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.
will consider the case A ≠ 0. We can divide through by A and obtain the
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
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
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
to the differential equation y\'\' − y = 0. (You should check this for yourself!)
Now here is a useful fact about linear differential equations: if
of the homogeneous differential equation y\'\' + by\' + cy = 0, then so is the
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
be constant for this to work. This means that for the differential equation in
(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
where p and q are constants.
4. If the discriminant b2 − 4 c is negative , then the equation
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
as a square k2, we let k2 = −(b2 − 4c). Then ik will be a square root of b2 −
(ik)2 = i2k2 = (−1)k2 = −k2 = b2 − 4c. The solutions of the associated algebraic
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
of the associated algebraic equation are
= i and
Example 4. If the differential equation is y\'\' + 2y\' + 2y = 0 (so b = 2 and c =
b2 − 4c = 4 − 8 = −4). In this case the solutions of the associated algebraic
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
number. Suppose λ has real part a and imaginary part ib, so that λ = a + ib with a
real numbers. Then
assuming for the moment that complex numbers can be exponentiated so as to
law of exponents. The factor does not cause a problem, but what is
will work out if we take
Example 3. For y\'\' + y = 0 we found = i and
= −i, so the solutions are
and . The formula gives us
Our earlier observation that if
are solutions of the linear
then so is the combination
for any numbers p and q holds even if p and
are complex constants.
Using this fact with the solutions from our example, we
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
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
Exactly as before we can take and
to get the real solutions
and . (Check that these functions both satisfy the differential
general solution will be .
7. Repeated roots. Suppose the discriminant is zero: b2 − 4c = 0. Then the
equation” has one root. In this case both
solutions of the
Example 5. Consider the equation y\'\' + 4y\' + 4y = 0. Here b = c = 4. The
is b2 − 4c = 42 − 4 × 4 = 0. The only root is λ = −2. Check that
solutions. The general solution is then .
8. Initial Conditions. For a first-order differential equation the undetermined
can be adjusted to make the solution satisfy the initial condition
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
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
down what the initial conditions mean in terms of the general solution, and then
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
the general solution . To find a solution satisfying
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.
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