Example 2 Solve
the following IVP and find the interval of validity for the 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 “ ”) and we only want a single
solution. In fact, only one of the
signs can be correct. So, to figure out which one is correct we can reapply
the initial condition to this. Only
one of the signs will give the correct value so we can use this to figure out
which one of the signs is correct. Plugging x = 1 into the solution
In this case it looks like the “+” is the correct sign for
our solution. Note that it is
completely possible that the “”
could be the solution so don’t always expect it to be one or the other.
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
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
as well as two complex values, but we can ignore complex values for interval
of validity computations. Finally a graph of the quantity under the radical
is shown below.
So, in order to get real solutions we will need to require
because this is the range of x’s for which the quantity is
positive. Notice as well that this
interval also contains the value of x
that is in the initial condition as it should.
Therefore, the interval of validity of the solution is .
Here is graph of the solution.