• Go To
• Notes
• Practice and Assignment problems are not yet written. As time permits I am working on them, however I don't have the amount of free time that I used to so it will take a while before anything shows up here.
• Show/Hide
• Show all Solutions/Steps/etc.
• Hide all Solutions/Steps/etc.
Paul's Online Notes
Home / Differential Equations / Higher Order Differential Equations / Basic Concepts
Show Mobile Notice Show All Notes Hide All Notes
Mobile Notice
You appear to be on a device with a "narrow" screen width (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 7-1 : Basic Concepts

We’ll start this chapter off with the material that most text books will cover in this chapter. We will take the material from the Second Order chapter and expand it out to $$n^{\text{th}}$$ order linear differential equations. As we’ll see almost all of the 2nd order material will very naturally extend out to $$n^{\text{th}}$$ order with only a little bit of new material.

So, let’s start things off here with some basic concepts for $$n^{\text{th}}$$ order linear differential equations. The most general $$n^{\text{th}}$$ order linear differential equation is,

$\begin{equation}{P_n}\left( t \right){y^{\left( n \right)}} + {P_{n - 1}}\left( t \right){y^{\left( {n - 1} \right)}} + \cdots + {P_1}\left( t \right)y' + {P_0}\left( t \right)y = G\left( t \right)\label{eq:eq1}\end{equation}$

where you’ll hopefully recall that,

${y^{\left( m \right)}} = \frac{{{d^m}y}}{{d{x^m}}}$

Many of the theorems and ideas for this material require that $${y^{\left( n \right)}}$$ has a coefficient of 1 and so if we divide out by $${P_n}\left( t \right)$$ we get,

$\begin{equation}{y^{\left( n \right)}} + {p_{n - 1}}\left( t \right){y^{\left( {n - 1} \right)}} + \cdots + {p_1}\left( t \right)y' + {p_0}\left( t \right)y = g\left( t \right)\label{eq:eq2}\end{equation}$

As we might suspect an IVP for an $$n^{\text{th}}$$ order differential equation will require the following $$n$$ initial conditions.

$\begin{equation}y\left( {{t_0}} \right) = {\overline{y}_0},\hspace{0.25in}y'\left( {{t_0}} \right) = {\overline{y}_1},\hspace{0.25in} \cdots ,\hspace{0.25in}{y^{\left( {n - 1} \right)}}\left( {{t_0}} \right) = {\overline{y}_{n - 1}}\label{eq:eq3}\end{equation}$

The following theorem tells us when we can expect there to be a unique solution to the IVP given by $$\eqref{eq:eq2}$$ and $$\eqref{eq:eq3}$$.

#### Theorem 1

Suppose the functions $${p_0},{p_1}, \ldots ,{p_{n - 1}}$$ and $$g\left( t \right)$$ are all continuous in some open interval $$I$$ containing $${t_0}$$ then there is a unique solution to the IVP given by $$\eqref{eq:eq2}$$ and $$\eqref{eq:eq3}$$ and the solution will exist for all $$t$$ in $$I$$.

This theorem is a very natural extension of a similar theorem we saw in the 1st order material.

Next we need to move into a discussion of the $$n^{\text{th}}$$ order linear homogeneous differential equation,

$\begin{equation}{y^{\left( n \right)}} + {p_{n - 1}}\left( t \right){y^{\left( {n - 1} \right)}} + \cdots + {p_1}\left( t \right)y' + {p_0}\left( t \right)y = 0\label{eq:eq4}\end{equation}$

Let’s suppose that we know $${y_1}\left( t \right),{y_2}\left( t \right), \ldots ,{y_n}\left( t \right)$$ are all solutions to $$\eqref{eq:eq4}$$ then by the an extension of the Principle of Superposition we know that

$y\left( t \right) = {c_1}{y_1}\left( t \right) + {c_2}{y_2}\left( t \right) + \cdots + {c_n}{y_n}\left( t \right)$

will also be a solution to $$\eqref{eq:eq4}$$. The real question here is whether or not this will form a general solution to $$\eqref{eq:eq4}$$.

In order for this to be a general solution then we will have to be able to find constants $${c_1},{c_2}, \ldots ,{c_n}$$ for any choice of $${t_0}$$ (in the interval $$I$$ from Theorem 1) and any choice of $${\overline{y}_1},{\overline{y}_2}, \ldots ,{\overline{y}_n}$$. Or, in other words we need to be able to find $${c_1},{c_2}, \ldots ,{c_n}$$ that will solve,

\begin{align*}{c_1}{y_1}\left( {{t_0}} \right) + {c_2}{y_2}\left( {{t_0}} \right) + \cdots + {c_n}{y_n}\left( {{t_0}} \right) & = {{\overline{y}}_0}\\ {c_1}{{y'}_1}\left( {{t_0}} \right) + {c_2}{{y'}_2}\left( {{t_0}} \right) + \cdots + {c_n}{{y'}_n}\left( {{t_0}} \right) & = {{\overline{y}}_1}\\ & \vdots \\ {c_1}y_1^{\left( {n - 1} \right)}\left( {{t_0}} \right) + {c_2}y_2^{\left( {n - 1} \right)}\left( {{t_0}} \right) + \cdots + {c_n}y_n^{\left( {n - 1} \right)}\left( {{t_0}} \right) & = {{\overline{y}}_{n - 1}}\end{align*}

Just as we did for 2nd order differential equations, we can use Cramer’s Rule to solve this and the denominator of each the answers will be the following determinant of an $$n$$ x $$n$$ matrix.

$\left| {\begin{array}{*{20}{c}}{{y_1}}&{{y_2}}& \cdots &{{y_n}}\\{{{y'}_1}}&{{{y'}_2}}& \cdots &{{{y'}_n}}\\ \vdots & \vdots & \ddots & \vdots \\{y_1^{\left( {n - 1} \right)}}&{y_2^{\left( {n - 1} \right)}}& \cdots &{y_n^{\left( {n - 1} \right)}}\end{array}} \right|$ As we did back with the 2nd order material we’ll define this to be the Wronskian and denote it by,

$W\left( {{y_1},{y_2}, \ldots {y_n}} \right)\left( t \right) = \left| {\begin{array}{*{20}{c}}{{y_1}}&{{y_2}}& \cdots &{{y_n}}\\{{{y'}_1}}&{{{y'}_2}}& \cdots &{{{y'}_n}}\\ \vdots & \vdots & \ddots & \vdots \\{y_1^{\left( {n - 1} \right)}}&{y_2^{\left( {n - 1} \right)}}& \cdots &{y_n^{\left( {n - 1} \right)}}\end{array}} \right|$

Now that we have the definition of the Wronskian out of the way we need to get back to the question at hand. Because the Wronskian is the denominator in the solution to each of the $${c_i}$$ we can see that we’ll have a solution provided it is not zero for any value of $$t = {t_0}$$ that we chose to evaluate the Wronskian at. The following theorem summarizes all this up.

#### Theorem 2

Suppose the functions $${p_0},{p_1}, \ldots ,{p_{n - 1}}$$ are all continuous on the open interval $$I$$ and further suppose that $${y_1}\left( t \right),{y_2}\left( t \right), \ldots {y_n}\left( t \right)$$ are all solutions to $$\eqref{eq:eq4}$$. If $$W\left( {{y_1},{y_2}, \ldots {y_n}} \right)\left( t \right) \ne 0$$ for every $$t$$ in $$I$$ then $${y_1}\left( t \right),{y_2}\left( t \right), \ldots {y_n}\left( t \right)$$ form a Fundamental Set of Solutions and the general solution to $$\eqref{eq:eq4}$$ is,

$y\left( t \right) = {c_1}{y_1}\left( t \right) + {c_2}{y_2}\left( t \right) + \cdots + {c_n}{y_n}\left( t \right)$

Recall as well that if a set of solutions form a fundamental set of solutions then they will also be a set of linearly independent functions.

We’ll close this section off with a quick reminder of how we find solutions to the nonhomogeneous differential equation, $$\eqref{eq:eq2}$$. We first need the $$n^{\text{th}}$$ order version of a theorem we saw back in the 2nd order material.

#### Theorem 3

Suppose that $${Y_1}\left( t \right)$$ and $${Y_2}\left( t \right)$$ are two solutions to $$\eqref{eq:eq2}$$ and that $${y_1}\left( t \right),{y_2}\left( t \right), \ldots {y_n}\left( t \right)$$ are a fundamental set of solutions to the homogeneous differential equation $$\eqref{eq:eq4}$$ then,

${Y_1}\left( t \right) - {Y_2}\left( t \right)$

is a solution to $$\eqref{eq:eq4}$$ and it can be written as

${Y_1}\left( t \right) - {Y_2}\left( t \right) = {c_1}{y_1}\left( t \right) + {c_2}{y_2}\left( t \right) + \cdots + {c_n}{y_n}\left( t \right)$

Now, just as we did with the 2nd order material if we let $$Y\left( t \right)$$ be the general solution to $$\eqref{eq:eq2}$$ and if we let $${Y_P}\left( t \right)$$ be any solution to $$\eqref{eq:eq2}$$ then using the result of this theorem we see that we must have,

$Y\left( t \right) = {c_1}{y_1}\left( t \right) + {c_2}{y_2}\left( t \right) + \cdots + {c_n}{y_n}\left( t \right) + {Y_P}\left( t \right) = {y_c}\left( t \right) + {Y_P}\left( t \right)$

where, $${y_c}\left( t \right) = {c_1}{y_1}\left( t \right) + {c_2}{y_2}\left( t \right) + \cdots + {c_n}{y_n}\left( t \right)$$ is called the complementary solution and $${Y_P}\left( t \right)$$ is called a particular solution.

Over the course of the next couple of sections we’ll discuss the differences in finding the complementary and particular solutions for $$n^{\text{th}}$$ order differential equations in relation to what we know about 2nd order differential equations. We’ll see that, for the most part, the methods are the same. The amount of work involved however will often be significantly more.