It’s time to start solving constant coefficient,
homogeneous, linear, second order differential equations. So, let’s recap how we do this from the last
section. We start with the differential
equation.
Write down the characteristic equation.
Solve the characteristic equation for the two roots, r_{1} and r_{2}. This gives
the two solutions
Now,
if the two roots are real and distinct (i.e. ) it will turn out that these two
solutions are “nice enough” to form the general solution
As with the last section, we’ll ask that you believe us when
we say that these are “nice enough”. You
will be able to prove this easily enough once we reach a later section.
With real, distinct roots there really isn’t a whole lot to
do other than work a couple of examples so let’s do that.
Example 1 Solve
the following IVP.
Solution
The characteristic equation is
Its roots are r_{1}
=  8 and r_{2} = 3 and so
the general solution and its derivative is.
Now, plug in the initial conditions to get the following
system of equations.
Solving this system gives and . The actual solution to the differential
equation is then

Example 2 Solve
the following IVP
Solution
The characteristic equation is
Its roots are r_{1}
=  5 and r_{2} = 2 and so
the general solution and its derivative is.
Now, plug in the initial conditions to get the following
system of equations.
Solving this system gives and . The actual solution to the differential
equation is then

Example 3 Solve
the following IVP.
Solution
The characteristic equation is
Its roots are r_{1}
= and r_{2}
= 2 and so the general solution and its derivative is.
Now, plug in the initial conditions to get the following
system of equations.
Solving this system gives c_{1} = 9 and c_{2}
= 3. The actual solution to the
differential equation is then.

Example 4 Solve
the following IVP
Solution
The characteristic equation is
The roots of this equation are r_{1} = 0 and r_{2}
= . Here is the general solution as well as its
derivative.
Up to this point all of the initial conditions have been
at and this one isn’t. Don’t get too locked into initial conditions
always being at and you just automatically use that instead
of the actual value for a given problem.
So, plugging in the initial conditions gives the following
system of equations to solve.
Solving this gives.
The solution to the differential equation is then.

In a differential equations class most instructors
(including me….) tend to use initial conditions at t = 0 because it makes the work a little easier for the students as
they are trying to learn the subject.
However, there is no reason to always expect that this will be the case,
so do not start to always expect initial conditions at t = 0!
Let’s do one final example to make another point that you
need to be made aware of.
Example 5 Find
the general solution to the following differential equation.
Solution
The characteristic equation is.
The roots of this equation are.
Now, do NOT get excited about these roots they are just
two real numbers.
Admittedly they are not as nice looking as we may be used
to, but they are just real numbers.
Therefore, the general solution is
If we had initial conditions we could proceed as we did in
the previous two examples although the work would be somewhat messy and so we
aren’t going to do that for this example.

The point of the last example is make sure that you don’t
get to used to “nice”, simple roots. In
practice roots of the characteristic equation will generally not be nice,
simple integers or fractions so don’t get too used to them!