# Notes on Second Order Linear Differential Equations

**1.** The general second order homogeneous linear differential
equation with constant coefficients looks like

where is an unknown function of the variable , and , , and are constants. If this becomes a first order linear equation, which we already know how to solve. So we will consider the case . We can divide through by and obtain the equivalent equation

where and .

``Linear with constant coefficients'' means that each term in the equation is a constant times or a derivative of . ``Homogeneous'' excludes equations like
which can be solved, in certain important
cases, by an extension of the methods we will study here.

**2.** In order to solve this equation, we guess that there is a
solution of the form

where is an unknown constant. Why? Because it works!

We substitute
in our equation. This
gives

Since is never zero, we can divide through and get the equation

Whenever is a solution of this equation, will automatically be a solution of our original differential equation, and if is not a solution, then cannot solve the differential equation. So the substitution transforms the differential equation into an algebraic equation!

*Example 1.* Consider the differential equation

Plugging in give us the associated equation

which factors as

this equation has and as solutions. Both and are solutions to the differential equation . (You should check this for yourself!)

*Example 2.* For the differential equation

we look for the roots of the associated algebraic equation

Since this factors as , we get both and as solutions to the differential equation. Again, you should check that these are solutions.

**3.** For the general equation of the form

we need to find the roots of , which we can do using the quadratic formula to get

If the

*discriminant*is positive, then there are two solutions, one for the plus sign and one for the minus.

This is what we saw in the two examples above.

Now here is a useful fact about linear differential equations: if
and are solutions of the homogeneous differential equation
, then so is the linear combination
for any numbers and . This fact is easy to check (just plug
into the equation and regroup terms; note that the
coefficients and do not need to be constant for this to
work.

This means that for the differential equation in Example 1 (), any
function of the form

is a solution. Indeed, while we can't justify it here,

*all*solutions are of this form. Similarly, in Example 2, the general solution of

is

**4.** If the discriminant
is negative, then the equation
has no solutions, unless we enlarge the number field to include ,
i.e. unless we work with complex numbers. If , then since we can write
any positive number as a square , we let . Then
will be a square root of ,
since
.
The solutions of the associated algebraic equation are then

*Example 3.* If we start with the differential equation
(so and ) the discriminant is ,
so is a square root of the discriminant and the solutions of the
associated algebraic equation are and
.

*Example 4.* If the differential equation is
(so and and
). In this
case the solutions of the
associated algebraic equation are
, i.e.
and
.

**5.** Going from the solutions of the associated algebraic equation
to the solutions of the differential equation involves interpreting
as a function of when is a complex number.
Suppose has real part and imaginary part , so that
with and real numbers. Then

assuming for the moment that complex numbers can be exponentiated so as to satisfy the law of exponents. The factor does not cause a problem, but what is ? Everything will work out if we take

and we will see later that this formula is a necessary consequence of the elementary properties of the exponential, sine and cosine functions.

**6.** Let us try this formula with our examples.

*Example 3.* For we found and
, so the solutions are and .
The formula gives us
and
.

Our earlier observation that if and are solutions of the
linear differential equation, then so is the combination
for any numbers and holds even if and are complex constants.

Using this fact with the solutions from our example, we notice that
and
are both solutions. *When we are given a problem with real
coefficients it is customary, and always possible, to exhibit real
solutions.* Using the fact about linear combinations again, we can say
that
is a solution for any and . This
is the general solution. (It is also correct to call
the general solution; which one you use depends on the context.)

*Example 4.* . We found
and
. Using the formula we have

Exactly as before we can take and to get the real solutions and . (Check that these functions both satisfy the differential equation!) The general solution will be .

**7.** *Repeated roots*. Suppose the discriminant is zero:
. Then the ``characteristic equation''
has one root. In this case both **and**
are solutions of the differential equation.

*Example 5.* Consider the equation . Here
. The discriminant is
.
The only root is . Check that **both** and
are solutions. The general solution is then
.

**8.** *Initial Conditions*. For a first-order
differential equation the undetermined constant can be adjusted to
make the solution satisfy the initial condition ; in the
same way the and the in the general solution of a second
order differential equation can be adjusted to satisfy initial
conditions. Now there are two: we can specify both the value and
the first derivative of the solution for some ``initial'' value
of .

*Example 5.* Suppose that for the differential
equation of Example 2, , we want a solution with
and . The general solution is
, since the two roots of the characteristic
equation are 1 and . The method is to write down what the
initial conditions mean in terms of the general solution, and then
to solve for and . In this case we have

This leads to the set of linear equations with solution . You should check that the solution

satisfies the initial conditions.

*Example 6.* For the differential equation of
Example 4, , we found the general solution
. To find a solution
satisfying the initial conditions and
we proceed as in the last example:

So and . Again check that the solution

satisfies the initial conditions.

Scott Sutherland 2006-02-08