**Warning: MathJax requires JavaScript to process the mathematics on this page.**

If your browser supports JavaScript, be sure it is enabled.

If your browser supports JavaScript, be sure it is enabled.

### Theoretical Considerations

#### Discussion

We are now going to consider higher order equations such as $y'' - y = 0$. While the techniques we use will generally work for any order equation, we will concentrate mainly on second order equations as they are the most common type in applications. The prominence of second order equations is a consequence of the fact that Newton's second law relates force to acceleration, the second derivative of position with respect to time. It may seem like the typical perversity of a math professor to start with theoretical considerations rather than solving problems, but I have my reasons. We will need to know what the general solution to a higher order equation looks like, however, before we can start searching for it. Accordingly we will start by discussing the theory of linear higher order equations, which will give us a clue about what we are looking for.One way to solve higher order equations is by reducing them to collections of first order equations. Unfortunately, this is easy only if the equation is linear and constant coefficient. On the other hand, linear constant coefficient equations are very common in applications so this defect isn't as troubling as it might first seem.

A differential equation is **linear** if it can be
written in the form
$$a_n(x)\frac{d^ny}{dx^n}+\cdots+a_1(x)\frac{dy}{dx}+a_0(x)y=f(x)$$
For linear differential equations we have the following theorem,
which we shall not prove.

#### Theorem

If $a_{n-1}(x),\ldots,a_1(x),a_0(x),f(x)$ are all continuously differentiable in an interval about $x_0$, then the initial value problem $$ \begin{align} &\frac{d^ny}{dx^n}+a_{n-1}(x)\frac{d^{n-1}y}{dx^{n-1}} +\cdots+a_1(x)\frac{dy}{dx}+a_0(x)y=f(x) \\ &y(x_0)=y_0,\quad y'(x_0)=y_1,\ldots,y^{(n-1)}(x_0)=y_{n-1} \end{align} $$ has a unique solution in some interval about $x_0$.Note that the conditions for the initial value problem in the theorem are the values of the function and its derivatives at a single point. While it is possible to consider other sorts of conditions, those other sorts of problems (prominently ``boundary value problems'') do not necessarily have solutions, and if they have solutions they aren't necessarily unique. In this class, we will concentrate on initial value problems. Fortunately in most physical situations, these are also the natural conditions to know for a problem.

While we are interested in understanding solutions of linear differential equations, much of the information we need holds for general linear equations. It will be convenient to introduce the general notion of a linear operator in what follows.

An **operator** is a mapping that takes functions to
functions. An operator, $L$, is **linear** if
$$L(f+g)=Lf+Lg \qquad\text{and}\qquad L(cf)=cL(f) $$
for all functions $f$ and $g$ and all constants $c$.

#### Examples

Let $D$ denote the differentiation operator. This is an operator because the derivative of a function is another function (though possibly the constant function or the zero function). This is also a linear operator since $$D(f+g)=Df+Dg\qquad\text{and}\qquad D(cf)=cDf $$ by the usual rules of differentiation.
A more complicated example is
$$Ly=\frac{d^ny}{dx^n}+a_{n-1}(x)\frac{d^{n-1}y}{dx^{n-1}}+
\cdots+a_1(x)\frac{dy}{dx}+a_0(x)y$$
This is also a linear operator because
$$ \begin{align}
L(y+z)&=
\frac{d^n(y+z)}{dx^n}+a_{n-1}(x)\frac{d^{n-1}(y+z)}{dx^{n-1}}+
\cdots+a_1(x)\frac{d(y+z)}{dx}+a_0(x)(y+z) \\
&=\frac{d^ny}{dx^n}+\frac{d^nz}{dx^n}+\cdots+a_0(x)y+a_0(x)z \\
&=\left(\frac{d^ny}{dx^n}+\cdots+a_0(x)y\right)
+\left(\frac{d^nz}{dx^n}+\cdots+a_0(x)z\right) \\
&=Ly+Lz
\end{align} $$
since the derivative of the sum is the sum of the derivatives and also
$$ \begin{align}
L(cy)&=\frac{d^n(cy)}{dx^n}+a_{n-1}(x)\frac{d^{n-1}(cy)}{dx^{n-1}}+
\cdots+a_1(x)\frac{d(cy)}{dx}+a_0(x)(cy) \\
&=c\frac{d^ny}{dx^n}+ca_{n-1}(x)\frac{d^{n-1}y}{dx^{n-1}}+
\cdots+ca_1(x)\frac{dy}{dx}+ca_0(x)y\\
&=c\left(\frac{d^ny}{dx^n}+a_{n-1}(x)\frac{d^{n-1}y}{dx^{n-1}}+
\cdots+a_1(x)\frac{dy}{dx}+a_0(x)y\right) \\
&=cLy
\end{align} $$
since the derivative of a constant times a function is the constant times the
derivative of the function.
This last example shows that a linear differential equation can be written
as
$$Ly=f$$
where $L$ is a linear operator. We call any operator of the form $L$ a
**linear differential operator**.
A linear differential equation is **homogeneous** if
it can be written in the form
$$Ly=0.$$
A linear differential equation of the form
$$Ly=f$$
where $f(x)$ is not identically 0 is called **inhomogeneous**,

#### Theorem

Suppose $L$ is a linear operator and $Ly=0$ and $Lz=0$. Then $L(c_1y+c_2z)=0$.
*Proof:* $L(c_1y+c_2z)=L(c_1y)+L(c_2z)=c_1Ly+c_2Lz=0+0=0$

So if we have two solutions to a linear homogeneous equation, all their linear combinations are also solutions. This is true for all linear equations, including linear differential equations of course. For inhomogeneous equations, the sum of two solutions will not be a solution but we do have the following.

#### Theorem

Suppose $L$ is a linear operator and $Ly=0$ and $Ly_p=f$. Then $L(y+y_p)=f$
*Proof:* $L(y+y_p)=Ly+Ly_p=0+f=f$.

So if we have a solution to a linear inhomogeneous equation and a solution to
the corresponding linear homogeneous equation their sum is another solution to
the linear inhomogeneous equation.
What is more, *every* solution to the linear inhomogeneous equation
can
be found from any *single* solution to the linear inhomogeneous
equation
in this fashion.

#### Theorem

Suppose $y_p$ is a solution to $Ly_p=f$ where $L$ is a linear operator. Then the set $$ \{y+y_p:Ly=0\}$$ is the set of all solutions to $Lz=f$.
*Proof:* Suppose $Lz=f$. Then $L(z-y_p)=Lz-Ly_p=f-f=0$. So
$z=(z-y_p)+y_p$
with $L(z-y_p)=0$ and so $z$ is in the specified set.

So to find the general solution of a linear inhomogeneous equation it is enough to find one particular solution and then find all the solutions to the corresponding homogeneous equation. Now just as for first order equations, we expect the general solution to a higher order equation to involve arbitrary constants. Since the initial value problem for an $n^{th}$ order equation has a unique solution when given $n$ conditions, we should expect the general solution to an $n^{th}$ order equation to involve $n$ arbitrary constants. Furthermore, given $n$ different solutions of a homogeneous equation, $y_1(x),\ldots,y_n(x)$, we have a theorem that assures us that $c_1y_1(x)+\cdots+c_ny_n(x)$ is also a solution to the homogeneous equation. This solution appears to involve $n$ arbitrary constants, but looks can be deceiving. Consider the following example.

*Example:* $y''+4y'+3y=0$

Two different solutions are $y_1(x)=e^{-x}$ and $y_2(x)=2e^{-x}$. So $y(x)=c_1e^{-x}+c_2(2e^{-x})$ is also a solution to the equation and it involves two arbitrary constants $c_1$ and $c_2$. But $e^{-3x}$ is also a solution and it can not be obtained by any choice of $c_1$ and $c_2$. So while $y(x)$ is always a solution of the equation for any choice of $c_1$ and $c_2$, it doesn't give all the solutions, despite its two arbitrary constants.

The difficulty here, of course, is that we really only have one
constant split into two pieces. $y(x)=(c_1+2c_2)e^{-x}$ and the one
arbitrary constant is $c_1+2c_2$. There was nothing wrong with our
general approach, the problem was that our two different solutions
just weren't different enough. We need to replace the term different
by the technical condition, linearly independent.
A collection of $n$ functions $y_1(x),\ldots,y_n(x)$ is
**linearly independent** if the only choice of constants
$c_1,\ldots,c_n$ for which
$$c_1y_1(x)+\cdots+c_ny_n(x)=0$$
for all $x$ is $c_1=0,\ldots,c_n=0$.

#### Theorem

If $y_1(x),\ldots,y_n(x)$ are linearly independent solutions to an $n^{th}$ order linear homogeneous differential equation $Ly=0$, then the general solution to $Ly=0$ is $y(x)=c_1y_1(x)+\cdots+c_ny_n(x)$.
Of course checking linear independence by using the definition is
very messy. There is a simple test for linear independence using the
Wronskian of the set of functions.
The **Wronskian** of a pair of functions $y_1(x)$ and
$y_2(x)$ is $W(y_1,y_2)=y_1(x)y_2'(x)-y_2(x)y_1'(x)$.
This definition only works for pairs of functions. The definition can
be extended to a collection of $n$ functions, but it depends on the
concept of a determinant of an $n\times n$
matrix. If you have seen determinants before, stop by my office and we will go
over Wronskians in the general case. If you haven't seen determinants, then
you ought to take Math 551 and make their acquaintance.

#### Theorem

If $W(y_1,y_2)\ne0$ for some value of $x$, then $y_1$ and $y_2$ are linearly independent.
*Proof:*
Suppose $$c_1y_1(x)+c_2y_2(x)=0$$ for all $x$. Differentiating this
equation we obtain $$c_1y_1'(x)+c_2y_2'(x)=0$$ for all $x$ as well.
Now if we multiply the second equation by $y_1(x)$ and the first
equation by $y_1'(x)$ and subtract we obtain
$$c_2[y_1(x)y_2'(x)-y_2(x)y_1'(x)]=0$$
or $c_2W(y_1,y_2)(x)=0$ for all $x$ by the definition of the Wronskian.
Since $W(y_1,y_2)(x)\ne0$ for some $x$, we must have $c_2=0$. On the other
hand, if we multiply the first equation by $y_2'(x)$ and the second
equation by $y_2(x)$ and subtract we obtain $c_1W(y_1,y_2)(x)=0$ for
all $x$. Again, since $W(y_1,y_2)(x)\ne0$ for some $x$, it must also
be the case that $c_1=0$. But then we have shown that the only choice
of constants $c_1$ and $c_2$ for which $c_1y_1(x)+c_2y_2(x)=0$ is
$c_1=c_2=0$ and so $y_1$ and $y_2$ are linearly independent.

Note that the theorem just says that if the Wronskian is non-zero then the functions are linearly independent. It doesn't say that if the functions are linearly independent then the Wronskian is non-zero. It doesn't say that because it isn't true, but we will not have occasion to worry about situations where linearly independent functions have a zero Wronskian in this course.

If you have any problems with this page, please contact bennett@math.ksu.edu.

©2010, 2014 Andrew G. Bennett