Implicit Solutions (Level 2) In this video we are going to illustrate how
to verify implicit solutions to ordinary differential equations, as well as finding explicit solutions
and defining appropriate intervals of definition. Let’s go ahead and jump straight into 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 y’ = x/y, the implicit solution to verify is y^2 = x^2-3,
alright let's first verify this implicit solution. Let’s start by taking the derivative of
this function implicitly with respect to x, carrying out the differentiation and applying
the chain rule along with the power rule we obtain the following expression. Next it’s
just a matter of solving for the derivative in this case y prime doing that we obtain
the following expression, simplifying the expression we see that the derivative is equal
to x over y. The next step is to verify that the left hand side of the differential equation
equals the right hand side, so we go ahead substitute this expression into the differential
equation as follows. Notice that both sides of the equation match this means that the
implicit solution is indeed a solution to the differential equation.
Now let’s determine an appropriate interval of definition, let’s start by plotting this
function. Recall that an equation of this form describes a hyperbola that opens along
the x axis and is centered at the origin. Inspecting the graph we see that the derivative
will not be defined at the vertices because the slope of the tangent line at these points
is vertical and therefore undefined, meaning that we need to exclude these points from
the interval of definition, next let’s check and see if we can find an explicit solution
by solving for the variable y and obtaining an expression in terms of x. Looking at the
implicit solution we see that solving for the variable y is fairly straight forward
it’s just a matter of taking the square root of both sides of the expression, doing
that we obtain the following, y = +-sqrt(x^2-3),
here we have two explicit solutions one represents
the upper and the other represents the lower half of the hyperbola.
Notice that the domain of each explicit solution
is defined by the inequality the absolute value of x is greater than or equal than the square root of 3,
however, because this implicit function is not differentiable
at the vertices we need to exclude these values from the domain so that we can define an appropriate
interval of definition. If we were to use the positive explicit solution our interval
of definition would be define from negative infinity to negative square root of 3 exclusive
or from the interval positive square root of 3 exclusive to positive infinity. If were
to use the negative explicit solution we would have the same interval of definition but this
time we would be using the bottom half the hyperbola. In this example we were able to
find an explicit solution and use it to define an appropriate interval of definition,
keep in mind that this would not always be the case, it might just be impossible to find
an explicit solution from the implicit relation. Alright let’s try the next example.
The differential equation is y’ = (y-1)(1-2y) and the implicit solution to verify is ln[(2y-1)/(y-1)] = x,
alright lets first verify this implicit solution. We first need to find the derivative
of this implicit relation. Before we proceed with the implicit differentiation step lets
first rewrite this expression in a more derivative friendly form. Let’s apply the laws or rules
of logarithms that you learned in your algebra 2 class. We are going to break apart the natural
logarithm into a difference of logarithms as follows. Now that we have the implicit
function in this form, lets go ahead and take the derivative implicitly with respect to x,
carrying out the derivative and applying the chain rule we obtain the following expression.
Next we need to solve for the first derivative in this case y prime, so we go ahead and factor
out this term as follows. Next let’s go ahead and add the fractions by finding a common
denominator and rewriting the numerators accordingly, next let’s simplify the terms in the numerator,
and finally let’s multiply both sides of the equation by the reciprocal of the expression
inside the bracket, doing this we obtain the following expression for the derivative.
After that, lets substitute this expression into the differential equation, and verify that
the both the left and right hand side of the equation match. Substituting the expression
into the equation and simplifying the left hand side of the equation we see that both
sides coincide. This means that this implicit function is indeed a solution to the differential equation.
Now let’s determine an appropriate interval of definition, the implicit relation is not
a conic section or an implicit function that you have encountered in your studies of mathematics,
so let’s first see if we can find an explicit solution, going back to the implicit solution
let’s try to solve for y in terms of x. We can achieve this by first exponentiating
both sides of the equation and using the number e as our base. Doing that we obtain the following,
next let’s multiply both sides of the equation by the quantity (y-1), after that we need
to distribute exponential term into the quantity (y-1) as follows. From here we need to move
all the terms that contain the variable y into one side of the equation and the rest
of the terms to the other side of the equation, Moving all the terms with the variable y to
the left and the rest to the right side of the equation we obtain the following expression.
At this point, it’s just a matter of factoring out the variable y from the left hand side
of the equation, then we solve for the variable y by dividing both sides by the quantity (2-e^x)
doing that we obtain the following explicit solution. So we were lucky that we actually
found an explicit solution to the differential equation. Now let’s take a look at the graph
of this explicit solution. Notice that this graph is not defined at x = ln(2),
so we would expect to see a vertical asymptote at this given value of x, as a result
the graph will be split into two regions, which translates into 2 distinct intervals
of definition to choose from. One interval of definition is from negative infinity to the
natural log of 2 exclusive, and the second interval is from the natural log of 2 exclusive to
positive infinity. Either of these two intervals are appropriate intervals of definition.
Alright in our next video we will continue verifying implicit solutions to differential
equations and go over slightly more challenging examples.