# University of Florida/Egm4313/s12.team7/Report2

## Background Theory: Solving ODEs with Constant Coefficients

Consider a second-order homogeneous linear ODE with constant coefficients a and b.

$y''+ay'+by=0$ (I-1)

To solve this problem, note that the solution to a first-order linear ODE of the form:

$y'+ky=0$ is an exponential function, yielding a solution of the form:

$y=e^{\lambda x}$ (I-2)

Based upon this, we will expand the solution to apply to second-linear homogeneous ODEs. To do so, we must first find equations describing $y'$ and $y''$ , and so we will take the derivatives (with respect to x) of (I-2 ).

$y'=\lambda e^{\lambda x}$ (I-3)

$y''=\lambda ^{2}e^{\lambda x}$ (I-4)

We will now substitute (I-2 ), (I-3 ), and (I-4 ) into (1-1 ) to obtain the relationship:

$\lambda ^{2}e^{\lambda x}+a\lambda e^{\lambda x}+be^{\lambda x}=0$ (I-5)

Simplifying, we have:
$(\lambda ^{2}+a\lambda +b)e^{\lambda x}=0$ $\lambda ^{2}+a\lambda +b=0$ (I-6)

Since (I-6 ) follows the same form as the quadratic equation, we can solve for $\lambda _{1}$ and $\lambda _{2}$ as follows:

$\lambda _{1}={\frac {1}{2}}(-a+{\sqrt {(a^{2}-4b)}}$ $\lambda _{2}={\frac {1}{2}}(-a-{\sqrt {(a^{2}-4b)}}$ Referring back to algebra, we know that the solution to these two equations can be one of three cases:

#### Case 1

Two real roots if...

$a^{2}-4b>0$ These two roots give us two solutions:

 $y_{1}=e^{\lambda _{1}x}$ and $y_{2}=e^{\lambda _{2}x}$ The corresponding general solution then takes the form of the following:

$y=c_{1}e^{\lambda _{1}x}+c_{2}e^{\lambda _{2}x}$ (I-7)

#### Case 2

A real double root if...

$a^{2}-4b=0$ This yields only one solution:

$y_{1}=e^{\lambda x}$ In order to form a basis, a set of two linearly independent solutions, we must find another solution. To do so, we will use the method of reduction of order, where:

$y_{2}=uy_{1}$ $y'_{2}=u'y_{1}+uy'_{1}$ and
$y''_{2}=u''y_{1}+2u'y'_{1}+uy''_{1}$ yeilding
$(u''y_{1}+2u'y'_{1}+uy''_{1})+a(u'y_{1}+uy'_{1})+buy_{1}=0$ $u''y_{1}+u'(2y'_{1}+ay_{1})+u(y''_{1}+a'y_{1}+by_{1})=0$ Since $y_{1}$ is a solution of (1-1), the last set of parentheses is zero.
Similarly, the first parentheses is zero as well, because

$2y'_{1}=-ae^{ax/2}=-ay_{1}$ From integration, we get the solution:

$u=c_{1}x+c_{2}$ If we set $c_{1}=1$ and $c_{2}=0$ $y_{2}=xy_{1}$ Thus, the general solution is:

$y=(c_{1}x+c_{2})e^{\lambda _{2}x}$ (I-8)

#### Case 3

Complex conjugate roots if...

$a^{2}-4b<0$ In this case, the roots of (1-6 ) are complex:

$y_{1}=e^{-ax/2}cos(\omega x)$ and
$y_{2}=e^{-ax/2}sin(\omega x)$ Thus, the corresponding general solution is of the form:

$y=e^{-ax/2}(Acos(\omega x)+Bsin(\omega x))$ (I-9)

### Contributions

Typed by - Mark James 2/6/2012 12:08 PM UTC
Reviewed By - Jennifer Melroy 2/6/2012 2:15 UTC
Edited By - N/A