  # Notes on Second Order Linear Differential Equations

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. Prev Next

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