First year: Introduction to differential equations
We will only talk about first and second order homogeneous and nonhomogeneous differential equations with constant coefficients.
(I)
First order linear differential equation
A first order linear differential equation with constant coefficients is a
differential equation of the form
ay 'by  0,
where a and b are constants. Solving such an equation means finding all the
functions y that satisfy the differential equation.
For example, let’s solve  y ' y  0 , where y is a function of x :
The differential equation can be rewritten as follows
y'
 1.
y
Integrating on both sides with respect to x , we get:
y'
 y dx   dx
 ln | y |  x  c  | y | e xc

y  Ae x ,
with A  e c .
A can be chosen to be any constant number, so that there is an infinity of
solutions.
Now, if you are given some additional information on the function you are
looking for, you can calculate the constant A .
For example, say you have to solve the initial value problem
 y ' y  0,
y (0)  3.
You know this differential equation has a solution of the form y  Ae x . In order
for the solution to satisfy y (0)  3 , we must have
3  Ae0

A3 .
So the unique solution to the initial value problem
 y ' y  0,
y (0)  3
is y  3e x .
You can also be asked to solve a non-homogeneous differential equation,
i.e. a differential equation of the form
ay 'by  f ( x),
where f (x) is a given function.
In this case, you first solve the homogeneous equation ay 'by  0 and then
find one particular solution of the non-homogeneous equation ay 'by  f ( x).
Add the general solution of the homogeneous equation to the particular
solution of the non-homogeneous equation to get all the solutions of the nonhomogeneous equation. It might sound complicated, but it’s not that bad,
actually:
Say you have to solve
 y' y  e x .
You know that the general solution to the homogeneous equation  y ' y  0 is
y  Ae x . We just need to find one particular solution of the non-homogeneous
equation  y' y  e 2 x .
To find this particular solution, you have to make an educated guess. Here
y  e 2 x works since
 
 y' y   e 2 x 'e 2 x  2e 2 x  e 2 x  e 2 x .
The general solution to the non-homogeneous problem is therefore
y  Ae x  e 2 x .
Exercises:
(1) Solve the following differential equations:
(i)
(ii)
3 y ' y  0.
2 y '3 y  0.
(2) Solve the following initial value problems:
(i)
(ii)
y ' y  0, y (0)  2.
2 y '3 y  0, y (0)  0.
(3) Solve the following non-homogeneous equations:
(i)
(ii)
(II)
y ' y  e 2 x .
y ' y  2 cos( x).
Second order linear differential equations
A second order linear differential equation with constant coefficients is a
differential equation of the form
ay ' 'by ' cy  0 ,
where a, b and c are constants.
The characteristic equation associated with the differential equation above
is
a2  b  c  0 .
This quadratic has two (real or imaginary) roots:  and  .
If    , then the general solution to the second order differential equation
above is of the form
y  Aex  Be x .
If    , then the general solution is of the form
y  Aex  Bxex .
Example:
Solve the following initial value problem:
y ' ' y  0,
y (0)  0, y ' (0)  1.
The characteristic equation is
2  1  0 ,
so that the characteristic equation has two roots:   i and     i.
The general solution to the second order linear differential equation is
therefore
y  Aeix  Be ix .
In order to solve the initial value problem, we need to find the constants A
and B :

 y (0)  0


 y ' (0)  1


 A  B  0


iA  iB  1


 A  B


 iB  iB  1

i

A  2
.

i
B
2

The solution to the initial value problem is therefore
y


i
i
 e ix  e ix   cos( x)  i sin( x)  cos( x)  i sin( x)   sin( x).
2
2
Exercises:
(1) Solve the following IVPs (Initial Value Problems):
(i)
(ii)
y ' '4 y  0, y (0)  1, y ' (0)  0.
y ' '2 y ' y  0, y (0)  y ' (0)  1.
(2) Solve the following non-homogeneous equations:
(i)
(ii)
y ' ' y '6 y  4.
y ' ' y ' y  1.
Remark: you solve a second order non-homogeneous differential equation the
way you’d solve a first order non-homogeneous equation: first you find all the
solutions of the homogeneous problem (i.e. RHS=0), then you calculate one
particular solution of the non-homogeneous problem and, finally, you add the
general solution and the particular equation.