Differential Equations (Part 4. Bernoulli Differential Equations)
Posted by Suhaivi Hamdan on 6:49 PM with No comments
Bernoulli Differential Equations
In this section we are going to take a look at differential
equations in the form,


where p(x) and q(x) are continuous functions on the
interval we’re working on and n is a
real number. Differential equations in
this form are called Bernoulli Equations.
First notice that if or then the equation is linear and we already
know how to solve it in these cases.
Therefore, in this section we’re going to be looking at solutions for
values of n other than these two.
In order to solve these we’ll first divide the differential
equation by to get,


We are now going to use the substitution to convert this into a differential equation
in terms of v. As we’ll see this will lead to a differential
equation that we can solve.
We are going to have to be careful with this however when it
comes to dealing with the derivative, . We need to determine just what is in terms of our substitution. This is easier to do than it might at first
look to be. All that we need to do is
differentiate both sides of our substitution with respect to x.
Remember that both v and y are functions of x and so we’ll need to use the chain rule on the right side. If you remember your Calculus I you’ll recall
this is just implicit differentiation. So, taking the derivative gives us,



Now, plugging this as well as our substitution into the
differential equation gives,


This is a linear differential equation
that we can solve for v and once we
have this in hand we can also get the solution to the original differential
equation by plugging v back into our
substitution and solving for y.
Let’s take a look at an example.
Example 1 Solve
the following IVP and find the interval of validity for the solution.
Solution
So, the first thing that we need to do is get this into
the “proper” form and that means dividing everything by
The substitution and derivative that we’ll need here is,
With this substitution the differential equation becomes,
So, as noted
above this is a linear differential equation that we know how to solve. We’ll do the details on this one and then
for the rest of the examples in this section we’ll leave the details for you
to fill in. If you need a refresher on
solving linear differential equations then go back to that section for a quick review.
Here’s the
solution to this differential equation.
Note that we
dropped the absolute value bars on the x
in the logarithm because of the assumption that
Now we need to
determine the constant of integration.
This can be done in one of two ways.
We can can convert the solution above into a solution in terms of y and then use the original initial
condition or we can convert the initial condition to an initial condition in
terms of v and use that. Because we’ll need to convert the solution
to y’s eventually anyway and it
won’t add that much work in we’ll do it that way.
So, to get the
solution in terms of y all we need
to do is plug the substitution back in.
Doing this gives,
At this point
we can solve for y and then apply
the initial condition or apply the initial condition and then solve for y.
We’ll generally do this with the later approach so let’s apply the
initial condition to get,
Plugging in for
c and solving for y gives,
Note that we
did a little simplification in the solution.
This will help with finding the interval of validity.
Before finding
the interval of validity however, we mentioned above that we could convert
the original initial condition into an initial condition for v.
Let’s briefly talk about how to do that. To do that all we need to do is plug
So, in this
case we got the same value for v
that we had for y. Don’t expect that to happen in general if
you chose to do the problems in this manner.
Okay, let’s now find the interval of validity for the
solution. First we already know that
The two
possible intervals of validity are then,
and since the
second one contains the initial condition we know that the interval of
validity is then
Here is a graph
of the solution.

Let’s do a couple more examples and as noted above we’re
going to leave it to you to solve the linear differential equation when we get
to that stage.
Example 2 Solve
the following IVP and find the interval of validity for the solution.
Solution
The first thing we’ll need to do here is multiply through
by
The substitution here and its derivative is,
Plugging the
substitution into the differential equation gives,
We rearranged a
little and gave the integrating factor for the linear differential equation
solution. Upon solving we get,
Now go back to y’s.
Applying the
initial condition and solving for c
gives,
Plugging in c and solving for y gives,
There are no
problem values of x for this
solution and so the interval of validity is all real numbers. Here’s a graph of the solution.

Example 3 Solve
the following IVP and find the interval of validity for the solution.
Solution
First get the differential equation in the proper form and
then write down the substitution.
Plugging the substitution into the differential equation
gives,
Again, we’ve
rearranged a little and given the integrating factor needed to solve the
linear differential equation. Upon
solving the linear differential equation we have,
Now back
substitute to get back into y’s.
Now we need to
apply the initial condition and solve for c.
Plugging in c and solving for y gives,
Next, we need
to think about the interval of validity.
In this case all we need to worry about it is division by zero issues
and using some form of computational aid (such as Maple or Mathematica) we
will see that the denominator of our solution is never zero and so this
solution will be valid for all real numbers.
Here is a graph
of the solution.

To this point we’ve only worked examples in which n was an
integer (positive and negative) and so we should work a quick example where n
is not an integer.
Example 4 Solve
the following IVP and find the interval of validity for the solution.
Solution
Let’s first get the differential equation into proper
form.
The
substitution is then,
Now plug the
substitution into the differential equation to get,
As we’ve done
with the previous examples we’ve done some rearranging and given the
integrating factor needed for solving the linear differential equation. Solving this gives us,
In terms of y this is,
Applying the
initial condition and solving for c
gives,
Plugging in for
c and solving for y gives us the solution.
Note that we
multiplied everything out and converted all the negative exponents to
positive exponents to make the interval of validity clear here. Because of the root (in the second term in
the numerator) and the x in the
denominator we can see that we need to require
Here is the
graph of the solution.

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