Diff eq. Interval of definition

In summary: I wasn't sure what you meant. Can you explain?The second one means that the function y must be defined and finite everywhere in the interval. Since the solution y uses log, sec and tan, none of which have the whole of ##\mathbb R## in their domain, you need to take that into account in determining your interval, so as to avoid the missing domain points.
  • #1
Rijad Hadzic
321
20

Homework Statement


In problems 11-14 verify that the indicated function is an explicit solution of the given differential equation. Assume an appropriate interval I of definition for each solution.

Question 13:
y'' + y = tanx ; y = -cos(x)ln(sec(x) + tan(x))

Anyways, next problem...

In problems 15-18 verify that the indicated function y = g(x) is an explicit solution of the given first order differential equation. Proceed by giving g(x) simply as a function and give its domain. Then by considering g(x) as a solution of the differential equation, give at least one interval I of definition.

Question 15:

(y-x)y' = y-x+8 ; y = x+( 4(x+2)^(1/2) )

Homework Equations

The Attempt at a Solution


My answer for question 13:

Here I verified that is is indeed an explicit solution by plugging in. I posted the derivatives because it might help answer my questions.

y=-cos(x)ln(sec(x) + tan(x))
y' = sin(x)ln(sec(x)+tan(x))-1
y'' = cos(x)ln(sec(x)+tan(x))+tan(x)

Moving on to the next part of the question, "Assume an appropriate interval I of definition for each solution."

This is how I understand this question: At the beginning of the question it says "verify that the indicated function is an explicit solution" and then "Assume an appropriate interval I of definition for each solution."

So the solution being mentioned here is just y(x). y(x) is the only function that I'm finding an interval of definition for. Even though the interval of definition is the same for y, y', and y'', I care about y(x) because that's my solution.

Can anyone tell me if my understanding is correct here?

Now my answer for question 15:
so

y = x+( 4(x+2)^(1/2) )

y' = ( (x+2)^(1/2) + 2 ) / (x+2)^(1/2)

I checked and it is indeed an explicit solution.

so for y = x+( 4(x+2)^(1/2) ) , you can plug in any number equal to or greater than -2 and you will still get a solution. So going with the logic of the question before this one, I chose my interval of definition to be [-2, infinity). But my book tells me that the largest interval of definition is actually (-2, infinity)
This leads me to believe my previous theory was wrong. The largest interval of definition would be [-2, infinity) if I am just taking y into account. But it looks like for y', it is (-2, infinity) and that's the answer.

and if I keep on taking derivatives of y after y', I always get a denominator of (x+2)^(n/2), meaning that its going to be (-2, infinity) for every derivative after y' as well.

If you guys can please answer the previous questions and this one final question:

so when finding the interval of definition, it is going to be the interval that satisfies every single dependent variable and all of its derivatives as well? I know the question was really long guys, but its a simple diff eq question but my book and instructor just haven't been able to get the concept across to me... I appreciate the time guys and I truly do care about not wasting anyones time hopefully I can contribute to this community one day as well.
 
Physics news on Phys.org
  • #2
Rijad Hadzic said:
So the solution being mentioned here is just y(x). y(x) is the only function that I'm finding an interval of definition for. Even though the interval of definition is the same for y, y', and y'', I care about y(x) because that's my solution.

Can anyone tell me if my understanding is correct here?
the exact meaning of 'Interval of Definition' may vary between texts. However the most natural interpretation is that it must be an open interval U such that:
  • the DE is well-defined on U
  • U is contained in the domain of the solution function
  • U contains any x values used in boundary conditions ('initial values')
You can forget the third one here, as the problems do not specify boundary conditions.

The second one means that the function y must be defined and finite everywhere in the interval. Since the solution y uses log, sec and tan, none of which have the whole of ##\mathbb R## in their domain, you need to take that into account in determining your interval, so as to avoid the missing domain points.

The first one means you also need to ensure that all components of the DE are well-defined everywhere in the interval. That means you need to avoid the points where tan blows up, and any points where the second derivative y'' does not exist.
 
  • #3
Thanks for the reply andrewkirk.

Your first point makes sense then. The DE has to be defined for every dependent variable. so if y= g(x), you have to find the domain for every y function and its derivatives that are in the DE.

On your second point, when you said "U is contained in the domain of the solution function," and "The second one means that the function y must be defined and finite everywhere in the interval."

that ties in with the first point, meaning its not just for y, but the highest order of y's derivative in the equation, so its also for y', y'', and so on.

Am I understanding this correctly?
 
  • #4
Rijad Hadzic said:

Homework Statement


In problems 11-14 verify that the indicated function is an explicit solution of the given differential equation. Assume an appropriate interval I of definition for each solution.

Question 13:
y'' + y = tanx ; y = -cos(x)ln(sec(x) + tan(x))

Anyways, next problem...

In problems 15-18 verify that the indicated function y = g(x) is an explicit solution of the given first order differential equation. Proceed by giving g(x) simply as a function and give its domain. Then by considering g(x) as a solution of the differential equation, give at least one interval I of definition.

Question 15:

(y-x)y' = y-x+8 ; y = x+( 4(x+2)^(1/2) )

Homework Equations

The Attempt at a Solution


My answer for question 13:

Here I verified that is is indeed an explicit solution by plugging in. I posted the derivatives because it might help answer my questions.

y=-cos(x)ln(sec(x) + tan(x))
y' = sin(x)ln(sec(x)+tan(x))-1
y'' = cos(x)ln(sec(x)+tan(x))+tan(x)

Moving on to the next part of the question, "Assume an appropriate interval I of definition for each solution."

This is how I understand this question: At the beginning of the question it says "verify that the indicated function is an explicit solution" and then "Assume an appropriate interval I of definition for each solution."

So the solution being mentioned here is just y(x). y(x) is the only function that I'm finding an interval of definition for. Even though the interval of definition is the same for y, y', and y'', I care about y(x) because that's my solution.

Can anyone tell me if my understanding is correct here?

Your first solution contains the logarithm of ##g(x) = \sec(x) + \tan(x)##, and so only exists (as a real quantity) where ##g(x) > 0##.
 
  • #5
Rijad Hadzic said:
that ties in with the first point, meaning its not just for y, but the highest order of y's derivative in the equation, so its also for y', y'', and so on.

Am I understanding this correctly?
Yes. Don't forget that the 'so on' includes tan x.
 

What is a differential equation?

A differential equation is a mathematical equation that describes the relationship between an unknown function and its derivatives. It is used to model various phenomena in physics, engineering, and other fields.

What is an interval of definition for a differential equation?

The interval of definition for a differential equation is the range of values for the independent variable for which the equation is valid. It is important to determine the interval of definition in order to find the solutions to the equation.

Why is it important to determine the interval of definition for a differential equation?

Determining the interval of definition is important because it helps to ensure that the solutions to the differential equation are valid and meaningful. It also helps to determine the range of values for which the equation can be used to model a particular phenomenon.

How do you determine the interval of definition for a differential equation?

The interval of definition for a differential equation can be determined by analyzing the coefficients and initial conditions of the equation. It is also important to consider any restrictions on the independent variable, such as physical or practical limitations.

Can the interval of definition for a differential equation change?

Yes, the interval of definition for a differential equation can change depending on the specific values of the coefficients and initial conditions. It is important to always check the interval of definition when solving a differential equation to ensure that the solutions are valid.

Similar threads

  • Calculus and Beyond Homework Help
Replies
5
Views
297
  • Calculus and Beyond Homework Help
Replies
2
Views
285
  • Calculus and Beyond Homework Help
Replies
9
Views
553
  • Calculus and Beyond Homework Help
Replies
8
Views
765
  • Calculus and Beyond Homework Help
Replies
3
Views
293
  • Calculus and Beyond Homework Help
Replies
14
Views
258
  • Calculus and Beyond Homework Help
Replies
6
Views
552
  • Calculus and Beyond Homework Help
Replies
6
Views
766
  • Calculus and Beyond Homework Help
Replies
2
Views
549
  • Calculus and Beyond Homework Help
Replies
15
Views
1K
Back
Top