Simple second order ordinary differential equations
Many differential equations we find in nature are second order equations. In this help
sheet we will demonstrate how the simplest form of the second order differential can
be solved, before outlining the standard procedure for solving such an equation.
It is easy to see why second order differential equations appear in engineering and
physics as the questions are often about how objects move when forces act. As force
directly relates to acceleration and acceleration is the second derivative of
displacement, a second order differential equation is needed to relate force and
position. It is surprising to find that these equations have properties which make them
useful in a far wider range of disciplines.
As integration is the opposite of differentiation, it would seem reasonable that we
would have to integrate twice to solve this sort of equation. Every time we integrate
we must add in a constant of integration. Therefore we expect there to be two
arbitrary constants in our general solution to the equation. Indeed, in many physical
systems we find that once we have described how the object moves under the action
of the forces we are still free to choose a starting position and velocity for the object.
However, as we shall see, we manage to form a solution without ever seeming to
perform an integration. Despite this, it is still necessary to maintain two arbitrary
constants in the general solution to allow us to solve for the initial or boundary
conditions given1.
Solving a second order differential equation
The simplest second order differential equation we come across is of the form:
d2y
dy
a 2  b  cy  0
dx
dx
where a, b and c are numbers. To solve this we start by assuming that the solution is
y  e mx . At this point this is just a guess but it will turn out to be a good one.
Assuming the solution is of this form, we can substitute it into the above equation and
find:
am 2e mx  bmemx  ce mx  0
where the m 2 and m have appeared because every time we take a derivative of
e mx we drop down a multiple of m . As every term is multiplied by e mx we can divide
everywhere by e mx , leaving us with:
1
The distinction between initial conditions and boundary conditions is one of semantics; initial
conditions are given when the problem is given in terms of time (t) and boundary conditions are given
when the problem is given in terms of space (x).
am 2  bm  c  0
which is known as the auxiliary equation. In a question we would be given the values
of a, b and c and we could attempt to solve this quadratic by factoring it. If we can’t
find a simple factorisation it is always possible to use the quadratic formula to solve
for values of m :
m1, 2
 b  b 2  4ac

2a
where m1 is given by calculating the formula with a plus and m2 is given by
calculating with a minus. Once we have solved this quadratic we will find ourselves
in one of three possible positions:
1) Distinct real roots
As long as b 2  4ac is a positive number the auxiliary equation will yield two real
numbers m1 and m2 . We can multiply both of these solutions by an arbitrary constant
and add them together, leading to a general solution:
y  Aem1x  Be m2 x .
If we have been supplied with boundary conditions we can solve for A and B to give
a particular solution.
2) Repeated real roots
If b 2  4ac is zero then m1 will be the same as m2 , i.e. m1  m2 , so we may as well
drop the subscript and call the single solution to the auxiliary equation simply m .
However, as this is a second order problem we will have two boundary conditions, so
we still require two constants in our solution. To make the two general solutions
different we multiply one by x and form a solution of the form:
y   A  Bx emx
3) Imaginary roots
If b 2  4ac is negative then the only way to solve the quadratic is with imaginary
numbers. The two roots will be complex conjugates of each other, that is, both
solutions have the same numbers in their real and imaginary parts but one will have a
positive imaginary part and the other a minus:
m1    i , m2    i .
Here  and  are just numbers once again; we have used Greek letters to avoid
confusion with the a and b used in the differential equation. This allows us to form
the solution as:
y  ex  A cosx   B sin x 
It is very important to notice that the real part  is used only in the exponential and
the imaginary part  is used only in the sin and cos parts. Also notice that we have
dropped the plus and minus.
Steps for solving
We summarize the steps to solving a simple second order equation below:
d2y
dy
a 2  b  cy  0
Starting from:
dx
dx
2
am  bm  c  0
Form the auxiliary equation:
 b  b 2  4ac
2a
Then form the solution according to the following possibilities:
Solve the auxiliary equation: m1, 2 
Solutions to the auxiliary equation
Two real distinct roots m1, m2
Real repeated roots m
Complex roots   i
Form of general solution
y  C1em1x  C2em2 x
y  C1  C2 x emx
y  ex C1 cosx   C2 sin x 
Finally, solve for constants if you have been given initial conditions.
Worked examples:
Example 1
Find the general solution to the differential equation:
d2y
dy
 2  3y  0
2
dx
dx
and also the particular solution given initial conditions y0  0 and y' 0  1 .
Solution
First we form the auxiliary equation:
m 2  2m  3  0.
In this instance it is simple to see that this can be factorised with ease giving:
m  3m  1  0,
so we can read off the solutions m1  3 and m2  1 . As we have two distinct real
roots we can say that the general solution is simply:
yx   Ae3 x  Be  x .
As we have been given initial conditions we can also find the particular values of A
and B. The first initial condition tells us that y  0 when x  0 ; substituting these
values into our general solution we get:
0  Ae30  Be 0  A  B  0
(as e 0  1 )
The second initial condition tell us that the differential of y is 1 when x is 0. The
differential of the general solution is:
y' x   3 Ae3 x  Be  x ,
substituting in the initial conditions we get:
1  3A B.
We can now solve these two equations simultaneously (see the simultaneous equation
help sheet and videos) to find:
A
1
1
,B  
4
4
Thus the particular solution is:
yx  
1 3x 1 x
e  e
4
4
Example 2
Find the general solution to the equation:
4 y ' '4 y ' y  0
and then find the particular solution that solves the boundary conditions y0  0,
y2  e .
Solution
First notice that the question has been given to us using dash notation; this question is
the same as:
d2y
dy
4
4  y  0
dx
dx
Therefore we have the auxiliary equation:
4m 2  4m  1  0
which isn’t obviously factorisable so we can use the quadratic formula:
m1, 2 
4
 42  4  4 1
24
we can see that  4  4  4  1  0 , therefore we only get one value for m which is
m  1 2 . As we have a repeated root we can form the general solution:
2
x
2
yx    A  Bx e .
Using the first boundary condition we find that:
0
2
y0   A  B  0e  A  0
i.e. we have found that A  0 . From the second boundary condition we find:
2
y 2  2 Be 2  e
Dividing both sides by e we find:
2 B  1  B
1
2
Leaving us with the particular solution:
x
y
1 2
xe
2
Example 3
Find the general solution to the equation:
2 y ' '6 y '5 y  0
and then find the particular solution for the initial conditions y0  1 and
1
y ' 0    .
2
Solution 3
We find the auxiliary equation to be:
2m 2  6m  5  0
and once again a simple factorisation does not seem possible, so we use the quadratic
formula:
 6  62  4  2  5
.
22
The first thing to notice is that 6 2  4  2  5  4 , so if we type this into a calculator we
will get an error. This means that the solutions are complex numbers:
m1, 2 
3 1
m1, 2    i .
2 2
This time we have complex roots with real part   
3
and the imaginary part
2
1
. Notice that we ignore the  sign when reading off the second part. Looking
2
back at the table we can form the general solution:

yx   e
3
 x
2

 x
 x 
 A cos   B sin   .
2
 2 

To use the initial conditions to solve for the particular solution we will also need the
differential of the general solution:
3  x
 x
 x   x A  x  B
 x 
y '  x    e 2  A cos   B sin     e 2   sin    cos  
2
2
 2 
 2 

 2 2 2
3
3
Note that we had to use the product rule. From the first initial condition, y0  1 , we
find:
y0  e0  A cos0  B sin 0  A  1
1
i.e. A  1 . From the second initial condition, y ' 0    , we find:
2
3
B
3 B
1
 1

y ' 0    e 0 1  cos0  B sin 0   e 0   sin 0   cos0       .
2
2
2 2
2
 2

We can now solve this giving B  2 , therefore the particular solution is:
yx   e
3
 x
2
  x
 x 
 cos   2 sin    .
 2 
 2
Example 4
Find the general solution to the equation:
d2y
dy
3  0
2
dx
dx
Solution 4
This equation looks a little different as it is missing a y term, but we can solve it in
the same way. First we form the auxiliary equation:
m2  3m  0,
which factorises to:
mm  3  0
thus the solutions are m1  0 and m2  3 . It is common for people to ignore or forget
to include the m1  0 in the general solution. The correct general solution is:
yx   Ae0x  Be 3 x
or simply:
yx   A  Be 3x .
Example 5
Find the general solution to the equation:
d 2 y dy
  2y  0
dx 2 dx
with the boundary conditions y0  1 and y  0 .
Solution 5
Solving the differential equation should be fairly easy in this case. The auxiliary
equation is:
m2  m  2  0
which is factorised into:
m 1m  2  0
and therefore the general solution is:
yx   Ae x  Be 2 x .
Now we have to think about the boundary conditions we have been given. The second
condition imposes that as x gets very large y must approach zero. Looking at the
two terms in the general solution it is clear that the first one, Ae x , simply gets larger
and larger as x increases so we cannot include this in the particular solution.
Therefore we are forced to take A  0 . The second term, Be 2 x , does indeed get
smaller as x increases, therefore we can use the first boundary term to find B :
y0  Be 0  B  1,
Giving the particular solution:
yx   e 2 x .