(New page: Back to the MA366 course wiki == Homogeneous Equations with Constant Coefficients == ---- Here, for the first time, I'll introduce the notion of a characteristic equation (whic...)
 
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Here, for the first time, I'll introduce the notion of a characteristic equation (which I will define later). This is an important concept, and this type of differential equation is important. You'll find yourself using the technique learned in this section frequently in the course. Also, before we start, I'd like to introduce a bit of notation. Instead of constantly referring to the given differential equation as, well, "the given differential equation", let's simplify things. Each given differential equation can be be written as a function of y. For instance, the differential equation
+
Here, for the first time, I'll introduce the notion of a characteristic equation (which I will define later). This is an important concept, and this type of differential equation is important. You'll find yourself using the technique learned in this section frequently in the course. Also, before we start, I'd like to introduce a bit of notation. Instead of constantly referring to the given differential equation as, well, "the given differential equation", let's simplify things. Each given differential equation can be written as a function of y. For instance, the differential equation
  
 
<math>xy''+y'=0</math>
 
<math>xy''+y'=0</math>
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<math>ay''+by'+cy=0</math>
 
<math>ay''+by'+cy=0</math>
  
where a, b and c are arbitrary real constants. Before I tell you the general form of the solution to this kind of equation, see if you can guess at it yourself. Suppose we tried a trigonometric function like sine. Sine goes "back and forth" between cosine and sine as you repeatedly differentiate it. So if we plugged
+
where a, b and c are arbitrary real constants. Using our new notation, we can say that:
 +
 
 +
<math>L(y)=ay''+by'+cy=0</math>
 +
 
 +
It turns out that exponential functions of the form exp(rt) satisfy the equation. If we assume that y=exp(rt), then we have
 +
 
 +
<math>
 +
\begin{align}
 +
y&=e^{rt}\\
 +
y'&=re^{rt}\\
 +
y''&=r^2e^{rt}
 +
\end{align}
 +
</math>
 +
 
 +
Therefore, assuming that y has the form exp(rt), and setting L(y)=0, we obtain
 +
 
 +
<math>
 +
\begin{align}
 +
L(y)=ar^2e^{rt}+bre^{rt}+ce^{rt}&=0\\
 +
(ar^2+br+c)e^{rt}&=0\\
 +
ar^2+br+c&=0
 +
\end{align}
 +
</math>
 +
 
 +
 
  
 
[[MA366|Back to the MA366 course wiki]]
 
[[MA366|Back to the MA366 course wiki]]

Revision as of 11:13, 26 October 2009

Back to the MA366 course wiki

Homogeneous Equations with Constant Coefficients


Here, for the first time, I'll introduce the notion of a characteristic equation (which I will define later). This is an important concept, and this type of differential equation is important. You'll find yourself using the technique learned in this section frequently in the course. Also, before we start, I'd like to introduce a bit of notation. Instead of constantly referring to the given differential equation as, well, "the given differential equation", let's simplify things. Each given differential equation can be written as a function of y. For instance, the differential equation

$ xy''+y'=0 $

can be written as

$ L(y)=xy''+y' $

and any solution α to the differential equation will have the property that

$ L(\alpha)=0 $

With that out of the way, let's discuss homogeneous equations with constant coefficients.


Suppose your differential equation is of the form

$ ay''+by'+cy=0 $

where a, b and c are arbitrary real constants. Using our new notation, we can say that:

$ L(y)=ay''+by'+cy=0 $

It turns out that exponential functions of the form exp(rt) satisfy the equation. If we assume that y=exp(rt), then we have

$ \begin{align} y&=e^{rt}\\ y'&=re^{rt}\\ y''&=r^2e^{rt} \end{align} $

Therefore, assuming that y has the form exp(rt), and setting L(y)=0, we obtain

$ \begin{align} L(y)=ar^2e^{rt}+bre^{rt}+ce^{rt}&=0\\ (ar^2+br+c)e^{rt}&=0\\ ar^2+br+c&=0 \end{align} $


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