I apologize for the outage on the site yesterday and today. Lamar University is in Beaumont Texas and Hurricane Laura came through here and caused a brief power outage at Lamar. Things should be up and running at this point and (hopefully) will stay that way, at least until the next hurricane comes through here which seems to happen about once every 10-15 years. Note that I wouldn't be too suprised if there are brief outages over the next couple of days as they work to get everything back up and running properly. I apologize for the inconvienence.

Paul

August 27, 2020

*i.e.*you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

### Section 7-2 : Homogeneous Differential Equations

As with 2^{nd} order differential equations we can’t solve a nonhomogeneous differential equation unless we can first solve the homogeneous differential equation. We’ll also need to restrict ourselves down to constant coefficient differential equations as solving non-constant coefficient differential equations is quite difficult and so we won’t be discussing them here. Likewise, we’ll only be looking at linear differential equations.

So, let’s start off with the following differential equation,

\[{a_n}{y^{\left( n \right)}} + {a_{n - 1}}{y^{\left( {n - 1} \right)}} + \cdots + {a_1}y' + {a_0}y = 0\]Now, assume that solutions to this differential equation will be in the form \(y\left( t \right) = {{\bf{e}}^{r\,t}}\) and plug this into the differential equation and with a little simplification we get,

\[{{\bf{e}}^{r\,t}}\left( {{a_n}{r^n} + {a_{n - 1}}{r^{n - 1}} + \cdots + {a_1}r + {a_0}} \right) = 0\]and so in order for this to be zero we’ll need to require that

\[{a_n}{r^n} + {a_{n - 1}}{r^{n - 1}} + \cdots + {a_1}r + {a_0} = 0\]This is called the **characteristic polynomial/equation** and its roots/solutions will give us the solutions to the differential equation. We know that, including repeated roots, an \(n\)^{th} degree polynomial (which we have here) will have \(n\) roots. So, we need to go through all the possibilities that we’ve got for roots here.

This is where we start to see differences in how we deal with \(n\)^{th} order differential equations versus 2^{nd} order differential equations. There are still the three main cases: real distinct roots, repeated roots and complex roots (although these can now also be repeated as we’ll see). In 2^{nd} order differential equations each differential equation could only involve one of these cases. Now, however, that will not necessarily be the case. We could very easily have differential equations that contain each of these cases.

For instance, suppose that we have an 9^{th} order differential equation. The complete list of roots could have 3 roots which only occur once in the list (*i.e.* real distinct roots), a root with multiplicity 4 (*i.e.* occurs 4 times in the list) and a set of complex conjugate roots (recall that because the coefficients are all real complex roots will always occur in conjugate pairs).

So, for each \(n\)^{th} order differential equation we’ll need to form a set of \(n\) linearly independent functions (*i.e.* a fundamental set of solutions) in order to get a general solution. In the work that follows we’ll discuss the solutions that we get from each case but we will leave it to you to verify that when we put everything together to form a general solution that we do indeed get a fundamental set of solutions. Recall that in order to this we need to verify that the Wronskian is not zero.

So, let’s get started with the work here. Let’s start off by assuming that in the list of roots of the characteristic equation we have \({r_{\,1}},{r_{\,2}}, \ldots ,{r_{\,k}}\) and they only occur once in the list. The solution from each of these will then be,

\[{{\bf{e}}^{{r_{\,1}}\,t}},\hspace{0.25in}{{\bf{e}}^{{r_{\,2}}\,t}},\hspace{0.25in} \cdots ,\hspace{0.25in}{{\bf{e}}^{{r_{\,k}}\,t}}\]There’s nothing really new here for real distinct roots.

Now let’s take a look at repeated roots. The result here is a natural extension of the work we saw in the 2^{nd} order case. Let’s suppose that \(r\) is a root of multiplicity \(k\) (*i.e.* \(r\) occurs \(k\) times in the list of roots). We will then get the following \(k\) solutions to the differential equation,

So, for repeated roots we just add in a \(t\) for each of the solutions past the first one until we have a total of \(k\) solutions. Again, we will leave it to you to compute the Wronskian to verify that these are in fact a set of linearly independent solutions.

Finally, we need to deal with complex roots. The biggest issue here is that we can now have repeated complex roots for 4^{th} order or higher differential equations. We’ll start off by assuming that \(r = \lambda \pm \mu \,i\) occurs only once in the list of roots. In this case we’ll get the standard two solutions,

Now let’s suppose that \(r = \lambda \pm \mu \,i\) has a multiplicity of \(k\) (*i.e.* they occur \(k\) times in the list of roots). In this case we can use the work from the repeated roots above to get the following set of 2\(k\) complex-valued solutions,

The problem here of course is that we really want real-valued solutions. So, recall that in the case where they occurred once all we had to do was use Euler’s formula on the first one and then take the real and imaginary part to get two real valued solutions. We’ll do the same thing here and use Euler’s formula on the first set of complex-valued solutions above, split each one into its real and imaginary parts to arrive at the following set of 2\(k\) real-valued solutions.

\[\begin{align*}&{{\bf{e}}^{\lambda t}}\cos \left( {\mu \,t} \right),\hspace{0.25in}{{\bf{e}}^{\lambda t}}\sin \left( {\mu \,t} \right),\hspace{0.25in}t{{\bf{e}}^{\lambda t}}\cos \left( {\mu \,t} \right),\hspace{0.25in}t{{\bf{e}}^{\lambda t}}\sin \left( {\mu \,t} \right),\hspace{0.25in} \cdots ,\\ & \hspace{3.25in}{t^{k - 1}}{{\bf{e}}^{\lambda t}}\cos \left( {\mu \,t} \right),\,\,\,\,\,\,\,{t^{k - 1}}{{\bf{e}}^{\lambda t}}\sin \left( {\mu \,t} \right)\end{align*}\]Once again, we’ll leave it to you to verify that these do in fact form a fundamental set of solutions.

Before we work a couple of quick examples here we should point out that the characteristic polynomial is now going to be at least a 3^{rd} degree polynomial and finding the roots of these by hand is often a very difficult and time consuming process and in many cases if the roots are not rational (*i.e.* in the form \(\frac{p}{q}\)) it can be almost impossible to find them all by hand. To see a process for determining all the rational roots of a polynomial check out the Finding Zeroes of Polynomials page in the Algebra notes. In practice however, we usually use some form of computation aid such as Maple or Mathematica to find all the roots.

So, let’s work a couple of example here to illustrate at least some of the ideas discussed here.

The characteristic equation is,

\[{r^3} - 5{r^2} - 22r + 56 = \left( {r + 4} \right)\left( {r - 2} \right)\left( {r - 7} \right) = 0\hspace{0.25in} \Rightarrow \hspace{0.25in}{r_{\,1}} = - 4,\,\,\,{r_{\,2}} = 2,\,\,\,{r_{\,3}} = 7\]So, we have three real distinct roots here and so the general solution is,

\[y\left( t \right) = {c_1}{{\bf{e}}^{ - 4t}} + {c_2}{{\bf{e}}^{2t}} + {c_3}{{\bf{e}}^{7t}}\]Differentiating a couple of times and applying the initial conditions gives the following system of equations that we’ll need to solve in order to find the coefficients.

\[\begin{array}{*{20}{l}}{\,\,\,\,\,1 = y\left( 0 \right) = {c_1} + {c_2} + {c_3}}\\{ - 2 = y'\left( 0 \right) = - 4{c_1} + 2{c_2} + 7{c_3}}\\{ - 4 = y''\left( 0 \right) = 16{c_1} + 4{c_2} + 49{c_3}}\end{array}\hspace{0.25in} \Rightarrow \hspace{0.25in}\begin{array}{*{20}{l}}{{c_1} = \frac{{14}}{{33}}}\\{{c_2} = \frac{{13}}{{15}}}\\{{c_3} = - \frac{{16}}{{55}}}\end{array}\]The actual solution is then,

\[y\left( t \right) = \frac{{14}}{{33}}{{\bf{e}}^{ - 4t}} + \frac{{13}}{{15}}{{\bf{e}}^{2t}} - \frac{{16}}{{55}}{{\bf{e}}^{7t}}\]So, outside of needing to solve a cubic polynomial (which we left the details to you to verify) and needing to solve a system of 3 equations to find the coefficients (which we also left to you to fill in the details) the work here is pretty much identical to the work we did in solving a 2^{nd} order IVP.

Because the initial condition work is identical to work that we should be very familiar with to this point with the exception that it involved solving larger systems we’re going to not bother with solving IVP’s for the rest of the examples. The main point of this section is the new ideas involved in finding the general solution to the differential equation anyway and so we’ll concentrate on that for the remaining examples.

Also note that we’ll not be showing very much work in solving the characteristic polynomial. We are using computational aids here and would encourage you to do the same here. Solving these higher degree polynomials is just too much work and would obscure the point of these examples.

So, let’s move into a couple of examples where we have more than one case involved in the solution.

The characteristic equation is,

\[2{r^4} + 11{r^3} + 18{r^2} + 4r - 8 = \left( {2r - 1} \right){\left( {r + 2} \right)^3} = 0\]So, we have two roots here, \({r_{\,1}} = \frac{1}{2}\) and \({r_{\,2}} = - 2\) which is multiplicity of 3. Remember that we’ll get three solutions for the second root and after the first one we add \(t\)’s only the solution until we reach three solutions.

The general solution is then,

\[y\left( t \right) = {c_1}{{\bf{e}}^{\frac{1}{2}t}} + {c_2}{{\bf{e}}^{ - \,2t}} + {c_3}t{{\bf{e}}^{ - \,2t}} + {c_4}{t^2}{{\bf{e}}^{ - \,2t}}\]The characteristic equation is,

\[{r^5} + 12{r^4} + 104{r^3} + 408{r^2} + 1156r = r{\left( {{r^2} + 6r + 34} \right)^2} = 0\]So, we have one real root \(r = 0\) and a pair of complex roots \(r = - 3 \pm 5\,i\) each with multiplicity 2. So, the solution for the real root is easy and for the complex roots we’ll get a total of 4 solutions, 2 will be the *normal* solutions and two will be the normal solution each multiplied by t.

The general solution is,

\[y\left( t \right) = {c_1} + {c_2}{{\bf{e}}^{ - 3t}}\cos \left( {5t} \right) + {c_3}{{\bf{e}}^{ - 3t}}\sin \left( {5t} \right) + {c_4}\,t\,{{\bf{e}}^{ - 3t}}\cos \left( {5t} \right) + {c_5}\,t\,{{\bf{e}}^{ - 3t}}\sin \left( {5t} \right)\]Let’s now work an example that contains all three of the basic cases just to say that we that we’ve got one work here.

The characteristic equation is

\[{r^5} - 15{r^4} + 84{r^3} - 220{r^2} + 275r - 125 = \left( {r - 1} \right){\left( {r - 5} \right)^2}\left( {{r^2} - 4r + 5} \right) = 0\]In this case we’ve got one real distinct root, \(r = 1\), and double root, \(r = 5\), and a pair of complex roots, \(r = 2 \pm i\) that only occur once.

The general solution is then,

\[y\left( t \right) = {c_1}{{\bf{e}}^t} + {c_2}{{\bf{e}}^{5t}} + {c_3}\,t\,{{\bf{e}}^{5t}} + {c_4}{{\bf{e}}^{2t}}\cos \left( t \right) + {c_5}{{\bf{e}}^{2t}}\sin \left( t \right)\]We’ve got one final example to work here that on the surface at least seems almost too easy. The problem here will be finding the roots as well see.

The characteristic equation is

\[{r^4} + 16 = 0\]So, a really simple characteristic equation. However, in order to find the roots we need to compute the fourth root of -16 and that is something that most people haven’t done at this point in their mathematical career. We’ll just give the formula here for finding them, but if you’re interested in seeing a little more about this you might want to check out the Powers and Roots section of the Complex Numbers Primer.

The 4 (and yes there are 4!) 4^{th} roots of -16 can be found by evaluating the following,

\[\sqrt[4]{{ - 16}} = {\left( { - 16} \right)^{\frac{1}{4}}} = \sqrt[4]{{16}}{{\bf{e}}^{\left( {\frac{\pi }{4} + \frac{{\pi k}}{2}} \right)i}} = 2\left( {\cos \left( {\frac{\pi }{4} + \frac{{\pi k}}{2}} \right) + i\sin \left( {\frac{\pi }{4} + \frac{{\pi k}}{2}} \right)} \right)\hspace{0.25in}k = 0,1,2,3\]

Note that each value of \(k\) will give a distinct 4^{th} root of -16. Also, note that for the 4^{th} root (and ONLY the 4^{th} root) of any negative number all we need to do is replace the 16 in the above formula with the absolute value of the number in question and this formula will work for those as well.

Here are the 4^{th} roots of -16.

\[\begin{align*}&k = 0:\hspace{0.25in}2\left( {\cos \left( {\frac{\pi }{4}} \right) + i\sin \left( {\frac{\pi }{4}} \right)} \right) = 2\left( {\frac{1}{{\sqrt 2 }} + \frac{1}{{\sqrt 2 }}i} \right) = \sqrt 2 + i\sqrt 2 \\ & k = 1:\hspace{0.25in}2\left( {\cos \left( {\frac{{3\pi }}{4}} \right) + i\sin \left( {\frac{{3\pi }}{4} + \frac{{\pi k}}{2}} \right)} \right) = 2\left( { - \frac{1}{{\sqrt 2 }} + \frac{1}{{\sqrt 2 }}i} \right) = - \sqrt 2 + i\sqrt 2 \\ & k = 2:\hspace{0.25in}2\left( {\cos \left( {\frac{{5\pi }}{4}} \right) + i\sin \left( {\frac{{5\pi }}{4}} \right)} \right) = 2\left( { - \frac{1}{{\sqrt 2 }} - \frac{1}{{\sqrt 2 }}i} \right) = - \sqrt 2 - i\sqrt 2 \\ & k = 3:\hspace{0.25in}2\left( {\cos \left( {\frac{{7\pi }}{4}} \right) + i\sin \left( {\frac{{7\pi }}{4}} \right)} \right) = 2\left( {\frac{1}{{\sqrt 2 }} - \frac{1}{{\sqrt 2 }}i} \right) = \sqrt 2 - i\sqrt 2 \end{align*}\]

So, we have two sets of complex roots : \(r = \sqrt 2 \pm i\sqrt 2 \) and \(r = - \sqrt 2 \pm i\sqrt 2 \). The general solution is,

\[y\left( t \right) = {c_1}{{\bf{e}}^{\sqrt 2 t}}\cos \left( {\sqrt 2 \,t} \right) + {c_2}{{\bf{e}}^{\sqrt 2 t}}\sin \left( {\sqrt 2 \,t} \right) + {c_3}{{\bf{e}}^{ - \sqrt 2 t}}\cos \left( {\sqrt 2 \,t} \right) + {c_4}{{\bf{e}}^{ - \sqrt 2 t}}\sin \left( {\sqrt 2 \,t} \right)\]So, we’ve worked a handful of examples here of higher order differential equations that should give you a feel for how these work in most cases.

There are of course a great many different kinds of combinations of the basic cases than what we did here and of course we didn’t work any case involving 6^{th} order or higher, but once you’ve got an idea on how these work it’s pretty easy to see that they all work pretty in pretty much the same manner. The biggest problem with the higher order differential equations is that the work in solving the characteristic polynomial and the system for the coefficients on the solution can be quite involved.