Stony Brook University Mathematics Department
1. The general second order homogeneous linear differential equation with
constant coefficients
looks like
Ay'' + By' + Cy = 0,
where y is an unknown function of the variable x, and A, B, and C are constants.
If A = 0
this becomes a first order linear equation , which we already know how to solve.
So we
will consider the case A ≠ 0. We can divide through by A and obtain the
equivalent
equation
y'' + by' + cy = 0
where b = B/A and c = C/A.
“Linear with constant coefficients” means that each term in the equation is a
constant
times y or a derivative of y. “Homogeneous” excludes equations like y'' +by' +cy
= f (x)
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 λ 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 λ is a solution of this equation ,
will automatically be a solution of our
original differential equation, and if λ 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
y'' − y = 0.
Plugging in give us the associated equation
which factors as
this equation has λ = 1 and λ = −1 as solutions. Both
and
are solutions
to the differential equation y'' − y = 0. (You should check this for yourself!)
Example 2. For the differential equation
y'' + y' − 2y = 0,
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
y'' + by' + cy = 0,
we need to find the roots of
, which we can do
using the quadratic formula
to get
If the discriminant b2 − 4c 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 y'' + by' + cy = 0, then so is the
linear combination
for any numbers p and q . This fact is easy
to check (just plug
into the equation and regroup terms; note that the coefficients b and c do not
need to
be constant for this to work. This means that for the differential equation in
Example 1
(y'' − y = 0), any function of the form
where p and q are any constants
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
y'' + y' − 2y = 0
is
where p and q are constants.
4. If the discriminant b2 − 4c 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 b2 − 4c < 0, then since we can write any positive
number
as a square k 2, we let k2 = −(b2 − 4c). Then ik will be a square root of b2 −
4c, since
(ik)2 = i2k2 = (−1)k2 = −k2 = b2 − 4c. The solutions of the associated algebraic
equation
are then
Example 3. If we start with the differential equation y'' + y = 0 (so b = 0 and
c = 1) the
discriminant is b2 − 4c = −4, so 2i is a square root of the discriminant and the
solutions
of the associated algebraic equation are
= i and
= −i.
Example 4. If the differential equation is y'' + 2y' + 2y = 0 (so b = 2 and c =
2 and
b2 − 4c = 4 − 8 = −4). In this case the solutions of the associated algebraic
equation are
λ= (−2 ± 2i)/2, i.e.
= −1 + i and
= −1 − i.
5. Going from the solutions of the associated algebraic equation to the
solutions of the
differential equation involves interpreting
as a function of x when λ is a
complex
number. Suppose λ has real part a and imaginary part ib, so that λ = a + ib with a
and b
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 y'' + y = 0 we found = i and
= −i, 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 p and q holds even if p and
q
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 y = p cos x+q sin x is a
solution for any
p and q. This is the general solution. (It is also correct to call
the general
solution; which one you use depends on the context.)
Example 4. y'' +2y' +2y = 0. We found
= −1+i and
= −1−i. 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: b2 − 4c = 0. Then the
“characteristic
equation” 
has one root. In this case both
and
are
solutions of the
differential equation.
Example 5. Consider the equation y'' + 4y' + 4y = 0. Here b = c = 4. The
discriminant
is b2 − 4c = 42 − 4 × 4 = 0. The only root is λ = −2. 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 p and the q 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 x.
Example 5. Suppose that for the differential equation of Example 2, y'' + y' −
2y = 0, we
want a solution with y(0) = 1 and y'(0) = −1. The general solution is
,
since the two roots of the characteristic equation are 1 and −2. The method is
to write
down what the initial conditions mean in terms of the general solution, and then
to solve
for p and q. In this case we have
This leads to the set of linear equations p+q = 1, p−2q = −1 with solution q =
2/3, p =
1/3. You should check that the solution
satisfies the initial conditions.
Example 6. For the differential equation of Example 4, y'' + 2y' + 2y = 0, we
found
the general solution 
. To find a solution satisfying
the initial
conditions y(0) = −2 and y'(0) = 1 we proceed as in the last example:
So p = −2 and q = −1. Again check that the solution
satisfies the initial conditions.
Problems cribbed from Salas-Hille- Etgen, page 1133
In exercises 1-10, find the general solution. Give the real form.
In exercises 11-16, solve the given initial-value problem.
