Example 1 Solve
the following IVP and find the interval of validity for the solution.
So, the first thing that we need to do is get this into
the “proper” form and that means dividing everything by . Doing this gives,
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.
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.
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 into the substitution and then use the
original initial condition. Doing this
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 and that means we’ll avoid the problems of
having logarithms of negative numbers and division by zero at . So, all that we need to worry about then is
division by zero in the second term and this will happen where,
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.