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 b^{2} − 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 b^{2} − 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 b^{2} − 4c < 0, then since we can write any positive
number

as a square k ^{2}, we let k^{2} = −(b^{2} − 4c). Then ik will be a square root of b^{2} −
4c, since

(ik)^{2} = i^{2}k^{2} = (−1)k^{2} = −k^{2} = b^{2} − 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 b^{2} − 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

b^{2} − 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: b^{2} − 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 b^{2} − 4c = 4^{2} − 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.