Linear Dependence and Linear Independence Linear Dependence
The
-number of functions
are said to be
linearly dependent on the interval
, provided that there exists constants
not all zero such that
on
; that is, when
.
Another way that a set of functions can be determined to be linear dependent is when the n x n
Wronksian of n-number of functions is
exactly equal to zero:
Example 10:
Show that the functions
are linearly dependent.
We will go about it two ways: finding the constants
such that
. The second way will be by using the Wronskian.
It is alright for
one of the constants
to be zero, as long as all of the constants aren't zero. If we choose
, we are left with
. Noting that
, we can pick the remaining constants such that
. If we chose
, we have:
.
The alternative way is by using the Wronskian:
Since
, this set of functions are linearly dependent.
Linear Independence
A function is
linearly independent when the n x n Wronskian of n-number of functions is
not equal to zero:
Example 11:
Show that the functions
are linearly independent.
Find the Wronskian:
Since
, this set of functions are linearly independent.
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Reduction of Order
When solving a homogeneous DE, we noted there were three cases:
1) Real and Distinct roots
2) Real repeated roots
3) Complex conjugate roots
Let us re-examine the second case.
We said that if we had real repeated roots, then the solutions would have the form
and
. We will now see where the x comes from.
We want the solutions to be linearly independent. In other words,
or
. But
where
is not a constant.
In solving
, we will assume that
is a solution to this DE.
Substitute these values into the differential equation:
Rewrite the DE so it's in terms of u (i.e. au"+bu'+cu=0)
The part highlighted in blue is equal to zero, since
is a solution to the DE.
Thus, we are left with:
If we let
, we get a first order DE:
Separate the variables:
Since
,
![\therefore \color{red}\boxed{u=\int \left[\frac{e^{-\int P(x)\,dx}}{y_{1}^{2}}\right]\,dx}](http://www.mathhelpforum.com/math-help/latex2/img/10d08b4c3cf2e5539a273aace0ea24e1-1.gif "\therefore \color{red}\boxed{u=\int \left[\frac{e^{-\int P(x)\,dx}}{y_{1}^{2}}\right]\,dx}")
![\therefore \color{red}\boxed{y_2=\left(\int \left[\frac{e^{-\int P(x)\,dx}}{y_{1}^{2}}\right]\,dx\right)y_1}](http://www.mathhelpforum.com/math-help/latex2/img/072bf9de4ca3293650f57331c5b84bcc-1.gif "\therefore \color{red}\boxed{y_2=\left(\int \left[\frac{e^{-\int P(x)\,dx}}{y_{1}^{2}}\right]\,dx\right)y_1}")
.
Example 12:
Solve
Assuming a solution of the form
, we have:
with a multiplicity of 2 (Thus, we have repeated roots).
The solutions will thus have the form
and
where
![u=\int \left[\frac{e^{-\int P(x)\,dx}}{y_{1}^{2}}\right]\,dx](http://www.mathhelpforum.com/math-help/latex2/img/d6cd363abcacfca44b216fbcac1396e8-1.gif "u=\int \left[\frac{e^{-\int P(x)\,dx}}{y_{1}^{2}}\right]\,dx")
.
Find u(x):
![u(x)=\int \left[\frac{e^{\int 2\,dx}}{(e^x)^{2}}\right]\,dx=\int\frac{e^{2x}}{e^{2x}}\,dx=\int\,dx=\color{red}\boxed{x}](http://www.mathhelpforum.com/math-help/latex2/img/df0095c99ee83868e8c911cea73e10e7-1.gif "u(x)=\int \left[\frac{e^{\int 2\,dx}}{(e^x)^{2}}\right]\,dx=\int\frac{e^{2x}}{e^{2x}}\,dx=\int\,dx=\color{red}\boxed{x}")
.
Therefore,
.
Therefore,
.
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Will post more when I'm done with finals...
