Welcome to a lesson on linear dependent
and linear independent functions.
In this video, we'll define linear dependent
and linear independent functions
and then use the Wronskian,
a special determinant, to determine if two functions
are linear independent.
So we mentioned this in the last lesson,
but if we have a set of functions
f sub one through f sub n,
the set is linear dependent on an interval I
if there exists constant c sub one through c sub n
not all zero such that they satisfy this equation here,
where we have the sum of multiples of the functions
where the sum is equal to zero.
And the functions are linear independent
if they are not linear dependent.
But in this video, we're going to focus on determining
whether functions are linear independent or dependent
using a special determinant called the Wronskian.
So if we have functions f sub one through f sub n
that have at least n minus one derivatives,
then the determinant given here
is called the Wronskian of the functions.
And the value of this determinant can help us determine
if the functions are dependent or independent.
Looking at this determinant, remember the number of
rows and columns must always be the same
where the first row will be original functions,
the second row will be the first derivatives,
the third row would be the second derivatives,
all the way down to the n minus oneth derivative
in the last row.
And in this video, we're going to be focusing on
just having two functions,
so we'll have a two by two determinant
where the first row will be the original functions
and the second row will be the first derivatives.
So for example, if we have two functions, f sub one
and f sub two,
that are differentiable on a given interval,
if the Wronskian doesn't equal zero
for some x in the in the interval,
then f sub one and f sub two
are linear independent on the interval.
So if we can show the Wronskian doesn't equal zero
for some x in the interval,
we can quickly determine the functions are
linear independent.
And if f sub one and f sub two are linear dependent
on the interval, then the Wronskian will equal zero
for all x in the interval.
But we still also want to find the values of c sub one
and c sub two
that satisfy the definition for linear dependent functions.
And now the reason we care so much about
linear independent functions is that if
y sub one and y sub two
are solutions to this linear second order
homogeneous differential equation
and the Wronskian doesn't equal zero,
then y sub one and y sub two are called the fundamental
set of solutions.
And y equals c sub one time y sub one
plus c sub two times y sub two
is the general solution to the differential equation.
So to find the general solution, we need to find
two solutions such that the Wronskian is nonzero.
Let's take a look at our first example.
Here we want to determine if the following functions
are linear independent.
So we'll start by calculating the Wronskian.
And because we have two functions,
this would be the Wronskian.
So the Wronskian of f sub one
comma f sub two
would be equal to a two by two determinant
where the first row would be the original functions,
so f sub one is x squared
plus two x,
and the second function is x squared minus two x.
So our second row would be the derivatives of our functions.
So here we would have two x plus two,
and here we'd have two x minus two.
And now the determinant is going to be equal to
this product minus this product.
So we would have the quantity
x squared plus two x
times two x minus two
minus x squared minus two x
times two x plus two.
Let's go ahead and multiply this out.
We'll have four products here
and four products here.
So we'll have two x to the third
minus two x squared
plus four x squared
minus four x
and then minus two x cubed
plus two x squared
minus four x squared
minus four x.
Let's go ahead and combine like terms.
We have two x cubed,
this should be plus x squared
minus four x.
Here we'll have two x cubed
minus two x squared
minus four x.
We'll notice how we're subtracting here,
so this difference would be zero
and so would this difference
leaving us with two x squared
minus negative two x squared
which would be four x squared.
We'll notice that four x squared
is only going to equal zero
when x equals zero.
So this doesn't equal zero unless
x equals zero.
And since we weren't given an interval,
we'll go ahead and assume the interval is from
negative infinity to positive infinity.
And this is enough information to determine
these two functions are linear independent.
Again, the Wronskian will not be zero
unless x equals zero.
So there are infinite number of values of x in interval
where the Wronskian is not zero.
So let's go ahead and make our conclusion
on the next slide.
Since the Wronskian
doesn't equal zero except at x equals zero,
f sub one of x and f sub two of x
are linear independent.
Again, because the interval was not given,
we're assuming the interval is from negative infinity
to positive infinity.
Let's take a look at another example.
We'll start by setting up the Wronskian.
So again, we have a two by two determinant
because we have two functions
where the first row will be sine two x
and cosine two x.
And the second row will be the derivatives.
The derivative sine two x would be
cosine two x times two,
or two cosine two x.
The derivative of cosine two x would be
negative sine two x times two,
or negative two sine two x.
Now the determinant will be this product
minus this product.
So we'd have negative two
sine squared two x
and then minus two cosine squared two x.
Let's go ahead and factor out negative two.
So we'd have negative two times
sine squared two x
plus cosine squared two x,
and we should recognize this identity here,
that would be equal to one.
So we have negative two times one
which is equal to negative two.
We'll notice how this is never going to equal zero,
and because the Wronskian is nonzero,
we know the given functions are linear independent.
So again, since the Wronskian
equals negative two, which is nonzero,
the two functions
are linear independent.
Let's go ahead and take a look at one more example.
We'll set up the Wronskian.
Again, the first row will be the functions
two to the power of x plus three
and two to the x.
The second row will be the derivatives.
Here's our derivative formula just in case
we need the review.
We'll have natural log two
times two to the x plus three,
and we'll have natural log two
times two to the x.
The determinant will be this product minus this product.
So we'll have two to the power of x plus three
times natural log two times two to the x
minus
two to the x natural log two
times two to the x plus three.
We'll notice how these three factors are the same
and therefore their difference would be zero.
So notice how the Wronskian is equal to zero
over the entire interval
which we assume is from negative infinity
to positive infinity.
So while this is not enough information to determine
that the functions are dependent,
it would be logical to think that they are.
But to verify that they are dependent,
we do want to find the value of c sub one and c sub two
that satisfy the definition.
Let's go ahead and do this on the next slide.
Again, because the Wronskian is zero
over the entire interval,
to show they are linear dependent,
we want to find c sub one where c sub one
times two to the power of x plus three
plus c sub two times two to the x
must be equal zero.
Let's apply a power property here
and write this as c sub one times two to the x
times two to the third.
Notice how, because we're adding the exponents,
we come at this as a product with the same base
plus c sub two times two to the x
equals zero.
Well two to the third is equal to eight,
so let's write this as eight c sub one
times two to the x
plus c sub two times two to the x
equals zero.
So if these two terms were opposites,
this sum would be zero.
So let's go ahead and let c sub one
be equal to one-eighth.
Notice if c sub one is equal to one-eighth,
this would just be one times two to the x,
and therefore if we let c sub two
be equal to negative one, this sum would be zero.
Let's go ahead and check this.
If we have eight
times one-eighth
times two to the x
plus negative one times two to the x,
this would be positive one two to the x
or just two to the x,
and we'd have plus,
this would be negative two to the x
which is equal to zero.
So we found a value of c sub one and c sub two
that satisfy this equation here
which verifies the two functions are linear dependent.
So f sub one of x
and f sub two of x
are linear dependent.
So it is important to recognize that just because
the Wronskian is equal to zero,
we still do want to verify we can find the constants
which satisfy the definition for linear dependent functions.
I hope you found this lesson helpful.