Second Order Differential Equations (Part 3. Complex Roots)
Posted by Suhaivi Hamdan on 7:57 PM with No comments
Complex Roots
In this section we will be looking at solutions to the
differential equation


in which roots of the characteristic equation,


are complex roots in the form .
Now, recall that we arrived at the characteristic equation
by assuming that all solutions to the differential equation will be of the form


Plugging our two roots into the general form of the solution
gives the following solutions to the differential equation.


Now, these two functions are “nice enough” (there’s those
words again… we’ll get around to defining them eventually) to form the general
solution. We do have a problem
however. Since we started with only real
numbers in our differential equation we would like our solution to only involve
real numbers. The two solutions above
are complex and so we would like to get our hands on a couple of solutions
(“nice enough” of course…) that are real.
To do this we’ll need Euler’s Formula.



A nice variant of Euler’s Formula that we’ll need is.


Now, split up our two solutions into exponentials that only
have real exponents and exponentials that only have imaginary exponents. Then use Euler’s formula, or its variant, to
rewrite the second exponential.


This doesn’t eliminate the complex nature of the solutions,
but it does put the two solutions into a form that we can eliminate the complex
parts.
Recall from the basics section that if two
solutions are “nice enough” then any solution can be written as a combination
of the two solutions. In other words,



will also be a solution.
Using this let’s notice that if we add the two solutions
together we will arrive at.


This is a real solution and just to eliminate the extraneous
2 let’s divide everything by a 2. This
gives the first real solution that we’re after.


Note that this is just equivalent to taking


Now, we can arrive at a second solution in a similar manner. This time let’s subtract the two original
solutions to arrive at.


On the surface this doesn’t appear to fix the problem as the
solution is still complex. However, upon
learning that the two constants, c_{1}
and c_{2} can be complex
numbers we can arrive at a real solution by dividing this by 2i.
This is equivalent to taking


Our second solution will then be


We now have two solutions (we’ll leave it to you to check
that they are in fact solutions) to the differential equation.


It also turns out that these two solutions are “nice enough”
to form a general solution.
So, if the roots of the characteristic equation happen to be
the general solution to the differential
equation is.


Let’s take a look at a couple of examples now.
Example 1 Solve
the following IVP.
Solution
The characteristic equation for this differential equation
is.
The roots of this equation are
Now, you’ll note that we didn’t differentiate this right
away as we did in the last section.
The reason for this is simple.
While the differentiation is not terribly difficult, it can get a
little messy. So, first looking at the
initial conditions we can see from the first one that if we just applied it
we would get the following.
In other words, the first term will drop out in order to
meet the first condition. This makes
the solution, along with its derivative
A much nicer derivative than if we’d done the original solution. Now, apply the second initial condition to
the derivative to get.
The actual solution is then.

Example 2 Solve
the following IVP.
Solution
The characteristic equation this time is.
The roots of this are
Notice that this time we will need the derivative from the
start as we won’t be having one of the terms drop out. Applying the initial conditions gives the
following system.
Solving this system gives

Example 3 Solve
the following IVP.
Solution
The characteristic equation this time is.
The roots of this are
Applying the initial conditions gives the following
system.
Do not forget to plug the t = Ï€ into the exponential! This is one of the more common mistakes
that students make on these problems.
Also, make sure that you evaluate the trig functions as much as
possible in these cases. It will only
make your life simpler. Solving this
system gives
The actual solution to the IVP is then.

Let’s do one final example before moving on to the next
topic.
Example 4 Solve
the following IVP.
Solution
The characteristic equation for this differential equation
and its roots are.
The general solution to this differential equation and its
derivative is.
Plugging in the initial conditions gives the following
system.
So, the constants drop right out with this system and the
actual solution is.

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