growth dx
dt
=
kx with k > 0. Find a solution x = x(t) satisfying the condition x(0) = a0 where a is a
constant. Assume that x > 0.
2
Newtonian Dynamics
Developed by Isaac Newton (1642—1727), the model discussed is one of the most
complete and successful mathematical models ever proposed. It forms the basis of
almost all branches of science and provides some of the best examples of the importance
of differential equations for understanding the natural world.
Newton (and Leibniz independently) developed the differential calculus to deal with
rates of change.
Newton’s law of motion: d
dt
(mv) = F
– F is the force acting on a body,
– m is the mass of the body and v its velocity.
Constant mass
When the mass is constant, Newton’s law becomes. Once the forces F and the mass m are specified, the motion may be found by integrating
to obtain v(t). Recall that the velocity v =
dx
dt
is the rate of change of position x(t).
Therefore having found v(t), we can then integrate again to find x(t).
3
Force as function of v m
dv
dt
= F (v)
NOTE: This is a separable equation
Recall: that an object falling through air close to the surface of the Earth:
– is accelerated by its weight force,
– and decelerated by air resistance.
The weight force is W = mg
– m is the mass of the object,
– g is the acceleration due to gravity.
Resistance force
Assume the object is large and light, therefore it falls slowly.
The resistance force is then given by R = kmv
– v is the velocity
– m is the mass
– k is a constant.
The value of k depends on the properties of the object and the medium through which it
falls.
Example An object falls from rest through air that provides a resistance kmv . k, m
and v, positive by assumption. Find the subsequent velocity and distance fallen as functions
of time.
4
Force as a function of distance x
Very often the forces acting on an object depend on the position of the object rather
than time or velocity. The equation of motion becomes m
dv
dt
= F (x) .
We cannot integrate directly because now there are three variables v, x, t. . We
must first reduce the number to two.
Newton’s law of gravitation
In this case the force depends on the distance only. The law says that the force varies:
inversely as the square of the distance between the object and the centre of the
attracting body. The equation of motion is
m
dv
dt
= GMm
r2
where r is the radial distance from the centre to the object, M is the mass of the attracting
body and G is the gravitational constant (6.67× 10?11 N m2/kg).
The sign of the RHS is negative because the radial distance r is measured outwards
and gravity acts inwards.
Example Given Newton’s law of gravitation m
dv
dt
= GMm
r2
, find v(r).
Deriving Galileo’s law from Newton’s law
We can use the above work to look at motion close to the surface of the Earth.