- [Voiceover] Welcome to three examples on how to solve
linear second order homogeneous differential
equations with constant coefficients.
For these examples, the characteristic equation
will have two distinct real roots.
So let's begin with a review.
A linear second order homogeneous differential
equation must fit this form here,
then if the coefficients are not
functions of x but constants, differential equation
would fit this form here, where a, b, and c
are the constant coefficients,
and we use the values of a, b, and c to form
what's called the characteristic equation,
which is this quadratic equation here,
a r squared plus b r plus c equals zero.
And then based upon the types of solutions,
or characteristic values to this equation,
determines the form of the general solution.
So for these examples, since the characteristic equation
will have two distinct real roots,
the general solution to the DE will be in this form here.
So we'll solve the characteristic equation for r,
and since we'll have two solutions,
r sub one and r sub two, we'll simply perform
substitution into this function here.
Let's take a look at our first example.
Again notice the given differential equation
fits this form here, so we can find the general
solution using a characteristic equation.
Notice a the coefficient of y double prime equals one,
b the coefficient of y prime is negative one,
and c equals negative six, the coefficient of y.
So the characteristic equation would be
r squared minus r minus six equals zero.
And this is factorable,
it'll factor into two binomial factors.
We have r and r, they want the factors of negative six
that add to negative one which are
negative three and positive two.
So if this part is equal to zero
then either r minus three equals zero or
r plus two equals zero, so here r can
equal three or r can equal negative two.
Let's let r sub one equal three,
and r sub two equal negative two.
And again because we have two distinct real roots,
or two distinct real characteristic values,
we're using this form for the general solution,
so now we'll substitute r sub one and r sub two,
to determine the general solution.
So our general solution would be
y of x equals c sub one, any constant,
times e raised to the power of r sub one times x,
which should be three x, and then plus c sub two
times e raised to the power of r sub two times x,
which should be negative two x.
This would be our general solution
to the given differential equation.
Let's take a look at a second example.
Again the first thing we should recognize
is that this does fit the form of a
linear second order homogeneous differential equation
with constant coefficients but notice how there is no
y term and therefore c would be zero.
Therefore, a would be equal to one
and b would be equal to negative four.
So the characteristic equation given here
would be r squared minus four r equals zero.
Now we'll factor and solve to determine
the solutions or the characteristic values.
Remember the first step in factoring is
factor out the greatest common factor,
which would be r, so we'll have r times the quantity
r minus four equals zero, so either r equals zero
or r minus four equals zero, so our second solution is
r equals positive four, so again we'll let
r sub one equal zero and r sub two equal four,
and perform substitution to our function here
to determine the general solution.
So we have y of x equals c sub one, any constant,
times e raised to the power of zero times x
which would just be zero, plus c sub two, another constant,
times e raised to the power of r sub two times x,
which would be four x.
Remember e to the zero is equal to one.
Let's go and rewrite this as y of x equals
c sub one plus c sub two times
e raised to the power of four x.
This would be our general solution.
And while we're here let's go and
take a look at one more example.
Notice in this case the differential equation does
fit the form given here except now
notice how there's no y prime and
therefore b would be equal to zero.
So for this differential equation,
a equals nine, b equals zero, and c equals negative four.
So the characteristic equation would be
nine r squared minus four equals zero.
And this factors because we have a difference of squares
so we have two binomial factors.
The factors of nine r squared would be three r and three r.
The factors of four would be two and two,
remember one's a sum and one's a difference,
so go ahead and put the difference here and the sum here.
So either three r minus two equals zero
or three r plus two equals zero.
So here we add two and divide by three
so let's say r sub one equals 2/3.
Here we would subtract two and divide by three
so let's set r sub two equal to negative 2/3.
And once again we're going to perform substitution
for r sub one and r sub two into
this form of a general solution.
So y of x is equal to c sub one times e
raised to the power of r sub one times x,
that would be 2/3 times x or two x divided by three,
plus c sub two times e raised to the power of
r sub two times x which would be
negative two x divided by three,
so this would be our general solution
to the given differential equation.
And this'll be the last example for this video
but just keep in mind that this is the form
of the general solution only when the
characteristic equation has two distinct real roots.
Take a look at some more examples that address when
we have two real equal roots as well as two complex roots.
I hope you found these examples helpful.