Solutions (level 1)
One of the main goals in an introductory level course is to solve, or find solutions of differential
equations. In this video we will explore the concept of a solution of an ordinary differential
equation. Lets first start with the following definition:
Any function defined on an interval I and possessing at least n derivatives that are
continuous on I, which when substituted into an nth-order ordinary differential equation (also known as an ODE) reduces the equation to an identity, is said to be a solution of the equation on
the interval I. In other words, a solution of an nth-order ordinary
differential equation is a function that possesses at least n derivatives and for which
the equation equals 0. For all x in I. If this is the case we
say that the function satisfies the differential equation on I. In these videos we shall also
assume that a solution is a real-valued function as oppose to an imaginary or complex
function. It might be confusing to use phi to represent the solution of an ODE at a given
interval. So we will also use y(x) to denote a solution of an ODE at a given interval.
It is important to notice right off the bat, that a solution to a differential equation
is a function, unlike the solution to an algebraic equation which is (usually) a number, or a
set of numbers. This makes differential equations much more interesting, and often more challenging
to understand, than algebraic equations. It is very important to associate a solution of an ordinary differential equation with
an interval. The interval I is known as the interval of definition, it is also referred
to as the interval of existence, the interval of validity, or the domain of the solution.
This interval can be an open interval, a closed interval, and infinite interval, or all real numbers. For example say that we are asked to verify if the following function y=(1/16)*x^4 is a solution to the given differential equation y’=x*y^(1/2), on the interval negative infinity to positive infinity in other words all real numbers.
One way of verifying that the given function is a solution is to check, after substituting,
whether each side of the equation is the same for every value of x in the interval. Let’s first
find the derivative of the function taking the derivative of 1/16 times x raised to the
power of 4 we obtain, 1/4 times x raised to the power of 3, You need to be comfortable taking derivatives if you need a refresher
you can always go back and check the derivative videos under Calculus I. Alright having found
the derivative lets proceed to substitute the function y and its derivative. Always
use parenthesis when carrying out a substitution it will help reduce any errors associated
with negative signs. Carrying out the substitution and reducing the expression we obtain the
following, notice that we reduced the equation into an identity, each side of the equation
is the same for every real number x in the interval. This means that the function is indeed a solution
to the differential equation at the given interval in this case all real numbers.
At times the interval might not be all real numbers like in the following example.
Show that the following function y=x^(-3/2) is a solution to the differential equation 4x^2*y” + 12*x*y’ + 3y=0, for x>0, Alright
looking at the ODE we see that we are going to need the first and second derivative of
the function. The first derivative is equal to the following expression, and second derivative
is equal to the following, it is just a matter of applying the power rule from calculus I
or differential calculus. Next we go ahead and substitute the function and its derivatives
into the ODE as follows, making sure we use parenthesis when carrying out the substitution
doing that we obtain the following, next we go ahead and simplify the expressions doing
that we obtain the following, notice that we can simplify the left hand side of the
equation by collecting like terms, doing that we obtain the following, 0=0 notice that
the left hand side is equal the right hand side of the equation hence the function is
a solution of the differential equation, we are not done yet we need to assign an interval
for this solution, notice that the interval was already given to us in this case the interval
is from 0 exclusive to positive infinity or all the values of x that are greater than
0. Why is this function a solution to the ODE in this interval only?
Recall that we can rewrite the solution to this ODE as follows: 1 over the square root of x cubed. In this form it
is clear that we need to avoid x=0 because division by zero is undefined. We also have
a radical function specifically a square root function. Recall that we are only interested
in real values so if our differential equation only contains real numbers then we don’t
want solutions that yield imaginary or complex numbers. So, in an effort to avoid imaginary
or complex numbers we will also need to avoid negative values of x. As a result our domain
is restricted to only positive values that are greater than 0, hence the interval of
definition. This is why it is important to define an interval when finding a solution
to a differential equation. Even though a function may symbolically satisfy
a differential equation, because of certain restrictions brought about by the solution
we cannot use all the values of the independent variable and hence, must make a restriction
on the independent variable. This will be the case with many solutions to differential
equations. Notice that each differential equation possesses
the constant solution y=0 in the interval negative infinity to positive infinity or
all real numbers. If we were to substitute this function and its derivatives into the
ODE we would technically satisfy the ODE. A solution of a differential equation that
is identically zero on an interval I is said to be a trivial solution. For the most part
we will want to find non-trivial solutions of ODE’s. With this in mind we will shift gears and
talk about solution curves. The graph of a solution of an ODE is called a solution
curve. Since the solution is a differentiable function, it is continuous on its interval
I of definition. With that said, there may be a difference between the graph of the function and the graph of the solution. In other words, the domain of the function
is not always the same as the interval I of definition (or domain) of the solution.
Let’s illustrate this with the following example. Verify that the function y=1/x is a solution to the differential equation xy’ + y=0. Alright notice
that this ODE is a first-order linear ODE, so we need to substitute the function and
the first derivative into the ODE. Let’s first find the derivative of the function,
applying the power rule we obtain the following, next we go ahead and substitute the function
and its derivative into the equation as follows, then we go ahead and simplify the expressions,
then we collect like terms doing that we obtain the following expression 0=0, so we have verified
that this function is indeed a solution to the ODE, now we need to assign an appropriate
interval of definition, to figure out the interval of definition let’s take a look
at the graph of y=1/x. Notice that the domain of this function is the set of all real numbers
x except 0. The rational function is discontinues at x=0, it essentially has a vertical asymptote
at x=0. Because of this discontinuity the function is not differentiable at x=0, recall
that a solution to a differential equation has be differentiable at the given interval
as well as satisfying the differential equation. If we look at this function as a solution
to the differential equation we need to assign an interval of definition, this interval can
be anywhere except at x=0. It can be the interval [-2, -1], or the interval [3, 4],
or the interval from negative infinity to 0 exclusive, or the interval from 0 exclusive
to positive infinity. In all of these intervals the function is differentiable and produces
real numbers as oppose to complex or imaginary numbers. For the most part we want the interval
of definition to be as large as possible so we want to I to be either negative infinity
to 0 exclusive or 0 exclusive to positive infinity. In this example it really does not matter which interval we
choose as long as the solution is differentiable in the given interval and yields real numbers.
In a much later video we will come across specific problems that require us to choose
one interval over the other for now either interval are acceptable answers.
Alright in our next video we will practice verifying solutions to differential equations
and determining appropriate intervals of definition.