**Notes on Second Order Differential Equations**

Anthony Phillips -- Fall 1998

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

where

with

**2.** The method of solution that we will work on involves
looking for an exponential-type solution

Why? Because it works! We substitute in our equation. This gives

Since we can divide through and get

If we solve this quadratic equation for , then will automatically be a solution of our differential equation. So the substitution transforms the differential equation into an algebraic equation! This equation is called the

**3.** The solution of
is
(``quadratic formula'')

In this case

If the

*Example 1.* If the differential equation is *y*''-*y*=0 (so *c*_{1}=0 and *c*_{0}=-1) the roots of the characteristic
equation
are
.
Both *y*=*e*^{x} and *y*=*e*^{-x} are solutions. (Check this!)

*Example 2.* If the differential equation is
*y*''+*y*'-2*y*=0 the roots of the characteristic
equation
are
,
so
and
.
Both *y*=*e*^{x} and *y*=*e*^{-2x} are solutions. (Check this!)

**4.** If the discriminant
*c*_{1}^{2}-4*c*_{0} 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
*c*_{1}^{2}-4*c*_{0}=-*k*^{2} (we can write
any positive number as *k*^{2}!) then *ik* is a square root of
*c*_{1}^{2}-4*c*_{0}since
(*ik*)^{2} = *i*^{2}*k*^{2} = (-1)*k*^{2} = -*k*^{2}. The solutions of the associated
algebraic equation are then

*Example 3.* If we start with the differential equation
*y*''+*y*=0 (so *c*_{1}=0 and *c*_{0}=1) the discriminant is
*c*_{1}^{2}-4*c*_{0}=-4,
so 2*i* is a square root of the discriminant and the solutions of the
characteristic equation are
and
.

*Example 4.* If the differential equation is
*y*''+2*y*'+2*y*=0 (so *c*_{1}=2 and *c*_{0}=2 and
*c*_{1}^{2}-4*c*_{0}=4-8 =-4). In this
case the solutions of the
characteristic equation are
,
i.e.
and
.

**5.** Going from the solutions of the characteristic equation
to the solutions of the differential equation involves interpreting
as a function of *x* when
is a complex number.
Suppose
has real part *a* and imaginary part *ib*, so that
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

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
and
,
so the solutions are
*y*_{1}=*e*^{ix} and
*y*_{2}=*e*^{-ix}.
The formula gives us
and
.

Now here is a useful fact about linear differential equations: if *y*_{1}and *y*_{2} are solutions of the homogeneous differential equation
*y*''+*c*_{1}*y*'+*c*_{2}*y*=0, then so is the linear combination
*p y*_{1} + *q y*_{2} for any numbers *p* and *q*. This fact is easy to check (just plug
*p y*_{1} + *q y*_{2} into the equation and regroup terms; note that the
coefficients *a*_{1} and *a*_{0} do not need to be constant for this to
work, that *p* and *q* can be complex as well as real constants, and
that this works for equations af any order).

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 *p* and *q*. This
is the general solution. (It is also correct to call
*y*=*p e*^{ix} +
*q e*^{-ix} the general solution; which one you use depends on the context.)

*Example 4.*
*y*''+2*y*'+2*y*=0. 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:
*c*_{1}^{2}-4*c*_{0} =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*''+4*y*'+4*y*=0. Here
*c*_{1} = *c*_{0} = 4. The discriminant is
.
The only root is
.
Check that **both** *e*^{-2x} and
*x e*^{-2x} are solutions. The general solution is then
*y* = *p e*^{-2x} + *q x e*^{-2x}.

**8.** *Initial Conditions*. For a first-order
differential equation the undetermined constant can be adjusted to
make the solution satisfy the initial condition *y*(0)=*y*_{0}; 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*'-2*y*=0, we want a solution with
*y*(0)=1 and *y*'(0)=-1. The general solution is
*y*=*p e*^{x} + *q e*^{-2x}, 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

*Example 6.* For the differential equation of
Example 4,
*y*''+2*y*'+2*y*=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

tony@math.sunysb.edu