Differential Equations (Part 2. Separable Differential Equations)
Posted by Suhaivi Hamdan on 6:46 PM with No comments
We are now going to start looking at nonlinear first order
differential equations. The first type of nonlinear first order differential
equations that we will look at is separable differential equations.
A separable differential equation is any differential
equation that we can write in the following form.

(1)

Note that in order for a differential equation to be separable
all the y's
in the differential equation must be multiplied by the derivative and all the x's in the
differential equation must be on the other side of the equal sign.
Solving separable differential equation is fairly easy. We
first rewrite the differential equation as the following


Then you integrate both sides.

So, after doing the integrations in (2) you
will have an implicit solution that you can hopefully solve for the explicit solution,
y(x). Note that it
won't always be possible to solve for an explicit solution.
Recall from the Definitions
section that an implicit solution is a
solution that is not in the form while an explicit
solution has been written in that form.
We will also have to worry about the interval of validity for many of these
solutions. Recall that the interval of validity was the range of the
independent variable, x
in this case, on which the solution is valid. In other words, we need to avoid
division by zero, complex numbers, logarithms of negative numbers or zero, etc. Most of the solutions that we will get
from separable differential equations will not be valid for all values of x.
Let’s start things off with a fairly simple example so we
can see the process without getting lost in details of the other issues that
often arise with these problems.
Example 1 Solve
the following differential equation and determine the interval of validity
for the solution.
Solution
It is clear, hopefully, that this differential
equation is
separable. So, let’s separate the differential
equation and integrate both sides. As
with the linear first order officially we will pick up a constant of
integration on both sides from the integrals on each side of the equal
sign. The two can be moved to the same side and absorbed into each
other. We will use
the convention that puts the single constant on the side with the x’s.
So, we now have an implicit solution. This solution is easy enough to get an
explicit solution, however before getting that it is usually easier to find
the value of the constant at this point.
So apply the initial condition and find the value of c.
Plug this into the general solution and then solve to get
an explicit solution.
Now, as far as solutions go we’ve got the solution. We do need to start worrying about
intervals of validity however.
Recall that there are two conditions that define an
interval of validity. First, it must
be a continuous interval with no breaks or holes in it. Second it must contain the value of the
independent variable in the initial condition, x = 1 in this case.
So, for our case we’ve got to avoid two values of x.
Namely,
However, only one of these will contain the value of x from the initial condition and so we
can see that
must be the interval of validity for this solution.
Here is a graph of the solution.
Note that this does not say that either of the other two
intervals listed above can’t be the interval of validity for any
solution. With the proper initial
condition either of these could have been the interval of validity.
We’ll leave it to you to verify the details of the
following claims. If we use an initial
condition of
we will get exactly the same solution however in this case
the interval of validity would be the first one.
Likewise, if we use
as the initial condition we again get exactly the same
solution and in this case the third interval becomes the interval of
validity.
So, simply changing the initial condition a little can
give any of the possible intervals.

Example 2 Solve
the following IVP and find the interval of validity for the solution.
Solution
This differential equation is clearly separable, so let's
put it in the proper form and then integrate both sides.
We now have our implicit solution, so as with the first
example let’s apply the initial condition at this point to determine the
value of c.
The implicit solution is then
We now need to find the explicit solution. This is actually easier than it might look
and you already know how to do it.
First we need to rewrite the solution a little
To solve this all we need to recognize is that this is
quadratic in y and so we can use
the quadratic formula to solve it.
However, unlike quadratics you are used to, at least some of the
“constants” will not actually be constant, but will in fact involve x’s.
So, upon using the quadratic formula on this we get.
Next, notice that we can factor a 4 out from under the
square root (it will come out as a 2…) and then simplify a little.
We are almost there. Notice that we’ve actually got two
solutions here (the “
In this case it looks like the “+” is the correct sign for
our solution. Note that it is
completely possible that the “
The explicit solution for our differential equation is.
To finish the example out we need to determine the
interval of validity for the solution. If we were to put a large negative
value of x
in the solution we would end up with complex values in our solution and we
want to avoid complex numbers in our solutions here. So, we will need to
determine which values of x
will give real solutions. To do this we will need to solve the following
inequality.
In other words, we need to make sure that the quantity
under the radical stays positive.
Using a computer algebra system like Maple or Mathematica
we see that the left side is zero at x
=
So, in order to get real solutions we will need to require
Therefore, the interval of validity of the solution is
Here is graph of the solution.

Example 3 Solve
the following IVP and find the interval of validity of the solution.
Solution
First separate and then integrate both sides.
Apply the initial condition to get the value of c.
The implicit solution is then,
Now let’s solve for y(x).
Reapplying the initial condition shows us that the “
Let’s get the interval of validity. That’s easier than it might look for this
problem. First, since
Note that we were able to square both sides of the inequality
because both sides of the inequality are guaranteed to be positive in this
case. Finally solving for x we see that the only possible range
of x’s that will not give division
by zero or square roots of negative numbers will be,
and nicely enough this also contains the initial condition
x=0. This interval is therefore our interval of
validity.
Here is a graph of the solution.

Example 4 Solve
the following IVP and find the interval of validity of the solution.
Solution
This differential equation is easy enough to separate, so
let's do that and then integrate both sides.
Applying the initial condition gives
This then gives an implicit solution of.
We can easily find the explicit solution to this
differential equation by simply taking the natural log of both sides.
Finding the interval of validity is the last step that we
need to take. Recall that we can't plug negative values or zero into a
logarithm, so we need to solve the following inequality
The quadratic will be zero at the two points
So, possible intervals of validity are
From the graph of the quadratic we can see that the second
one contains x = 5, the value of
the independent variable from the initial condition. Therefore the interval of validity for this
solution is.
Here is a graph of the solution.

Example 5 Solve
the following IVP and find the interval of validity for the solution.
Solution
This is actually a fairly simple differential equation to
solve. I’m doing this one mostly
because of the interval of validity.
So, get things separated out and then integrate.
Now, apply the initial condition to find c.
So, the implicit solution is then,
Solving for r
gets us our explicit solution.
Now, there are two problems for our solution here. First we need to avoid Î¸ = 0 because of the natural log. Notice that because of the absolute value
on the Î¸ we don’t need to worry about Î¸ being negative. We will also need to avoid
division by zero. In other words, we
need to avoid the following points.
So, these three points break the number line up into four
portions, each of which could be an interval of validity.
The interval that will be the actual interval of validity
is the one that contains Î¸ = 1.
Therefore, the interval of validity is
Here is a graph of the solution.

Example 6 Solve
the following IVP.
Solution
This problem will require a little work to get it
separated and in a form that we can integrate, so let's do that first.
Now, with a little integration by parts on both sides we
can get an implicit solution.
Applying the initial condition gives.
Therefore, the implicit solution is.
It is not possible to find an explicit solution for this
problem and so we will have to leave the solution in its implicit form.
Finding intervals of validity from implicit solutions can often be very
difficult so we will also not bother with that for this problem.

As this last example showed it is not always possible to
find explicit solutions so be on the lookout for those cases.
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