- [Instructor] We're asked to consider
the initial value problem shown here,
and notice how the given differential equation
is a linear second order homogeneous
differential equation with constant coefficients,
meaning the differential equation fits this form here.
And we now know we can solve this type of
differential equation using a
characteristic equation shown here.
Where the solutions of this quadratic equation
give us the information needed to form
the general solution to the differential equation.
So for a quick review,
if the characteristic equation has two distinct
real roots or solutions,
the general solution is in this form.
If it has two real equal roots,
the general solution is in this form.
And if it has two complex roots,
in the form alpha plus or minus beta,
the general solution is in this form.
So let's first find the general solution
and then we'll work on determining the value of alpha
that satisfies this condition.
So notice how looking at the given differential equation,
a is equal to one,
b is equal to negative one,
and c is equal to negative twenty.
So the characteristic equation would be one r squared
minus one r minus twenty equals zero.
We can solve this by factoring.
We'll have two binomial factors.
The factors of r squared are r and r.
The factors have negative twenty that add to negative one
are negative five and positive four.
So notice how we have two distinct real solutions or roots.
We have r sub one equals five,
r sub two equals negative four.
Which means the general solution is y of t
equals c sub one times e,
raised to the power of five t,
plus c sub two times e
raised to the power of negative four t.
And our initial conditions are y of zero equals alpha,
and y prime of zero equals one.
So in order to use this initial condition,
we'll have to find y prime of t.
Let's go ahead and find that now.
So y prime of t is equal to derivative of y of t
with respect to t,
so we'd have c sub one times either the five t times five,
or five c sub one e to the five t,
and then plus c sub two times e to the negative four t,
times negative four or minus four times c sub two
e to the negative four t.
Now let's continue like we normally would if we were
solving an initial value problem.
So if y of zero equals alpha.
We'd substitute zero for t and y of t and alpha for y of t.
So if we substitute zero for t we'd have
e to the zero here and e to the zero here,
e to the zero equals one,
so that would give us equation c sub one
plus c sub two equals alpha.
And if y prime of zero must equal one,
using the derivative function,
we'd have e to the zero here and here.
So we'd have the equation five times c sub one
minus four times c sub two equals positive one.
Now remember our goal is to have the limit
as t approaches infinity of y of t equals zero.
Where y of t is this function here.
Let's write this as c sub one times e
raised to the power of five t.
And let's write this second term as plus c sub two
over e raised to the power of four t.
The reason I wrote this as a fraction
is because notice how for any constant c sub two,
as t approaches infinity,
this term here is going to approach zero.
So really our question is,
what would c sub one have to be in order
for the limit as t approaches infinity of this term
to be equal to zero?
Well if we take a look at the graph
of y equals e raised to the power of five t,
and y equals e raised to the power of negative four t,
we can quickly see the limit as t approaches infinity
of this function approaches zero,
with the limit as t approaches infinity of this function
obviously does not approach zero.
It approaches positive infinity.
And therefore the only way that the limit as t
approaches infinity of this term is going to approach zero
is if c sub one equals zero.
So this is the main idea of this question.
We have to recognize that in order for
this limit to equal zero,
this term must approach zero.
And that's only going to happen if c sub one equals zero.
So if c sub one equals zero,
notice how this equation just becomes negative four
times c sub two equals positive one.
And therefore c sub two equals negative 1/4.
And now going back up to this equation here,
if we know c sub one equals zero,
and c sub two equals negative 1/4,
we now know that alpha must equal negative 1/4
in order for the solution to this initial value problem
to approach zero as t approaches infinity.
So the answer to this question is alpha equals negative 1/4.
It doesn't ask but let's also give the particular solution.
It would be y of t equals,
and again we'll substitute zero for c sub one,
and negative 1/4 for c sub two.
So if c sub one equals zero,
this term is going to be zero,
and if c sub two is equal to negative 1/4,
we would just have y of t equals negative 1/4,
e raised to the power of negative 14.
Again it doesn't ask for the particular solution.
But this is what it would be.
I hope you found this helpful.