Implicit Solutions (Level 1) In this video we will learn about implicit
solutions to differential equations. You might have heard the term implicit function from
your study of precalculus and differential calculus. For example in precalculus you learn
about conic sections such as circles, ellipses, parabolas and hyperbolas these functions are
typically represented by an implicit relation. In your differential calculus class you learned
how to take the derivative of implicit functions. For the most part we usually deal with explicit
functions that is, a function in which the dependent variable is expressed solely in
terms of the independent variable along with some constants. For example the following
functions are explicit functions, in the same manner an explicit solution to a differential
equation, is one in which the dependent variable is expressed solely in terms of the independent
variable along with some constants. In other words, the variable y appears once either
on the left or right side of the equation and is also raised to the first power. In
these videos we can think of an explicit solution as an explicit formula of the form y equals
phi of x that we can manipulate, evaluate, and differentiate using standard rules.
We recently seen various explicit solutions in the previous videos, every single example
from the previous videos involved explicit solutions. Unfortunately, once we start solving
ordinary differential equations our methods of solution do not always lead directly to
an explicit solution. This is particularly true when attempting to solve nonlinear first-order
differential equations. As a result we have to be content with a relation or expression
that defines a solution implicitly. In general a relation of the form G(x,y) = 0
is said to be an implicit solution of an ordinary differential equation on an interval I, provided
there exists at least one function that satisfies the relation as well as the differential equation
on I. If you happened to solve a differential equation
and found a solution of the form G(x,y) = 0, then there exist at least one function that
satisfies both the relation (that is G(x, phi(x)) = 0) and the differential equation
on an interval I. At times if the implicit solution
is fairly simple, we may be able to solve for y in terms of x and obtain one or more
explicit solutions. For the most part there is usually no need or as a matter of fact
it’s just plain impossible to try and solve an implicit solution for y explicitly
in terms of x. The best we can do is to numerically generate the graph of the solution
and determine an appropriate interval of definition.
For example lets verify the following implicit solution x^2 + y^2 = 9,
and the differential equation to verify is y’ = -x/y, just the way we verified explicit solutions of
ODE’s we first need to find the derivative of the solution, notice that we are going
to have to find the derivative of this function by implicit differentiation a technique first
introduced way back in your differential calculus course. If you need a refresher you can always
look back at the calculus I videos. Alright taking the derivative of this function implicitly
with respect to x we obtain the following expression, now if we were to solve for y
prime we would obtain the following, simplifying the expression we see that y prime is equal
to negative x over y, if we substitute y prime into the differential equation we see that
both the left and right hand side match hence this implicit function is a solution to the
differential equation. Next we need to find an interval I of definition. To find this
interval lets go ahead and plot this implicit solution, recall that this equation describes
a circle with center located at (0,0) and with radius equal to 3. At first glance you
might be tempted to define the entire circle as your interval of definition unfortunately
this would be incorrect. Recall that a solution to a differential equation has to be differentiable
at the given interval I of definition. Notice that this function would not be differentiable
at x = -3 and x = 3 because the tangent line at these points would be vertical and therefore
undefined. Meaning the implicit function is not differentiable at these points, as a result
we would need label these points with open circles. Notice that this solution is now
broken up into two semi circles. It turns out that we can actually find an explicit
solution to this differential equation, let’s take a look at the implicit solution, notice
that we are able to solve for y in terms of x, all we need to do is to isolate the variable
y doing that we obtain the following expression: y = +- sqrt(9-x^2) notice that we were able
to find two explicit solutions to the differential equation the first is the positive square
root of nine minus x squared, and the second is the negative square root of 9 minus x squared.
Notice that these two functions satisfy the relation (that is, x squared plus the first explicit solution squared = 9 and
x^2 plus the second explicit solution squared equals 9.
as well as the differential equation. Moreover each explicit solution
represents a segment of the implicit solution. The first explicit solution forms the top
semi circle, and the negative explicit solution forms the bottom semicircle. In addition notice
that the domain of the explicit solution is defined from negative 3 inclusive to positive
3 inclusive. But as we just saw in order to satisfy the differential equation these solutions
can’t be defined at x = 3 and x = -3. As a result we have two options for our interval
of definition, we can use the first explicit solution and define our interval of definition
from negative 3 exclusive to positive 3 exclusive. Or we can use the second explicit solution
and define our interval of definition from negative 3 exclusive to positive 3 exclusive.
Either of these intervals can be used as our interval of definition. In a much later video
we will be required to choose one interval over the other for now either of the explicit
solutions along with their intervals are acceptable answers.
The final thing I want you be aware of when dealing with implicit solutions is to make
sure that any explicit solution that you happen to find has to be a real valued function.
For example say that we are asked to verify the following implicit solution x^2+y^2 = -9.
Notice that the only difference between this implicit function and the one from the previous
example is that this one is equal to negative 9, if we were to solve for an explicit solution
we would obtain the following expression. y = +- sqrt(-9-x^2). Notice that any value of x will force us to
take the square root of a negative number, and therefore we are unable to draw a graph
of this functions since it is not a real valued function.
Although the explicit solutions are not real valued functions, they nevertheless satisfy
the relation and they also satisfy the differential equation. So we have to be careful when finding
explicit solutions of implicit relations. The implicit solution might satisfy the differential
equation but an interval of definition cannot be determined because the explicit solution
is not a real valued function. When this occurs we say that the implicit relation is a formal
solution to the differential equation. Keep in mind that the implicit relation is not
a solution to the ODE since no interval of definition can be defined in any interval.
But the relation just happens to formally satisfy the differential equation.
Alright, to sum up, a solution of the form y = phi(x) is called an explicit solution
of an ordinary differential equation on the interval I.
Many differential equations don’t have explicit solution. The best we can do for
these equations is to eliminate the derivatives that appear in the equation and to simplify
the equation into a derivative free equation of the form G(x,y) = 0, this relation is called
an implicit solution of the differential equation. Implicit solutions are perfectly acceptable
and in some cases, necessary, as long as the equation actually defines y as a function
of x. Keep in mind that explicit solutions of implicit relations are sometimes difficult
if not impossible to solve for algebraically, in any case explicit solutions are preferable
and you are encouraged to find them whenever possible.
Alright in our next video we are going to cover a couple of examples illustrating how
to verify implicit solutions to Ordinary differential equations, as well as defining appropriate
intervals of definition.