Dr. Hatim Dirar
Department of Physics, College of Science
Imam Mohamad Ibn Saud Islamic University
Scopes
 Differential equations lead to:
 Definition
 Interpretation
 Forecasting
of any physical system
of any dimension
Scopes in Physics
 Mechanics
 Free Fall

let us considering free fall system defined by the following
equation of motion
ÿ=-g


The above equation is second order DE
Classification of Differential Equations
 DE is defined as an equation contains derivative of an unknown




function, which express the relationship we seek
the equation includes total derivative like
is called
ordinary differential equation (ODE)
If the function is of multivariables , the differential equation is
known as partial differential equation (PDE)
Order of DE indicates the order of the highest derivative appear
in the equation
Degree of the DE equations indicates the power of the derivative
of highest order after the equation has been rationalized
e.g.
is of second order and first degree
Classification of DE
 Example 2
The above equation has said to be of a first order and second
degree since it will contains
after it is rationalized
 DE has said to be linear if each term of it is such that the
dependant variable or its derivative occur only once and only to
the first power
e.g.
whereas, the following equation is nonlinear
Classification of DE
 If the right hand side of the DE is equal to zero it has
said to be homogenous, otherwise it is inhomogeneous
 If you know two solutions of any homogenous
equation, other solution can be constructed as the
linear combination of these solutions
 Trivial change of unfamiliar DE might convert it to a
recognizable DE
 Many DEs are too difficult to be solved and few can be
solved in closed form
Solution of DE
Solution of first-order DE
 DE of general form
or
is clearly first-order DE
Such equation can be solved by separation of variable
method
If
are reducible to
then we have
Solution of DE
 The above equation can be solved easily by integrating
directly
Example:
Solve the following DE
Solution:
Let us rearrange the equation or to separate the variables
hence it will be
Now let us integrate as;
Solution of DE
the solution will be
Where, C is constant
 Some times, the variable may not be separable. In such
case we need to change these variable so as to be
separable, the following is the general form;
Where f is an arbitrary function and a and b are
constants. If we let
Solution of DE
Then,
then the DE will be
From which we obtained,
such DE became separable
Solution of DE
Example:
Solve the following DE
Solution:
Let,
then
and the DE will be
Solution of DE
or,
hence the equation is separable can easy be solved by
integrating the variables
Solution of DE
 Homogenous DE, which has the general form may also
be reduced to an equation with separable variables
Example:
Solve the equation
Solution:
The right hand side can be written as
Solution of DE
hence we get function of single variable, let
and our equation become
from which we have
Integration gives
Solution of DE
 Ex:
Solve the following DE:
12-
3-
Exact equation
 Example:
Solve the equation
Solution:
The above equation has said to be homogenous if we will
be able to change the variable, such that to eliminate -5 and
-1 from numerator and denominator, respectively. Let,
Exact equation
 The above equation has said to be homogenous if we
will be able to change the variable, such that to
eliminate -5 and -1 from numerator and denominator,
respectively.
 Where, 𝛂 and 𝛃 are constants, chosen in such away our
equation will be homogenous. Substitute for both x
and y we get.
 Now let,
then.
 Hence our equation will be,
 herefore, the above equation becomes separable and hence
solvable.
 Example 2.5
Fall of a skydiver.
Solution: Assuming the parachute opens at the beginning
of the fall, there are
two forces acting on the parachute: the downward force of
gravity mg, and the
upward force of air resistance. If we choose a coordinate
system that has at the earth's surface and increases upward,
then the equation of motion of the falling diver, according
to Newton's second law, is
where m is the mass, g the gravitational acceleration, and k
a positive constant. In general the air resistance is very
complicated, but the power-law approximation is useful in
many instances in which the velocity does not vary
appreciably.
Experiments show that for subsonic velocity up to 300 m/s,
the air resistance is approximately proportional to .
The equation is separable
To make the equation more simple let
Assume that,
Therefore,
Let us apply the above reduction we get,
Upon integration
Where C is the constant of integration
let,
Fro1m the above solution it is obviously seen that, as; ,
This implies that if the fall is from sufficient high, the fallen
particle will fall with constant velocity known as terminal
velocity. To determine C we need to know the value of k,
which is being 30Kg/m
Exact DE
The following equation has said to integrable because it is
homogenous
The solution of the above equation is of the form
The above equation has said to be exact if
To check this let us let us consider the above equation,
By performing differential we get,
which the general properties of partial derivative of any well
behave f unction
If our exact DE is of the form,
Then it is easy to identify that,
Then it follows from the above equation that,
Hence, it is the same relation we had obtained previously
Examples:
Show that the equation,
general solution
is exact and find its
Solution:
Let us rearrange the above equation so as to look like the
following standard form
Since,
Therefore, the given equation is exact and its general
solution will be,
This implies that,
Where,
are the integration constants,
comparing the above solution it is clearly seen that,
Thus the general solution is,
It is very interesting to consider a DE of this type
The left hand side is an exact differential equation
and
on the right hand side is the function of x only.
Then the general solution can be written as,
Alternatively we can write the above solution as;
Since the LHS of the above equation is exact, then
Let us test the exactness of the above equation;
Hence, the above equation satisfies the requirement for
being exact; we can write the solution as:
Where,
Integrating Factor (IF)
Inexact DE can be converted to exact by multiplying it by
an IF. Different DE have different Ifs, which are rather
difficult to find.
Let us consider the following DE,
Let,
be the IF of the above DE. Now multiply the DE
by IF, we get
Now the RHS of the above equation has said to be
integrable; the condition is that the LHS is exactHence the
equation becomes,
Which yield,
Or,
Hence it is possible to write the general solution as;
Where,
Therefore, the solution will be;
Example:
Show that the given equation
is not exact; then
find a suitable IF that could makes the DE exact and look
for its general solution
Solution:
First of all let us write the above equation in the standard
form .i.e.
It is obviously seen that,
This implies that, the given equation is not exact
Let us divide the equation by x, we get;
From the above equation we found that,
Hence, the required IF is;
Now apply this IF to the given equation we get,
Integrating the above equation we get,
and
Application in Physics:
For the given LCR circuit, shown below, find the current
in the circuit as the function of time t
Solution:
First of all we need to setup the differential equation for the
current flowing in the circuit. R and L represents the constant
resistance and inductance, respectively.
and
where,
is the voltage dropped across R
is the voltage dropped across L
Let us apply Kirchhoff’s second law
here t represents the independent variable and stand for
the dependent. The general solution will be,
E and k are constants, and the evaluation of the above
integral gives,
Regardless the value of k, we see that,
At
, the solution will be,