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.

just amazing explanation, seen on youtube ever

Thank you so much for these clear videos.

Liked the video before watching. Your vids are awesome and easy to understand. 🙂

I can't thank you enough for the elucidating explanation. You videos precisely follow my text book.

Best diff eq video i've ever seen

Can you explain the singular solution