Second Order Differential Equations
As discussed in the introduction to differential equations, a vibrational system can be represented with a second-order differential equation of the form where is the mass of the body, is the damper constant, is the spring constant, and is some external force. In general, a second-order differential equation looks like
In this course, we're going to stick to simpler cases, where the following two assumptions hold:
- The equation is homogeneous: In the case of the vibration application, this means there's no external force on the object; it is simply deflected and left to vibrate.
- The coefficients are constants: and are constant.
General Solution
Theorem
For a differential equation of the form solutions are given by where and are two linearly independent solutions of the differential equation. When we solve these equations, we'll get two solutions, and then write the full solution this way. To get a particular solution, we'll need two given conditions, so that we can find and
Definition: Linear Independence
Two functions are linearly independent if neither is a constant multiple of the other.
For example, and are not linearly independent, but and are linearly independent. This will be significant in one case below.
Finding the Solution
Start with the differential equation: Follow me for a moment on a seeming leap: we are looking for this unknown function and all that we know is that if we add it, its derivative, and its second derivative together, they all cancel each other out; they must therefore look like the original function. The only function we know of whose derivatives look like it is the exponential function. So let's assume that the solution looks like Following that logic,
so when we substitute these into the differential equation, we get the following:
Since is never equal to it follows that
If we solve this equation, called the characteristic equation, to find we'll have the solutions.
Solution Possibilities
There are three possibilities for the forms of the solutions, based on the values of which are in turn based on the form of the characteristic equation. If we use the quadratic formula to find we find that
The three possibilities depend on the value of the discriminant of the quadratic equation.
Case 1: Two Distinct Real Roots
If the discriminant is positive, we'll get two real answers for that are different. In that case, the two solutions look like
which makes the full solution
Case 2: One Real Root
If the discriminant is zero, everything after the plus/minus is zero, so there will only be one real answer for This is the case where it is important to account for linear independence. Here, the two solutions look like
which makes the full solution
Case 3: A Conjugate Pair of Complex Roots
If the discriminant is negative, we'll get a complex conjugate pair of solutions, of the form In that case, the two solutions look like
which makes the full solution
Solve the following differential equation.
Solution
The characteristic equation is
Notice that we simply replace with with and with to get the characteristic equation. We can solve for using the quadratic formula, but this one also can be factored nicely:
which means that This is the second case; there is only one value for Therefore, the solution is
Solve the following differential equation.
Solution
The characteristic equation is
Again, we can solve for by factoring:
which means that This is the first case; there are two distinct real values for Therefore, the general solution is
To find the particular solution, we need to use the given initial conditions:
When you solve this system of two equations, you'll find that and so the particular solution is
Solve the following differential equation.
Solve the following differential equation.
Solution
The characteristic equation is
This one doesn't factor nicely, so we have to find using the quadratic formula:
Matching this to the form we find that and Therefore, the solution (according to Case 3) looks like