- [James] Welcome to a lesson
on linear second order homogeneous differential equations.
In this video we'll define a second order
ordinary differential equation,
define a linear second order differential equation,
and then find a general solution
to a linear second order homogeneous differential equation
with constant coefficients when the characteristic equation
has two real distinct roots.
A second order, ordinary, differential equation
has the form given here,
where we have the second derivative of y with respect to x
is equal to a function in terms of
x the independent variable, y the dependent variable,
and the first derivative of y with respect to x.
If the second order differential equation
can be written in this form here,
where we have a function of x times the second derivative,
plus the function of x times the first derivative,
plus the function of x times y equals g of x,
then the differential equation is a linear
second order differential equation.
If it doesn't fit this form, then we say it's nonlinear.
And then finally if g of x,
the function on the right, is equal to zero,
the linear second order differential equation
is homogeneous, or homogeneous.
And if g of x doesn't equal zero
we say it's nonhomogeneous, or nonhomogeneous.
But in our case the coefficients will be constants,
not functions of x.
So p of x will be a constant, which we'll call a,
q of x is a constant which we'll call b,
and r of x is a constant which we'll call c.
So all the linear, second order,
homogeneous differential equations that we'll be solving
will have constant coefficients.
Let's begin by looking at an example.
Let's say you want to find the general solution
to y double prime minus four y equals zero.
Notice if you compare it to this form here,
a, the coefficient of y square would be one,
b would be zero because there is no y prime in the equation,
and c would be negative four, the coefficient of y.
So if I took this equation and added four y to both sides
we could write y double prime equals four y,
which means we're looking for functions
such that the second derivative
is equal to four times the original function.
This should remind us of the exponential function
where the derivative of e to the u
is equal to e to the u times u prime.
Because you want the second derivative
to be four times the original function,
notice if you let y equal e to the power of two x,
we'd have to apply the chain rule twice
to find the second derivative.
For example the first derivative
would be equal to e to the power two x times two,
again applying the chain rule,
giving us two times e the power of two x,
and then I'll find the second derivative,
applying the chain rule again
we'd have two times e to the power of two x times two,
or four times e to the power of two x.
Notice how our second derivative
is four times the original function,
therefore this would be one solution
to our differential equation.
But there's also many others.
For example if we let y equal
e to the power of negative two x,
notice the first derivative would be
e to the power of negative two x times negative two,
or negative two e to the negative two x.
Then the second derivative would be
negative two e to the negative two x times negative two,
or four e to the power of negative two x.
Again, notice how y double prime
is four times the original function.
So to find the general solution to the differential equation
we'll apply a special principle
called the principle of superposition,
which states if y sub one of x and y sub two of x
are two solutions to a linear, homogeneous,
second order differential equation,
then y of x will be the general solution
where y of x equals c sub one, which is a constant,
times y sub one of x, plus c sub two, another constant,
times y sub two of x.
So if we know that our functions
are going to be exponentials,
we can use the principle of superposition
to find a general solution in this form here.
Again, this works when we find
two solutions to our differential equations.
So if we apply this to our previous example,
since we found these two solutions,
we can now apply the principle of superposition
to find our general solution
which should be y of x equals c sub one, a constant,
times e raised to the power of two x,
plus c sub two times e to the power of negative two x.
Notice in this form r sub one is equal to two,
and r sub two is equal to negative two.
Now we're going to discus a method
for finding the values of r sub one and r sub two
that will make solving
these types of differential equations much easier.
The method involves something
called the characteristic equation.
Let's begin by assuming
that our differential equation in this form here
has a solution of y of x equals e raised to the power of rx.
Then if we find the first derivative and second derivative
we should be able to perform substitution into this equation
and verify this is equal to zero.
So this y here is e to the power of rx,
the first derivative, or y prime,
would be equal to r times e to the power of rx,
and then finally the second derivative
would be equal to r squared e to the power of rx.
Because we know that y of x is a solution,
we know it must satisfy this equation,
so now if we focus on the left side,
and factor out the exponential part,
or e raised to the power of rx,
notice how we'd be left with this quadratic in terms of r,
or a, r squared plus br plus c.
Since this exponential will never equal zero,
we can just cancel it out or divide it out,
so this equation is going to be equal to zero
when this quadratic is equal to zero.
And this quadratic equation here
is called the characteristic equation
which we can use to find r sub one and r sub two
which will be the solution to this quadratic.
Once we find r sub one and r sub two
we'll be able to use the principle of superposition
discussed earlier to find our general solution,
as long as the original differential equation
fits this form here.
Let's go ahead and apply this to our previous example
to see if it works.
Going back to our original differential equation,
we first need to recognize our given DE is a linear,
second order, homogeneous differential equation
with constant coefficients, so we can solve it
using the characteristic equation given here.
We already said earlier that a is equal to one,
the coefficient of y double prime,
b is equal to zero because there is no y prime,
and c is equal to negative four,
which means our characteristic equation would be
one r squared, or just r squared,
and since b is equal to zero,
we just have minus four equals zero.
And now we'll solve this for r,
notice how we have a difference of squares
so this factors nicely.
We would have the factors of r squared which are r and r.
Factors of four, which are two and two,
one binomial as a sum, and one as a difference.
So let's go ahead and put a minus here and a plus here.
Notice that here, r minus two is equal to zero
when r equals two, let's call this r sub one.
and r minus two is equal to zero when r equals negative two.
Let's go ahead and call this r sub two.
Now that we know the values of r,
notice how we have two distinct real roots,
the general solution will be in this form here.
So just as we found before
by using what we know about derivatives,
our general solution y of x
is equal to c sub one times e to the power of two x,
plus c sub two times e to the power of negative two x.
But now before we go I do want to discuss one more thing
about using the characteristic equation
to solve these types of differential equations.
The types of solutions to the characteristic equation
will affect the general solutions
to our differential equation.
As discussed in this video,
if the characteristic equation has two distinct real roots
this will be the form of the general solution.
However if r sub one equals r sub two,
or we have two real and equal roots,
the general solution will be in this form,
where notice we have an extra factor of x
here in this second term,
and we'll discuss where this comes from in the next lesson.
And then finally if r sub one and r sub two
are complex roots in this form here,
then the general solution will be in this form.
And again, we'll discuss this in another lesson.
So there are three lessons discussing the three outcomes
when solving a DE using this characteristic equation
and there will also be several example-only videos
to support the lessons.
I hope you've found this helpful.