Implicit Solutions (Level 3) In this video we are going to continue illustrating
how to verify implicit solutions to ordinary differential equations, this time around we
are going to go over slightly more challenging examples. Let’s go ahead and take a look
at the first example. Verify that the indicated function is an implicit
solution to the differential equation. If possible, find an explicit solution.
Give an interval I of definition for each explicit solution.
The differential equation is 2xy + (x^2-y)y’ = 0 and the implicit solution to verify is
-2x^2*y+y^2 = 1, alright like in the previous video lets go ahead and take the derivative
of this implicit function. Taking the derivative and applying product rule along with the power
rule we obtain the following expression. Next we need to isolate the derivative term in
this case y prime, in order to accomplish this we need to move all the terms containing
y prime and move them to one side of the equation and the rest of the terms to the other side
of the equation. Doing that we obtain the following. The whole point of moving all the
terms to one side is so that we can factor out the y prime term, as follows, from here
it’s just a matter of solving for y prime, doing that we obtain the following expression.
Having found the derivative we proceed with the substitution step. Let’s verify that
the left and right hand side of the differential equation are equal to each other.
Carrying out the substitution we obtain the following expression. Next we go ahead and simplify
the expression, lets first distribute the quantity x squared minus y with 4xy as follows.
Then, let’s go ahead and add both terms together,
we are going to need to find a common denominator
and then proceed by adding the numerators. Doing that we obtain the following expression,
simplifying the numerator we see that all the terms cancel out leaving us with
0 over the quantity -2x^2+2y, which simplifies to 0, both sides of the differential equation
are equivalent hence this implicit relation is a solution to the differential equation.
Next let’s go ahead and assign an interval of definition. We first need to generate the
graph of this implicit solution; this implicit function is probably not familiar to you since
it does not have the form of any conic section. So let’s see if we can find an explicit
solution. Notice that at first glance it seems like it is plain impossible to solve for y
since we have two terms with this variable and one of them is raised to the power of
1 and the other is raised to the power of 2. How can we possibly solve for y?
It turns out that we can actually accomplish this, first of all notice that the implicit
solution contains a quadratic term a linear term and a constant term. Let’s rewrite
this expression by moving all terms to one side of the equation and setting the other
side of the equation equal to zero. Lastly let’s rewrite the terms with the powers
in descending order as follows. This implicit function is essentially a quadratic equation
of the form ax^2+bx+c = 0 but in this case instead of x representing the variable we
have y taking the place of the variable x. This means that we can use the all mighty
quadratic formula and find an explicit solution.
The coefficient a, b and c will be equal to 1, -2x^2 and -1 respectively.
Substituting these expressions into the quadratic formula we obtain the following expression, simplifying
though out, and dividing both terms by 2 we obtain the following explicit solutions.
y = x^2+-sqrt(x^4+1)
Next let’s go ahead and plot both of these solutions. The positive solution generates
the following graph, the domain of this function is essentially all real numbers, from negative
infinity to positive infinity, and this will also serve as our interval of definition.
If we use the second explicit solution we would obtain the following graph. This function
is also defined from negative infinity to positive infinity and this interval also serves
as an appropriate interval of definition. In this example we had an implicit solution
that was quadratic in form and therefore we were able to find an explicit solution by
solving for y in terms of x via the quadratic formula. Alright let’s go over the final example.
The differential equation is y’ = (-2x-y)/(x+2y)
and the implicit solution to verify is y^2+xy+x^2 = 0,
alright like always let's first find the
derivative of this function, taking the derivative implicitly and applying the power rule along
with the product rule we obtain the following expression. From here it’s just a matter
of solving for the derivative, to accomplish this we are going to need to move all the
terms containing y prime to one side of the equation and the rest of the terms to the
other side of the equation as follows. Next we go ahead and factor out the y prime
term, from here we go ahead and solve for y prime by dividing both sides of the equation
by the quantity 2y+x as follows. Having found the derivative of the implicit relation let's
go ahead and substitute this expression into the differential equation. Substituting the
expression we see that this implicit relation is a solution to the differential equation.
Next, let’s check to see if we can find an explicit solution. Notice that this implicit
relation is quadratic in form it contains a quadratic term, a linear term and a constant term.
In order to solve for y in terms of x we are going to need the quadratic formula
so we can find an explicit solution in terms of x.
The coefficient a, b and c will be equal to 1, x, and x squared respectively.
Substituting these expressions into the quadratic formula we obtain the following.
Simplifying the expression
inside the square root results in the following expression.
y = [-x +- sqrt(-3x^2)]/2
Having found an explicit solution
we are ready to graph the expression and find an appropriate interval of definition. Unfortunately,
we have a problem, notice that the domain of the radicand is always going to yield a
negative number except at x = 0 and as a result we are forced to take the square root of a
negative number. This explicit solution is not a real valued function. Since we do not
have a real valued function we cannot assigned an interval of definition. In the end this
implicit relation is a formal solution to the differential equation. We cannot say that
this implicit relation is a solution to the differential equation since we cannot assign
an interval of definition. Keep on the lookout for implicit solutions that formally satisfy
a differential equation.
To truly determine if an implicit relation is a solution we need to be able
to determine an appropriate interval of definition. So keep this little detail in mind.
Alright in our next video we will go over the concept of a family of solutions.