Example 2 Solve
the following IVP and find the interval of validity for the solution.
First identify M
and N and check that the
differential equation is exact.
So, the differential equation is exact according to the
test. However, we already knew that as
we have given you Ψ(x,y).
It’s not a bad thing to verify it however and to run through the test
at least once however.
Now, how do we actually find Ψ(x,y)?
Well recall that
We can use either of these to get a start on finding Ψ(x,y) by integrating as follows.
However, we will need to be careful as this won’t give us
the exact function that we need. Often it doesn’t matter which one you choose
to work with while in other problems one will be significantly easier than
the other. In this case it doesn’t
matter which one we use as either will be just as easy.
So, I’ll use the first one.
Note that in this case the “constant” of integration is
not really a constant at all, but instead it will be a function of the
remaining variable(s), y in this
Recall that in integration we are asking what function we
differentiated to get the function we are integrating. Since we are working with two variables
here and talking about partial differentiation with respect to x, this means that any term that
contained only constants or y’s
would have differentiated away to zero, therefore we need to acknowledge that
fact by adding on a function of y
instead of the standard c.
Okay, we’ve got most of Ψ(x,y) we just need to determine h(y) and we’ll be done. This is actually easy to do. We used to find most of Ψ(x,y) so we’ll use to find h(y). Differentiate our Ψ(x,y) with respect to y and set this equal to N (since they must be equal after
all). Don’t forget to “differentiate” h(y)!
Doing this gives,
From this we can see that
Note that at this stage h(y) must be only a function of y and so if there are any x’s
in the equation at this stage we have made a mistake somewhere and it’s time
to go look for it.
We can now find h(y)
You’ll note that we included the constant of integration, k, here. It will turn out however that this will end
up getting absorbed into another constant so we can drop it in general.
So, we can now write down Ψ(x,y).
With the exception of the k this is identical to the function that we used in the first
example. We can now go straight to the
implicit solution using (4).
We’ll now take care of the k. Since both k and c are unknown constants all we need to do is subtract one from
both sides and combine and we still have an unknown constant.
Therefore, we’ll not include the k in anymore problems.
This is where we left off in the first example. Let’s now apply the initial condition to
The implicit solution is then.
Now, as we saw in the separable differential equation
section, this is quadratic in y and
so we can solve for y(x) by using
the quadratic formula.
Now, reapply the initial condition to figure out which of
the two signs in the that we need.
So, it looks like the “-” is the one that we need. The explicit solution is then.
Now, for the interval of validity. It looks like we might well have problems
with square roots of negative numbers.
So, we need to solve
Upon solving this equation is zero at x = 11.81557624
and x = 1.396911133. Note that you’ll need to use some form of
computational aid in solving this equation.
Here is a graph of the polynomial under the radical.
So, it looks like there are two intervals where the
polynomial will be positive.
However, recall that intervals of validity need to be
continuous intervals and contain the value of x that is used in the initial condition. Therefore the interval of validity must be.
Here is a quick graph of the solution.